130 free, high-quality past-paper-style questions on Simple harmonic motion for IB Physics HL, each with a full mark scheme. Covers: Mass–spring and pendulum periods; Acceleration in SHM; Conditions for SHM; and more. For when you have finished the past papers, or want to practise one topic at a time.C.1 at Standard Level →Read the C.1 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.581 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1C-1A-02Mass–spring and pendulum periodsPaper 1AEasy1 mark
A mass on a spring oscillates with period T. The mass is quadrupled and the spring is replaced by one with twice the spring constant.
What is the new period?
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Step 1T = 2π√(m/k), so T ∝ √(m/k).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The ratio m/k becomes (4m)/(2k) = 2 times its original value.
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Step 3New period = √2 × T.
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Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis inverts the ratio, as though k were quadrupled and m doubled.
BThe two changes do not cancel: the mass increases by a larger factor than the spring constant.
A particle performs simple harmonic motion. When its displacement from equilibrium is 0.10 m its acceleration has magnitude 4.0 m s−2.
What is the period of the motion?
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Step 1The defining equation is a = −ω²x, so ω² = 4.0/0.10 = 40 s−2.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2ω = 6.3 rad s−1.
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Step 3T = 2π/ω = 2π/6.3 = 1.0 s.
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Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis is 1/ω = 0.16 s — the reciprocal of the angular frequency, not the period. (Omitting the square root, ω = 40 rad s−1 and 2π/40, gives the same wrong value.)
BThis is π/ω = 0.50 s — half the period, from using π instead of 2π.
CCorrect: T = 2π/√(a/x) = 2π/√40.
DThis inverts the ratio, taking ω² = x/a = 0.025 s²: ω = 0.158 rad s−1 and T = 2π/0.158 = 40 s.
Four objects oscillate about an equilibrium position. Resistive forces are negligible.
Which object performs simple harmonic motion?
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Step 1SHM requires a restoring force (and so an acceleration) that is proportional to the displacement from equilibrium and always directed towards the equilibrium position: a = −ω²x.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2For the trolley, a displacement x stretches one spring further and relaxes the other, giving a net force −(k1 + k2)x: proportional to x and opposite to it.
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Step 3In each of the other systems the restoring force is either constant in size or not proportional to the displacement.
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThe ball has a constant acceleration g while in the air, whatever its height: the motion is periodic but the force is not proportional to displacement.
BAt 60° the restoring force is proportional to sin θ, not to θ; the small-angle condition (sin θ ≈ θ) is not met, so the motion is not simple harmonic.
COn each face of the V the component of gravity along the slope is constant in size, so the restoring force does not grow with displacement.
DCorrect: the net spring force is proportional to the displacement and directed towards equilibrium.
A glider on a horizontal air track is held between two stretched springs that obey Hooke's law. It is displaced and released, and it oscillates with simple harmonic motion. Because of air resistance, the amplitude decreases slowly over many oscillations. The damping is light.
Which row describes how the period and the maximum speed of the glider change over these oscillations?
Diagram not to scale
PeriodMaximum speed
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Step 1T = 2π√(m/k) does not depend on the amplitude, so the period stays the same as the amplitude falls (light damping has a negligible effect on the period).
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The maximum speed is ωx0; ω is unchanged and x0 decreases, so the maximum speed decreases.
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis assumes that a smaller oscillation takes less time. The glider travels a shorter distance but at a proportionally smaller speed, so the period is unchanged.
BThe period is unchanged, but energy is removed by the air resistance, so the glider cannot keep the same maximum speed.
CThis assumes that the damping slows the oscillation down progressively. Light damping removes energy (amplitude) but does not change the period as the amplitude decays.
DCorrect: the period is independent of the amplitude, while the maximum speed ωx0 falls with the amplitude.
5IBP2H-019-01-01Phase and timing in SHMPaper 1AEasy1 mark
An engineer is designing a vibration isolator for a laboratory balance. The isolator should make the balance pan perform simple harmonic motion (SHM) when it is disturbed. Which condition must the motion of the pan satisfy for it to be SHM?
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SHM is defined by a = −ω2x. The acceleration is proportional to the displacement from equilibrium and always points back towards equilibrium, so it is opposite to the displacement. Option A describes a runaway motion. Option C uses velocity instead of acceleration. Option D has a restoring force of constant size. That gives periodic motion, but not SHM, because the force does not depend on x.
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Answer B
Answer: B
Common error: Saying only 'there is a restoring force' is not enough. For SHM the restoring force, and so the acceleration, must be proportional to displacement and opposite to it.
A child on a playground swing performs small oscillations that are approximately simple harmonic, with a period of 3.0 s. What is the period of the variation of the child's kinetic energy?
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The kinetic energy is greatest each time the child passes through the lowest point. This happens twice per oscillation, once in each direction. Kinetic energy depends on v2, so it is the same whichever way the child moves. The kinetic energy therefore repeats every half period: 3.0/2 = 1.5 s. Option C is the period of the displacement. Options A and D halve or double the period in the wrong way.
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Answer B
Answer: B
Exam tip: Kinetic energy and potential energy in SHM both vary at twice the frequency of the oscillation, because each is proportional to a squared quantity.
7IBP2H-019-00-01Conditions for simple harmonic motionPaper 1AEasy1 mark
Which of the following motions is (approximately) simple harmonic?
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SHM needs a restoring force, and so an acceleration, that is proportional to displacement from equilibrium and directed towards equilibrium. In a U-tube, a level difference of 2x gives an unbalanced weight proportional to x that acts back towards equilibrium, so a ∝ −x. The bouncing ball has a constant acceleration g between bounces. The puck has zero acceleration between impacts. The lift's acceleration has constant magnitude. None of these last three has an acceleration proportional to displacement.
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Answer B
Answer: B
Common error: Repeating (periodic) motion is not automatically SHM. Check that the acceleration is proportional to displacement and opposite in direction.
8IBP2H-019-00-02Mass–spring and pendulum periodsPaper 1AEasy1 mark
On a space station, an astronaut's mass is measured with a device in which the astronaut, strapped to a light seat, oscillates horizontally on a spring. An astronaut of mass m gives an oscillation period T. Ignore the mass of the seat. What happens for an astronaut of mass 4m?
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For a mass–spring system T = 2π√(m/k). The period depends on mass (inertia), not on weight, so the device still works in orbit. Multiplying m by 4 multiplies T by √4 = 2, giving 2T. The answer 4T forgets the square root. T/2 inverts the dependence. The 'weightless' option confuses mass with weight.
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Answer D
Answer: D
Exam tip: A mass–spring period depends only on m and k. A pendulum period depends only on L and g. Keep the two formulae separate.
9IBP3H-C1-20r7-01Phase and timing in SHMPaper 1AEasy1 mark
A pendulum clock hangs in a hospital corridor. Which statement must be true for the bob to be moving with simple harmonic motion?
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B is the defining condition: a = −ω²x. A is wrong because a constant acceleration gives uniformly accelerated motion, not oscillation about a point. C describes a drag force, which causes damping, not SHM. D has force ∝ x², which is not linear in x, so the motion would not be SHM.
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Answer B
Answer: B
Common error: Do not confuse a force that depends on speed (damping) with a restoring force that depends on displacement.
10IBP3H-C1-20r7-08SHM velocity and displacementPaper 1AEasy1 mark
A mass moves with simple harmonic motion on a horizontal spring and friction is negligible. At the equilibrium position, which statement is correct?
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At x = 0 the speed is maximum and a = −ω²x = 0, so C is correct. A and B each give a maximum acceleration, which occurs at the turning points. A also has zero kinetic energy, which also happens at the turning points. D is wrong because total energy ½kA² is constant and non-zero.
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Answer C
Answer: C
Exam tip: Speed is maximum where acceleration is zero, and acceleration is maximum where speed is zero.
11IBP3H-C1-20r10-01Acceleration in SHMPaper 1AEasy1 mark
A book trolley in a library runs on a horizontal track between two identical springs. It is displaced slightly along the track and released. Friction is negligible. Which statement must be true for the trolley's motion to be simple harmonic?
Diagram not to scale
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B is correct: SHM requires a restoring force F = −kx, so acceleration is proportional to displacement and opposite to it. A is wrong because a constant-magnitude force gives constant acceleration, not a=−ω2x. C describes a damping (drag) force, not a restoring force. D is wrong because the force must vary linearly, not with the square, of the displacement.
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Answer B
Answer: B
Common error: A force that points towards equilibrium is not enough for SHM; its magnitude must also be directly proportional to the displacement.
12IBP3H-C1-20r10-02Energy in SHMPaper 1AEasy1 mark
The top of a wind-turbine tower sways with simple harmonic motion of amplitude 0.20 m and period 2.0 s. What is the maximum magnitude of the acceleration of the tower top?
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ω = 2π/T = π rad s−1, so amax = ω2x0 = π2 × 0.20 = 1.97 m s−2, so the third option is correct. 0.628 m s−2 is ωx0, the maximum speed in m s−1, not acceleration. 7.90 m s−2 uses ω = 2π instead of 2π/T. 0.0500 m s−2 comes from x0/T2, which omits the factor 4π2.
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Answer C
Answer: C
Exam tip: Maximum acceleration is ω²x₀, where ω must first be found from 2π/T.
13IBP3H-C1-20r10-03SHM velocity and displacementPaper 1AEasy1 mark
The end of a diving springboard at a swimming pool oscillates vertically with simple harmonic motion after a diver leaves it. At the instant its displacement from equilibrium is half the amplitude, what fraction of the total energy of the oscillation is kinetic?
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Total energy ∝ x02 and potential energy ∝ x2, so at x = x0/2 the potential energy is 0.25 of the total and the kinetic energy is 0.75, so the second option is correct. 0.25 is the potential fraction. 0.50 wrongly assumes an equal share at half amplitude. 0.87 is the ratio of speeds v/vmax (√3/2), not of energies.
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Answer B
Answer: B
Common error: Energy fractions go as the square of displacement, so half the amplitude does not mean half the potential energy.
14IBP3H-C1-20r20-01Energy in SHMPaper 1AEasy1 mark
A vibration sensor fixed to the wall of a beehive moves with simple harmonic motion of frequency 50 Hz and amplitude 2.0 mm. What is the maximum acceleration of the sensor?
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Correct: B. amax = ω2x0 = (2π × 50)2 × 2.0×10−3 = 197 m s−2. A is ωx0, the maximum speed in m s−1, not the acceleration. C uses ω = f without the factor 2π. D uses 2πf2x0, squaring f but not 2π.
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Answer B
Answer: B
Common error: Angular frequency is ω = 2πf, and it must be squared as a whole in a = −ω²x.
15IBP3H-C1-20r20-02SHM velocity and displacementPaper 1AEasy1 mark
A mass on a spring in a hospital weighing device oscillates with simple harmonic motion of amplitude A and total energy E. What is the kinetic energy of the mass when its displacement from equilibrium is A/2?
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Correct: C. Potential energy is proportional to x2, so at x = A/2 it is E/4, leaving a kinetic energy of E − E/4 = 3E/4. A is the potential energy, not the kinetic energy. B assumes energy is shared equally at half amplitude. D is the ratio of the speed to the maximum speed, not an energy ratio.
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Answer C
Answer: C
Common error: Energy depends on the square of the displacement, so half the amplitude means a quarter of the total energy stored as potential energy.
16IBP3H-C1-20r20-03Damped oscillationsPaper 1AEasy1 mark
The suspension of a Formula 1 test rig is modelled as a lightly damped oscillator. Which row describes how the oscillations change over successive cycles?
A. amplitude decreases, total energy constant, period constant B. amplitude constant, total energy decreases, period constant C. amplitude and total energy decrease, period progressively decreases D. amplitude and total energy decrease, period approximately constant
Which is correct?
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Correct: D. Light damping removes energy through resistive forces, so amplitude falls, while the period stays approximately constant. A is wrong because a smaller amplitude means less energy. B is wrong because energy cannot be lost while the amplitude stays constant. C is wrong because light damping hardly changes the period.
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Answer D
Answer: D
Exam tip: Link amplitude and energy: energy is proportional to amplitude squared, so a falling amplitude always means energy is being lost.
17IBP3H-C1-20r39-01Acceleration in SHMPaper 1AEasy1 mark
Four trolleys move along a straight track in a library demonstration. For each trolley the acceleration a (in m s−2) is given in terms of its displacement x (in m) from a fixed point O. Which trolley moves with simple harmonic motion about O?
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Simple harmonic motion requires an acceleration that is proportional to the displacement and always directed towards O, which means a = −(constant)x. The relation a = −3x satisfies both conditions. The relation a = +3x has the acceleration directed away from O, so the motion runs away from O. The relation with x2 is not proportional to x and always points in the negative direction. The relation with 1/x is not proportional to x and becomes infinite at O.
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Answer C
Answer: C
Common error: Proportional to displacement is not enough; the minus sign (acceleration towards equilibrium) is part of the definition.
18IBP3H-C1-20r39-02Mass–spring and pendulum periodsPaper 1AEasy1 mark
A mass hanging from a light vertical spring performs small oscillations with a period of 1.2 s. The mass is replaced by one that is four times as large. The spring obeys Hooke's law. What is the new period?
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For a mass–spring system T = 2π√(m/k), so T is proportional to √m. Four times the mass gives √4 = 2 times the period, so 2.4 s. The value 4.8 s comes from treating T as proportional to m. The value 0.60 s uses an inverse square-root relation. The value 1.2 s wrongly assumes the period is independent of mass, which is true for a pendulum but not for a spring.
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Answer A
Answer: A
Exam tip: Identify whether the period depends on mass: it does for a mass–spring system (as √m) but not for a simple pendulum.
19IBP3H-C1-20r39-03Mass–spring and pendulum periodsPaper 1AEasy1 mark
A pendulum clock in a library has a simple pendulum that swings through a small angle with a period of 2.00 s. What is the length of the pendulum? (g = 9.8 m s−2)
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From T = 2π√(l/g), l = gT2/(4π2) = 9.8 × 4.00 / 39.5 = 0.99 m. The value 0.25 m results from using a period of 1.00 s, the time for a single swing. The value 3.1 m comes from using l = gT/(2π), with T not squared. The value 39 m comes from omitting the factor 2π altogether.
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Answer D
Answer: D
Common error: A clock's 'tick' interval can be half the period; use the full period of one complete oscillation in the formula.
21C-1A-03Phase and timing in SHMPaper 1AMedium1 mark
A particle performs SHM of period T and amplitude A.
What is the shortest time for the particle to move from the equilibrium position to a displacement of A/2?
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Step 1Starting at equilibrium, x = A sin ωt with ω = 2π/T.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2sin ωt = ½ gives ωt = π/6.
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Step 3t = (π/6)/(2π/T) = T/12.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: sin(2πt/T) = ½ when 2πt/T = π/6, i.e. t = T/12.
BThis is the time to reach A/√2, half the energy.
CT/6 is the time to go from maximum displacement A to A/2 (cos 2πt/T = ½) — the answer if timing had started at the extreme position, not at equilibrium.
Two horizontal mass–spring oscillators P and Q have masses that are equal. The spring constant of P is four times the spring constant of Q. The oscillators perform simple harmonic motion with the same total energy.
Which row gives (amplitude of P)/(amplitude of Q) and (maximum speed of P)/(maximum speed of Q)?
Amplitude of P/amplitude of QMaximum speed of P/maximum speed of Q
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Step 1At maximum displacement all the energy is elastic potential energy: ET = ½kx0² (equivalently ½mω²x0² with ω² = k/m). With equal energies, x0 ∝ 1/√k, so the amplitude ratio is √(1/4) = 1/2.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2At equilibrium all the energy is kinetic: ET = ½mvmax². Equal energies and equal masses give equal maximum speeds: ratio 1.
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Step 3Check: vmax = ωx0; P has twice the angular frequency (√4) and half the amplitude, so the product is the same.
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis uses x0 ∝ 1/k without the square root; ET depends on x0².
BThe amplitude ratio is right, but this takes vmax ∝ ω = √(k/m) and forgets that P also has the smaller amplitude.
CThis inverts the amplitude ratio, as if a stiffer spring allowed a larger amplitude for the same energy; a stiffer spring stores the same energy at a smaller extension.
DCorrect: x0 ∝ 1/√k for equal energy, and ½mvmax² is the same for both.
23C-1A-17Mass–spring and pendulum periodsPaper 1AMedium1 mark
A simple pendulum is used on the surface of a planet. The length L of the pendulum is varied and the period T is measured for each length. The graph shows L against T², with the line of best fit.
What length of pendulum has a period of 2.0 s on this planet?
Length of the pendulum against the square of its period, with the line of best fit (drawn to scale).Show mark scheme
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Step 1T = 2π√(L/g) gives L = (g/4π²)T², so L is proportional to T² and the line passes through the origin.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2From the line, gradient = 0.95 m/10.0 s² = 0.095 m s−2.
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Step 3For T = 2.0 s, T² = 4.0 s², so L = 0.095 × 4.0 = 0.38 m.
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Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis divides the gradient by T² (0.095/4.0) instead of multiplying: L increases with T².
BThis uses T instead of T²: 0.095 × 2.0 = 0.19 m. The graph is linear in T², not in T.
CCorrect: L = gradient × T² = 0.095 × 4.0 = 0.38 m.
DThis ignores the graph and uses g = 9.81 m s−2 for the Earth: L = 9.81 × 4.0/4π² = 0.99 m.
The graph shows how the velocity v of an object performing simple harmonic motion varies with time t. X is a point on the graph.
Which statements about the object at the instant X are correct?
I. Its displacement has its maximum magnitude.
II. Its acceleration has its maximum magnitude.
III. Its kinetic energy is zero.
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Step 1At X the velocity is zero, so the object is momentarily at rest at an extreme position: the displacement has its maximum magnitude (I) and the kinetic energy is zero (III).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The acceleration is the gradient of the v–t graph. At X the graph crosses the axis where it is steepest, so the acceleration has its maximum magnitude (II), as a = −ω²x also requires at maximum |x|.
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Step 3All three statements are correct.
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AIII is also correct: zero velocity means zero kinetic energy.
BII is also correct: zero velocity does not mean zero acceleration — the gradient of the v–t graph is greatest where v = 0.
CI is also correct: the object is at rest only at the extremes of its motion, where |x| = x0.
DCorrect: at X the object is at an extreme position, at rest, with the largest restoring force.
25C-1A-33Phase and timing in SHMPaper 1AMedium1 mark
An object performs simple harmonic motion with displacement x = x0 sin ωt.
Which row describes the phase of the velocity v and of the acceleration a relative to the displacement?
Velocity relative to displacementAcceleration relative to displacement
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Step 1v = ωx0 cos ωt = ωx0 sin(ωt + π/2): the velocity reaches its maximum a quarter of a cycle before the displacement does, so it leads by π/2.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2a = −ω²x = ω²x0 sin(ωt + π): the acceleration has its positive maximum when the displacement has its negative maximum, a phase difference of π.
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Step 3So v leads x by π/2 and a is in antiphase with x.
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThe lead and lag are reversed: cos ωt peaks at t = 0, a quarter-period before sin ωt peaks, so the velocity leads.
BThis ignores the minus sign in a = −ω²x: a is proportional to x but always opposite in sign.
CThis treats the velocity as always opposite to the displacement; in fact the velocity is zero when the displacement is maximum, which requires a phase difference of π/2.
DCorrect: v ∝ cos ωt leads by π/2; a ∝ −sin ωt is in antiphase.
26C-1A-34Phase difference between oscillatorsPaper 1AMedium1 mark
The graph shows how the displacement x varies with time t for two identical oscillators P and Q.
What is the magnitude of the phase difference between P and Q?
Displacement–time graphs of P (solid) and Q (dashed), drawn to scale. P crosses zero moving upwards at t = 0 and t = 1.20 s; Q does so at t = 0.15 s and t = 1.35 s.Show mark scheme
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Step 1From the graph the period of each oscillator is T = 1.20 s, and Q passes through each point of its cycle 0.15 s after P.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The time shift is 0.15/1.20 = 0.125 of a cycle.
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Step 3One cycle corresponds to 2π rad, so the phase difference = 2π × 0.125 = π/4 rad.
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Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
A0.125 is the fraction of a cycle; it has not been converted to radians by multiplying by 2π.
BThis multiplies the fraction of a cycle by π instead of 2π (treating one cycle as π rad).
27C-1A-66Mass–spring and pendulum periodsPaper 1AMedium1 mark
A simple pendulum hangs from the ceiling of a lift. When the lift is at rest, the period of small oscillations of the pendulum is T. The lift then moves with a constant downward acceleration of magnitude g/4.
What is the period of small oscillations while the lift accelerates?
Diagram not to scale
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Step 1With the bob at rest relative to the lift, mg − Fstring = m(g/4), so the tension is m(3g/4): the pendulum behaves as if the gravitational field strength were geff = 3g/4.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2T ∝ 1/√geff, so T′/T = √(g/geff) = √(4/3).
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Step 3T′ = (2/√3)T ≈ 1.15T.
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Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis adds the accelerations (geff = 5g/4), which is correct only for an upward acceleration: (2/√5)T ≈ 0.89T.
BThis assumes that the period depends only on the length of the pendulum. The restoring force depends on the tension, which is smaller in a lift accelerating downwards.
CCorrect: geff = g − g/4 = 3g/4, so T′ = T√(4/3) = (2/√3)T.
DThis uses T ∝ 1/geff, without the square root: (4/3)T.
28IBP2S-018-00-07SHM velocity and displacementPaper 1AMedium1 mark
A tuned mass damper at the top of a skyscraper oscillates in simple harmonic motion with amplitude A. What fraction of the total energy of the oscillation is kinetic energy when the displacement of the damper is A/2?
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The potential energy is \(E_p = \frac{1}{2}m\omega^2x^2\). At x = A/2 this is (1/2)2 = 1/4 of the total energy \(\frac{1}{2}m\omega^2A^2\). The kinetic energy is therefore 1 − 1/4 = 3/4 of the total. The option 1/4 is the potential energy fraction. The option √3/2 is the ratio of the speed to the maximum speed, not an energy ratio. The option 1/2 assumes energy varies linearly with displacement.
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Answer D
Answer: D
Common error: Energy depends on x², so being halfway out does not mean half the energy is potential.
29IBP2S-018-00-08SHM velocity and displacementPaper 1AMedium1 mark
The tip of an ultrasonic dental scaler vibrates in simple harmonic motion with amplitude A. At what magnitude of displacement is the speed of the tip equal to half its maximum speed?
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The speed is \(v = \omega\sqrt{A^2 - x^2}\) and the maximum speed is ωA. Setting \(\sqrt{A^2-x^2} = A/2\) gives x2 = 3A2/4, so x = (√3/2)A ≈ 0.87A. The option 0.50A assumes speed falls linearly with displacement. The option 0.71A is where the kinetic energy, not the speed, is half its maximum.
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Answer D
Answer: D
Exam tip: Use v = ω√(x₀² − x²) directly. Speed changes slowly near equilibrium and quickly near the turning points.
30IBP2H-019-01-02SHM velocity and displacementPaper 1AMedium1 mark
A tiny proof mass inside a smartphone MEMS accelerometer oscillates with simple harmonic motion of amplitude x0. What fraction of the total energy of the oscillator is kinetic energy when the displacement of the proof mass is x0/2?
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The elastic potential energy is proportional to x2. At x = x0/2 it is (1/2)2 = 1/4 of the total energy. Kinetic energy plus potential energy is constant, so the kinetic energy is 1 − 1/4 = 3/4 of the total. Option A is the potential energy fraction. Option B wrongly assumes energy shares linearly with displacement. Option D is the ratio v/vmax, not the energy ratio.
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Answer C
Answer: C
Exam tip: Find the potential energy fraction first using (x/x₀)², then subtract it from 1 to get the kinetic fraction.
31IBP2H-019-01-04Energy in SHMPaper 1AMedium1 mark
A wearable energy harvester contains a mass on a spring. The mass oscillates with SHM when the wearer walks. On a different walk, the amplitude of the oscillation doubles. The spring obeys Hooke's law throughout. Which row correctly describes the change in the maximum speed, the period and the total energy of the oscillator?
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The angular frequency is ω = √(k/m), which does not depend on amplitude, so the period is unchanged. The maximum speed is vmax = ωx0, so it doubles. The total energy is E = ½mω2x02, so it increases by a factor of 22 = 4. Option B assumes a larger amplitude takes longer. Options C and D confuse which quantity depends on the square of the amplitude.
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Answer A
Answer: A
Common error: For an ideal oscillator the period does not depend on amplitude. A bigger swing simply means a faster swing.
32IBP2H-019-01-05SHM velocity and displacementPaper 1AMedium1 mark
A tuned mass damper at the top of a tall tower oscillates with SHM of amplitude x0. What is the magnitude of the displacement of the damper when its speed is half of its maximum speed?
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The speed is v = ω√(x02 − x2). Setting this equal to ωx0/2 gives x02 − x2 = x02/4, so x2 = 3x02/4 and x = √3x0/2 ≈ 0.87x0. Option B assumes speed falls linearly with displacement. Option C is the displacement where kinetic and potential energies are equal. Option A has no physical basis.
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Answer D
Answer: D
Exam tip: Speed in SHM is not a linear function of displacement. The mass is still moving at half its maximum speed when it is about 87% of the way to maximum displacement.
33IBP2H-019-01-06Mass–spring and pendulum periodsPaper 1AMedium1 mark
After a car drives over a bump, its suspension performs lightly damped oscillations. The amplitude of the oscillation decreases by 10% in each complete cycle. What percentage of the oscillation energy is lost in each cycle?
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The energy of an oscillator is proportional to the square of the amplitude. After one cycle the amplitude is 0.90 of its previous value, so the energy is 0.902 = 0.81 of its previous value. The energy lost is therefore 1 − 0.81 = 0.19, or 19%. Option A assumes energy is proportional to amplitude. Option C simply doubles 10%. Option D is the fraction of energy that remains, not the fraction lost.
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Answer B
Answer: B
Common error: Check whether the question asks for the fraction lost or the fraction remaining. In damping questions these two answers are often both offered.
34IBP2H-019-00-03Phase and timing in SHMPaper 1AMedium1 mark
Two pistons in an engine move with SHM of the same frequency and amplitude. At one instant piston P is at its maximum positive displacement. At the same instant piston Q is at its equilibrium position, moving in the negative direction. Which statement about the phase relationship is correct?
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Take xP = x0 cos ωt. Q starts at zero and moves negative, so xQ = −x0 sin ωt = x0 cos(ωt + π/2). Q therefore leads by π/2. Check: Q is moving towards its negative maximum, which it reaches a quarter-cycle before P does. A difference of π would need Q at its negative maximum. Choosing 'P leads' is the common sign error.
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Answer A
Answer: A
Common error: To find which oscillator leads, ask which one reaches the same point in its cycle (for example, the positive maximum) first.
35IBP2H-019-00-04Acceleration in SHMPaper 1AMedium1 mark
An accelerometer records the motion of a vibrating machine part. A graph of acceleration against displacement is a straight line through the origin with gradient −400 s−2. What is the frequency of the oscillation?
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Comparing with a = −ω²x, the gradient is −ω², so ω² = 400 s−2 and ω = 20 rad s−1. Then f = ω/2π = 20/2π ≈ 3.2 Hz. The option 400 Hz takes the gradient as the frequency. The option 20 Hz confuses ω with f. The option 0.31 Hz is actually the period in seconds (2π/20).
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Answer B
Answer: B
Exam tip: On an a–x graph for SHM the gradient is −ω², not −ω and not −f.
36IBP2H-019-00-05SHM velocity and displacementPaper 1AMedium1 mark
A vibration-isolation table in a laboratory oscillates with SHM of amplitude x0. What fraction of the total energy is kinetic energy when the displacement is x0/2?
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Ek = ½mω²(x0² − x²) and ET = ½mω²x0². The ratio is 1 − (x/x0)² = 1 − 1/4 = 3/4. The option 1/4 is the potential-energy fraction. The option 1/2 assumes energy varies linearly with x. The option √3/2 is the speed ratio v/vmax, not the energy ratio.
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Answer D
Answer: D
Common error: Energy depends on x², so halving the displacement leaves only a quarter of the energy as potential energy.
37IBP2H-019-00-06Phase and timing in SHMPaper 1AMedium1 mark
The tip of a diving board oscillates with SHM of period T and amplitude x0. What is the shortest time for the tip to move from its equilibrium position to a displacement of x0/2?
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With x = x0 sin ωt, setting x = x0/2 gives sin ωt = ½, so ωt = π/6. Then t = (π/6)/(2π/T) = T/12. The option T/8 assumes constant speed over the quarter-cycle. The option T/6 is the time from the maximum displacement to x0/2. The option T/4 is the time to reach the full amplitude.
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Answer C
Answer: C
Exam tip: The tip moves fastest near equilibrium, so it covers the first half of the amplitude in less than half of the quarter-period.
38IBP2S-018-02-07Energy in SHMPaper 1AMedium1 mark
The displacement of an oscillating component in an electric toothbrush model is given by x = 0.040 cos(5.0t), where x is in metres and t is in seconds. What is the speed of the component when x = 0.020 m?
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Use v = ω√(x0² − x²) with ω = 5.0 rad s−1 and x0 = 0.040 m. This gives v = 5.0 × √(0.0016 − 0.0004) = 5.0 × 0.0346 = 0.17 m s−1. The answer 0.20 is the maximum speed ωx0. The answer 0.10 is ωx. The answer 0.035 omits ω.
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Answer B
Answer: B
Exam tip: Read ω directly from the argument of the cosine. Then use v = ω√(x₀² − x²), which gives the speed at any displacement.
39IBP3H-C1-20r7-02SHM velocity and displacementPaper 1AMedium1 mark
A sensor of mass m on a beehive scale oscillates horizontally with simple harmonic motion of amplitude A. Ignore friction. At the instant its displacement from equilibrium is A/2, what fraction of the total energy is kinetic?
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Ek = ½k(A² − x²) and E = ½kA², so Ek/E = 1 − x²/A² = 1 − 1/4 = 3/4, so C is correct. A (1/4) is the fraction that is potential. B assumes energy is shared equally at half amplitude (it is equal only at x = A/√2). D confuses the displacement ratio at which energies are equal with the fraction asked.
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Answer C
Answer: C
Common error: Energy depends on the square of displacement, so halving the displacement does not halve the potential energy.
40IBP3H-C1-20r7-03Energy in SHMPaper 1AMedium1 mark
A mass on a spring oscillates with simple harmonic motion. The amplitude of the oscillation is doubled, with the same mass and spring. Which row describes the change?
A period doubles; maximum kinetic energy doubles B period unchanged; maximum kinetic energy doubles C period doubles; maximum kinetic energy increases four times D period unchanged; maximum kinetic energy increases four times
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The period T = 2π√(m/k) does not depend on amplitude. The maximum kinetic energy equals the total energy ½kA², which is proportional to A², so it increases four times: D is correct. A and C wrongly make the period depend on amplitude. A and B treat energy as proportional to A rather than A².
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Answer D
Answer: D
Exam tip: In SHM the period is independent of amplitude while the energy goes as amplitude squared.
41IBP3H-C1-20r7-04Phase and timing in SHMPaper 1AMedium1 mark
Two identical oscillators have displacements x1 = x0 sin ωt and x2 = x0 sin(ωt + π/2). At the instant oscillator 1 is at its maximum positive displacement, oscillator 2 is
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x2 = x0 cos ωt. When ωt = π/2, x1 is maximum and x2 = 0. The velocity of oscillator 2 is −ωx0 sin ωt = −ωx0, which is negative, so C is correct. A would need zero phase difference. B has the sign of the velocity reversed. D would need a phase difference of π.
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Answer C
Answer: C
Common error: A phase difference of π/2 means one oscillator is at an extreme when the other passes through equilibrium; check the direction of motion using the sign of the velocity.
42IBP3H-C1-20r7-05Phase difference between oscillatorsPaper 1AMedium1 mark
Two vibrating bee-wing models have the same period of 0.80 s. Model B reaches its maximum displacement 0.10 s after model A does. What is the magnitude of the phase difference between them?
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Phase difference = 2π × (time lag/period) = 2π × 0.10/0.80 = π/4 rad, so B is correct. A omits a factor of 2 (π × 0.125). C treats 0.10 s as a quarter of the period. D gives the fraction 0.10/0.80 = 0.125 without multiplying by 2π.
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Answer B
Answer: B
Exam tip: Convert a time lag to a phase difference by multiplying the fraction of a period by 2π.
43IBP3H-C1-20r7-06Conditions for SHMPaper 1AMedium1 mark
A student releases a simple pendulum of length 0.50 m from an angle of 40° to the vertical and finds its motion is not accurately simple harmonic. What is the best explanation?
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The restoring force is mg sin θ along the arc. It is proportional to the displacement only when sin θ ≈ θ, which holds for small angles, so B is correct. A is wrong because the tangential component of the weight still provides a restoring force. C is wrong because mass cancels in the equation of motion. D is wrong because damping reduces amplitude but does not change the force law.
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Answer B
Answer: B
Common error: SHM for a pendulum requires small angles; the mass of the bob does not matter.
44IBP3H-C1-20r10-04Energy in SHMPaper 1AMedium1 mark
In a library demonstration a mass m hangs from a light spring and oscillates vertically with period T. The mass is replaced by a mass 4m and the spring by one with four times the force constant. What is the new period of oscillation?
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T = 2π√(m/k). Replacing m by 4m and k by 4k leaves m/k unchanged, so the period stays T, so the second option is correct. T/2 follows from changing only the spring constant. 2T follows from changing only the mass. 4T takes the mass factor without the square root and ignores the spring.
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Answer B
Answer: B
Exam tip: Substitute both changes into T = 2π√(m/k) together; the ratio m/k may be unchanged.
45IBP3H-C1-20r10-05Phase and timing in SHMPaper 1AMedium1 mark
A float on a swimming-pool lane rope bobs vertically with simple harmonic motion. Its acceleration a and displacement x from equilibrium are related by a = −(160 s−2)x. What is the period of the motion?
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Comparing with a = −ω2x gives ω2 = 160 s−2, so ω = 12.65 rad s−1 and T = 2π/ω = 0.497 s, so the second option is correct. 0.0393 s uses 2π/160, treating ω2 as ω. 0.0791 s is 1/ω, omitting the 2π. 2.01 s is the frequency in Hz, not the period.
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Answer B
Answer: B
Common error: The coefficient in a = −(…)x is ω², not ω; take the square root before using T = 2π/ω.
46IBP3H-C1-20r10-06SHM velocity and displacementPaper 1AMedium1 mark
A vibration sensor on a wind-turbine blade moves with simple harmonic motion of amplitude 0.12 m and angular frequency 5.0 s−1. What is its speed when its displacement from equilibrium is 0.072 m?
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v = ω√(x02 − x2) = 5.0 × √(0.0144 − 0.005184) = 5.0 × 0.096 = 0.48 m s−1, so the third option is correct. 0.60 m s−1 is the maximum speed ωx0. 0.36 m s−1 is ωx, which uses the displacement instead of the square-root expression. 0.24 m s−1 is ω(x0 − x), which subtracts amplitudes instead of squares.
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Answer C
Answer: C
Common error: In v = ω√(x₀² − x²) the squares are subtracted inside the root; do not subtract x from x₀ first.
47IBP3H-C1-20r10-07Energy in SHMPaper 1AMedium1 mark
A mass–spring oscillator in a library demonstration has total energy 0.80 J and amplitude 0.10 m. What is its kinetic energy when its displacement is 0.060 m?
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Potential energy = 0.80 × (0.060/0.10)2 = 0.288 J, so kinetic energy = 0.80 − 0.288 = 0.512 J, so the fourth option is correct. 0.288 J is the potential energy. 0.320 J takes energy ∝ (x0 − x)/x0. 0.480 J takes the kinetic energy as 0.80 × x/x0, treating energy as linear in displacement.
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Answer D
Answer: D
Exam tip: Potential energy in SHM is proportional to x²; find it first, then subtract from the total.
48IBP3H-C1-20r20-04Energy in SHMPaper 1AMedium1 mark
A beehive of mass m hangs from a spring of spring constant k and oscillates vertically with period T. Honey is added so that the total mass becomes 4m, and the spring is replaced by one of spring constant 2k. What is the new period?
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Correct: B. T = 2π√(m/k), so T′/T = √((4m/2k)/(m/k)) = √2. C ignores the change in the spring. D treats the spring constant as halved instead of doubled, giving √8. A inverts the ratio.
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Answer B
Answer: B
Exam tip: Period is proportional to √(m/k). Take the ratio of the new to the old value of m/k first, then the square root.
49IBP3H-C1-20r20-05Phase difference between oscillatorsPaper 1AMedium1 mark
Two vibration sensors A and B on the front wing of a Formula 1 car oscillate with simple harmonic motion at the same frequency of 8.0 Hz. Sensor B reaches its maximum positive displacement 0.025 s after sensor A reaches its maximum positive displacement. What is the smallest phase difference between the sensors, and which one lags?
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Correct: B. T = 1/8.0 = 0.125 s, so the phase difference is 2π × 0.025/0.125 = 1.26 rad. B reaches its maximum later, so it lags A. A has the wrong direction. C uses π instead of 2π per cycle. D is the fraction of a cycle, 0.20, not an angle in radians.
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Answer B
Answer: B
Common error: Convert a time delay to a phase difference using φ = 2πΔt/T. The sensor that reaches the same point later is the one that lags.
50IBP3H-C1-20r20-06Energy in SHMPaper 1AMedium1 mark
The piston of a hospital ventilator moves with simple harmonic motion of amplitude 5.0 cm and frequency 2.0 Hz. What is the speed of the piston when its displacement from equilibrium is 3.0 cm?
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Correct: C. ω = 2π × 2.0 = 12.57 rad s−1 and v = ω√(x02 − x2) = 12.57 × 0.040 = 0.50 m s−1. A uses f instead of ω. B is ωx = 12.57 × 0.030, which ignores the amplitude. D is the maximum speed ωx0, not the speed at 3.0 cm.
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Answer C
Answer: C
Exam tip: In v = ω√(x₀² − x²), use the same units for x₀ and x and convert to metres at the end.
51IBP3H-C1-20r20-08SHM velocity and displacementPaper 1AMedium1 mark
The piston of a hospital infusion pump moves with simple harmonic motion. At one instant its speed is half of its maximum speed. What is the magnitude of its acceleration at that instant as a fraction of its maximum acceleration?
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Correct: C. Half the maximum speed means KE = E/4, so PE = 3E/4. Since PE is proportional to x2, x/x0 = √3/2, and acceleration is proportional to x. A wrongly assumes acceleration varies linearly with speed. B is the fraction of energy that is potential. D is the fraction that is kinetic.
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Answer C
Answer: C
Common error: Speed and acceleration are not proportional. Use energy to find the displacement first, then use a ∝ x.
52IBP3H-C1-20r39-04Energy in SHMPaper 1AMedium1 mark
A buoy in a swimming-pool lane rope bobs vertically with simple harmonic motion of amplitude 0.050 m and angular frequency 8.0 rad s−1. What is the speed of the buoy when its displacement from equilibrium is 0.030 m?
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In SHM v = ω√(x02 − x2) = 8.0 × √(0.0502 − 0.0302) = 8.0 × 0.040 = 0.32 m s−1. The value 0.24 m s−1 is ωx, using the displacement instead of the amplitude-dependent expression. The value 0.40 m s−1 is the maximum speed ωx0, which occurs only at equilibrium. The value 0.16 m s−1 uses ω(x0 − x).
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Answer B
Answer: B
Exam tip: Speed in SHM at displacement x uses the difference of squares inside the root, not the difference of the lengths.
53IBP3H-C1-20r39-05Acceleration in SHMPaper 1AMedium1 mark
The top of a wind-turbine tower sways with simple harmonic motion of frequency 0.25 Hz and amplitude 0.40 m. What is the maximum acceleration of the top of the tower?
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The angular frequency is ω = 2πf = 2π × 0.25 = 1.57 rad s−1, and the maximum acceleration is ω2x0 = 1.572 × 0.40 = 0.99 m s−2. The value 0.025 m s−2 uses f2x0, forgetting the 2π. The value 0.63 m s−2 is ωx0, the maximum speed. The value 0.16 m s−2 squares only the frequency and keeps a single factor of 2π.
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Answer B
Answer: B
Common error: Convert frequency to angular frequency with ω = 2πf before using the SHM equations, and square ω for acceleration.
54IBP3H-C1-20r39-07Phase and timing in SHMPaper 1AMedium1 mark
A sensor in a wind-turbine nacelle has displacement x = x0 sin(ωt) from equilibrium, where T is the period. What is its acceleration at time t = T/4?
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At t = T/4 the phase is ωt = π/2, so x = +x0. The acceleration is a = −ω2x = −ω2x0, directed back towards equilibrium. Zero is the value of the velocity at this instant, not the acceleration. The value +ω2x0 ignores the minus sign. The value −ωx0 has the wrong power of ω and is not an acceleration.
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Answer A
Answer: A
Common error: At maximum displacement the speed is zero but the acceleration is at its largest; do not confuse the two.
A block of mass 0.20 kg on a horizontal spring performs simple harmonic motion. The graph shows how the kinetic energy of the block varies with time t.
What is the amplitude of the oscillation?
Kinetic energy of the block against time (drawn to scale).Show mark scheme
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Step 1The kinetic energy is a maximum each time the block passes through equilibrium, which happens twice per cycle. The graph repeats every 0.40 s, so the period of the oscillation is T = 0.80 s and ω = 2π/0.80 = 7.85 rad s−1.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2At the peak ½mvmax² = 0.090 J, so vmax = √(2 × 0.090/0.20) = 0.949 m s−1.
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Step 3x0 = vmax/ω = 0.949/7.85 = 0.12 m.
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis makes both errors: no ½ in the kinetic energy (vmax = 0.671 m s−1) and T = 0.40 s (ω = 15.7 rad s−1), giving 0.043 m.
BThis takes the repeat time of the kinetic-energy graph (0.40 s) as the period of the oscillation: 0.949/15.7 = 0.060 m.
CThis omits the ½ in Ek = ½mv², giving vmax = 0.671 m s−1 and 0.671/7.85 = 0.085 m.
DCorrect: T = 0.80 s, vmax = 0.949 m s−1, x0 = vmax/ω = 0.12 m.
The test mass of a seismometer has mass 0.50 kg and performs simple harmonic motion along a horizontal line. The graph shows how its potential energy Ep varies with displacement x for displacements up to 4.0 cm. The amplitude of the oscillation is 8.0 cm.
What is the maximum speed of the test mass?
Potential energy of the test mass against displacement, for displacements up to 4.0 cm (drawn to scale).Show mark scheme
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Step 1From the graph, Ep = 0.040 J at x = 0.040 m. Using Ep = ½mω²x²: ω² = 2 × 0.040/(0.50 × 0.040²) = 100 s−2, so ω = 10 rad s−1.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The amplitude is x0 = 0.080 m, so vmax = ωx0 = 10 × 0.080.
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Step 3vmax = 0.80 m s−1 (equivalently ET = ½mω²x0² = 0.16 J = ½mvmax²).
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Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis takes the 0.040 J read at the edge of the graph as the total energy: √(2 × 0.040/0.50) = 0.40 m s−1. The total energy is at x = 8.0 cm, where Ep is four times larger.
BThis omits the ½ in Ep = ½mω²x², giving ω² = 50 s−2 and vmax = 7.07 × 0.080 = 0.57 m s−1.
CCorrect: ω = 10 rad s−1 and vmax = ωx0 = 0.80 m s−1.
DThis is ω²x0 = 100 × 0.080 = 8.0, the maximum acceleration in m s−2, not the maximum speed.
57C-1A-70Phase and timing in SHMPaper 1AHard1 mark
A particle performs simple harmonic motion of period T and amplitude x0. At time t = 0 its displacement is +x0/2 and it is moving towards the equilibrium position.
What is the earliest time after t = 0 at which the kinetic energy of the particle equals its potential energy?
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Step 1EK = EP when ½mω²x² = ½ × ½mω²x0², i.e. at |x| = x0/√2.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Write x = x0 sin(ωt + φ): x = x0/2 and moving towards equilibrium (decreasing) gives φ = 5π/6.
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Step 3The particle passes through equilibrium before reaching |x| = x0/√2, at x = −x0/√2, where ωt + φ = 5π/4. So ωt = 5π/4 − 5π/6 = 5π/12 and t = 5T/24.
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis is the time to go from x0/2 to x0/√2 moving away from equilibrium (phase π/6 to π/4); the particle is moving the other way.
BThis is the time from the equilibrium position to x0/√2, ignoring the starting position x0/2.
CThis takes EK = EP at |x| = x0/2 (from x0/2 through equilibrium to −x0/2, a phase change of π/3). At x0/2 the potential energy is only one quarter of the total.
DCorrect: the particle goes from phase 5π/6 to 5π/4, a phase change of 5π/12, which takes 5T/24.
58IBP2H-019-01-03Phase and timing in SHMPaper 1AHard1 mark
The tip of a vibrating surgical ultrasonic scalpel moves with SHM described by x = x0 sin(ωt + φ). At t = 0 the displacement of the tip is +x0/2 and the tip is moving in the negative direction. What is the value of φ?
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At t = 0 the displacement is x0 sin φ = x0/2, so sin φ = 1/2. This gives φ = π/6 or φ = 5π/6. The velocity is v = ωx0 cos(ωt + φ). For the velocity to be negative at t = 0, cos φ must be negative, so φ = 5π/6. Option A matches the displacement but gives positive velocity. Options B and C come from solving cos φ = 1/2 or cos φ = −1/2, which mixes up the sine and cosine forms.
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Answer D
Answer: D
Exam tip: A sine equation has two solutions between 0 and 2π. Always use the sign of the velocity to decide which phase angle is correct.
59IBP2H-019-01-08Mass–spring and pendulum periodsPaper 1AHard1 mark
On a space station, an astronaut's mass is measured by strapping them into a chair attached to springs and letting the chair oscillate with SHM. During one measurement, the maximum speed of the chair is 0.50 m s−1 and its maximum acceleration is 2.0 m s−2. What is the period of the oscillation?
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The maximum speed is vmax = ωx0 and the maximum acceleration is amax = ω2x0. Dividing gives ω = amax/vmax = 2.0/0.50 = 4.0 rad s−1. The period is T = 2π/ω = 2π/4.0 ≈ 1.6 s. Option A is vmax/amax without the factor of 2π. Option C gives ω itself but labels it as a period. Option D is 2π × amax/vmax, which inverts the ratio.
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Answer B
Answer: B
Exam tip: The amplitude cancels when you divide amax by vmax. This gives ω directly, so you do not need to know x₀.
A mass hung on a spring of spring constant k oscillates with period T. The spring is cut into two equal halves. Both halves are attached side by side (in parallel) and support the same mass. What is the new period?
Diagram not to scale
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Each half stretches half as much under the same force, so each half has spring constant 2k. Two halves in parallel share the load, giving 4k. Since T ∝ 1/√k, the new period is T/√4 = T/2. The option T/√2 accounts for cutting but forgets the parallel arrangement. The option T assumes cutting does not change k. The option 2T inverts the dependence.
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Answer C
Answer: C
Common error: A shorter piece of the same spring is stiffer. The spring constant is inversely proportional to the length of the coil.
61IBP2H-019-00-08Phase and timing in SHMPaper 1AHard1 mark
The displacement of an oscillator is x = x0 sin(ωt + φ), with −π < φ ≤ π. At t = 0 the displacement is −x0/2 and the oscillator is moving in the positive direction. What is φ?
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At t = 0, sin φ = −½, which gives φ = −π/6 or φ = −5π/6. The velocity is v = ωx0 cos(ωt + φ), which must be positive at t = 0, so cos φ > 0. Hence φ = −π/6. The value −5π/6 has the right displacement but negative velocity. The value π/6 gives positive displacement. The value −π/3 gives x = −(√3/2)x0.
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Answer A
Answer: A
Exam tip: Displacement alone gives two possible phase angles. Use the direction of motion (the sign of cos) to choose between them.
62IBP2S-018-02-10SHM velocity and displacementPaper 1AHard1 mark
An oscillator moves with displacement x = x0 sin(ωt) and period T. What is the earliest time after t = 0 at which its kinetic energy equals its elastic potential energy?
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The kinetic and potential energies are equal when each is half the total energy. Potential energy is proportional to x², so x² = x0²/2 and x = x0/√2. Then sin(ωt) = 1/√2, so ωt = π/4 and t = (π/4)/(2π/T) = T/8. The answer T/12 is when x = x0/2. The answer T/6 is when x = (√3/2)x0. At T/4 all the energy is potential.
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Answer B
Answer: B
Common error: Equal energies does not occur at half the amplitude. Find the displacement from x² = x₀²/2 first, then convert it to a time.
A mass on a light spring of spring constant k oscillates with period T. The same mass is now hung on a spring made from one half of the same original spring, cut at its midpoint. What is the new period?
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Notes
Cutting the spring in half doubles the spring constant, because each half is twice as stiff for the same force. T = 2π√(m/k) so T′ = 2π√(m/2k) = T/√2: B. A forgets the square root. C and D are built on halving k instead of doubling it.
✓ 1
Answer B
Answer: B
Common error: A shorter piece of spring is stiffer, not softer.
64IBP3H-C1-20r10-08Phase and timing in SHMPaper 1AHard1 mark
A float in a swimming pool oscillates with displacement x = x0 sin(ωt + φ). At t = 0 the float is at x = +x0/2 and moving towards the equilibrium position. What is a possible value of the phase constant φ?
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Notes
x(0) = x0 sin φ = x0/2 gives sin φ = 1/2, so φ = π/6 or 5π/6. The velocity is proportional to cos φ and must be negative (towards equilibrium from a positive displacement), so φ = 5π/6, the third option. π/6 gives positive velocity (moving away). 7π/6 and 11π/6 give sin φ = −1/2, a negative starting displacement.
✓ 1
Answer C
Answer: C
Common error: sin φ fixes the starting displacement but leaves two angles; use the direction of motion (sign of cos φ) to choose between them.
65IBP3H-C1-20r20-07Phase and timing in SHMPaper 1AHard1 mark
A hive-scale platform oscillates horizontally with simple harmonic motion of period T. At time t = 0 it is at its maximum displacement. At what time is its kinetic energy first equal to its potential energy?
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Notes
Correct: B. With x = x0cos ωt, KE = PE when x2 = x02/2, so cos ωt = 1/√2, ωt = π/4 and t = T/8. D is the time at which the platform reaches equilibrium, where all the energy is kinetic. C comes from cos ωt = 1/2 (x = x0/2). A comes from taking ωt = π/6.
✓ 1
Answer B
Answer: B
Exam tip: Equal energies means each is half the total, so x² = x₀²/2, not x = x₀/2.
66IBP3H-C1-20r39-06SHM velocity and displacementPaper 1AHard1 mark
The free end of a diving board vibrates with simple harmonic motion after a diver leaves it. Damping may be ignored. At the instant when the displacement of the end from equilibrium is half the amplitude, what fraction of the total energy of the oscillation is kinetic?
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Notes
The total energy is proportional to x02 and the potential energy to x2. At x = x0/2 the potential energy is (1/2)2 = 1/4 of the total, so the kinetic energy is 3/4 of the total. The fraction 1/2 assumes energy varies linearly with displacement. The fraction 1/4 is the potential-energy fraction, not the kinetic one. The value √3/2 is the ratio of the speed to the maximum speed, not the energy ratio.
✓ 1
Answer D
Answer: D
Exam tip: Energy in SHM depends on the square of displacement and the square of speed; do not use the ratios of x or v directly as energy fractions.
67IBP3H-C1-20r39-08Acceleration in SHMPaper 1AHard1 mark
A trolley of mass m is attached to a fixed support by a spring of spring constant k and oscillates on a frictionless track with period T. A second identical spring is added alongside the first so that both springs are attached to the trolley and the support and are stretched equally. What is the new period?
Diagram not to scale
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Notes
Two identical springs side by side each provide a restoring force kx, so the effective spring constant is 2k. Then T′ = 2π√(m/2k) = T/√2. The value T/2 treats T as inversely proportional to k. The value 2T and the value √2T come from taking the effective spring constant to be smaller, as for springs joined end to end.
✓ 1
Answer C
Answer: C
Exam tip: Springs side by side add their spring constants; springs end to end halve it. Then use T proportional to 1/√k.
68C-1B-01Mass–spring and pendulum periodsPaper 1BMedium7 marks
A spring hangs from a clamp and a load of mass m is attached to its lower end. The load is pulled down a little and released, and a student times 10 complete vertical oscillations with a hand-held stopwatch. The uncertainty in each timing, dominated by the student's reaction time, is ±0.2 s. The procedure is repeated for five loads.
The period is expected to be T = 2π√(m/k), where k is the spring constant, so the student plots T² against m. The graph shows the data with error bars, the line of best fit, and the lines of maximum and minimum gradient that pass through all the error bars.
m / kg
0.100
0.200
0.300
0.400
0.500
10T / s
4.5
5.8
7.1
8.2
9.0
T² / s²
0.203
0.336
0.504
0.672
T² against m with error bars and the lines of best fit, maximum and minimum gradient (drawn to scale).
(a)
(i)
Calculate T² for the 0.500 kg load, together with its absolute uncertainty.
(2)
(b)
(i)
Use the graph to determine the gradient of the line of best fit and its absolute uncertainty.
(2)
(c)
(i)
Determine k with its absolute uncertainty.
(2)
(ii)
In a separate experiment, the 0.400 kg load hanging at rest extends the spring by 0.157 m. Deduce whether this result is consistent with your answer to (c)(i).
(1)
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Notes
Part (a)(i)
T = 9.0/10 = 0.90 s, so T² = 0.81 s²; ΔT = 0.2/10 = 0.02 s
✓ 1
Award this mark for the value 0.81 s² (0.810).
Fractional uncertainty in T² = 2 × 0.02/0.90 = 0.044, so Δ(T²) = 0.044 × 0.81 = ±0.04 s²
✓ 1
Accept ±0.036 s². Using ΔT = 0.2 s (not divided by 10) is a named error: [0] for this mark.
Part (b)(i)
Best-fit gradient = 1.55 s² kg−1
✓ 1
Accept 1.50–1.60. Large triangle expected.
Maximum 1.65 and minimum 1.40 s² kg−1, so uncertainty = (1.65 − 1.40)/2 = ±0.13 s² kg−1
✓ 1
Accept ±0.1 to ±0.15. Accept (max − best) or (best − min). The point plotted for 0.500 kg should match (a): allow ECF from (a).
Part (c)(i)
k = 4π²/gradient = 4π²/1.55 = 25.5 N m−1
✓ 1
Accept 24.6–26.4 N m−1. Allow ECF from (b).
Fractional uncertainty in k = that of the gradient = 0.13/1.55 = 8 %, so Δk = ±2 N m−1
✓ 1
Accept ±1.5 to ±2.5 N m−1, or the range 4π²/max to 4π²/min (23.9 to 28.2). Allow ECF from (b).
Part (c)(ii)
k = mg/x = 0.400 × 9.81/0.157 = 25.0 N m−1; this lies within the range from (c)(i), so the two methods are consistent
✓ 1
Calculation and a conclusion referring to the range are both needed. Allow ECF from (c)(i).
Answers: (a)(i) 0.81 ± 0.04 s² · (b)(i) 1.55 ± 0.13 s² kg−1 · (c)(i) 25.5 ± 2 N m−1 · (c)(ii) 25.0 N m−1; consistent (the remaining parts are explanations — see the table above)
69C-1B-07Mass–spring and pendulum periodsPaper 1BMedium7 marks
A student in a mountain laboratory, 3.0 km above sea level, measures the free-fall acceleration g with a simple pendulum. The bob is a metal sphere. For several lengths L the student times 20 small-amplitude oscillations. The length L is measured with a metre rule from the point of suspension to the top of the bob, which is quicker to do accurately.
The graph shows T² against L with the line of best fit.
L / m
0.300
0.400
0.500
0.600
0.700
0.800
20T / s
22.48
25.72
28.79
31.37
33.90
36.18
T² / s²
1.263
1.654
2.072
2.460
2.873
T² against L with the line of best fit (drawn to scale).
(a)
(i)
Calculate the missing value of T² and determine the gradient of the line of best fit, including its unit.
(2)
(b)
(i)
Determine g.
(1)
(c)
(i)
The line of best fit does not pass through the origin. Use the graph to estimate the radius of the bob, explaining your method.
(2)
(d)
(i)
The student wants to know whether the experiment can detect the effect of the altitude. The uncertainty in g from this experiment is ±1.5 %. Using data from the data booklet, determine the expected value of g at this altitude and comment.
(2)
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Notes
Part (a)(i)
T² = (36.18/20)² = 3.272 s²
✓ 1
Accept 3.27.
Large triangle on the line, e.g. (0, 0.052) to (0.90, 3.675): gradient = 4.03 s² m−1
✓ 1
Accept 3.95–4.10 s² m−1. Unit required. The point from (a)(i) lies on the line.
Part (b)(i)
g = 4π²/gradient = 4π²/4.03 = 9.80 m s−2
✓ 1
Accept 9.6–10.0 m s−2. Allow ECF from (a).
Part (c)(i)
T² = (4π²/g)(L + r), so the line meets the L axis at L = −r
✓ 1
Or: the true length is L + r, so every L is too small by the bob radius.
r = T² intercept/gradient = 0.052/4.03 = 0.013 m (≈ 1.3 cm)
✓ 1
Accept 0.010–0.016 m. Allow ECF from (a). ALT: extend the line to the L axis.
Part (d)(i)
g = GM/(R + h)² = 6.67 × 10−11 × 5.97 × 1024/(6.373 × 106)² = 9.804 m s−2
✓ 1
Accept 9.80 m s−2. Sea-level value 9.81 m s−2.
The change from sea level (≈ 0.01 m s−2, ≈ 0.1 %) is far smaller than the ±1.5 % (≈ ±0.15 m s−2) uncertainty, so the experiment cannot detect the altitude; its result agrees with both values
✓ 1
Allow ECF from (b): the agreement must use their value and the ±1.5 %.
Answers: (a)(i) 3.272 s²; 4.03 s² m−1 · (b)(i) 9.80 m s−2 · (c)(i) ≈ 1.3 cm · (d)(i) 9.80 m s−2; altitude not detectable (the remaining parts are explanations — see the table above)
70C-1B-11SHM with a phase anglePaper 1BMedium7 marks
A glider of mass 0.200 kg on a horizontal air track is held between two springs. A motion sensor records its displacement x from the equilibrium position. The data logger was started at an arbitrary instant during the oscillation. The graph shows the record for the first 3.0 s.
The motion is modelled by x = x0 sin(ωt + φ).
Displacement x of the glider against time t (drawn to scale).
(a)
(i)
Determine the period of the oscillation, explaining how you use the graph to obtain the most precise value.
(2)
(b)
(i)
Show that the phase angle φ is π/6.
(2)
(c)
(i)
Determine the velocity of the glider at t = 0.
(2)
(d)
(i)
Calculate the total energy of the oscillation.
(1)
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Mark
Notes
Part (a)(i)
Use the time spanned by as many whole cycles as possible: maxima at ≈ 0.13 s and ≈ 2.53 s are 3 cycles apart
✓ 1
Timing one cycle only: [0] for this mark.
T = (2.53 − 0.13)/3 = 0.80 s
✓ 1
Accept 0.79–0.81 s.
Part (b)(i)
At t = 0, x = 2.5 cm = x0 sin φ with x0 = 5.0 cm: sin φ = 0.50
✓ 1
φ = π/6 rather than 5π/6 because x is increasing at t = 0 (cos φ > 0)
✓ 1
The rejection of 5π/6 is needed for this mark.
Part (c)(i)
ω = 2π/T = 7.85 rad s−1; v = ωx0 cos φ = 7.85 × 0.050 × cos(π/6)
✓ 1
ALT: v = ±ω√(x0² − x²). Allow ECF from (a) and (b).
v = +0.34 m s−1 (away from equilibrium, in the positive direction)
✓ 1
Accept 0.33–0.35 m s−1. Sign (or direction) required.
Part (d)(i)
ET = ½ × 0.200 × 7.85² × 0.050² = 1.5 × 10−2 J
✓ 1
Accept 1.5–1.6 × 10−2 J. Allow ECF from (a) and (c).
Answers: (a)(i) 0.80 s · (c)(i) +0.34 m s−1 · (d)(i) 1.5 × 10−2 J (the remaining parts are explanations — see the table above)
A smartphone is fixed to a trolley of total mass 0.600 kg that is held between two stretched springs on a long track. The trolley is displaced and released. A motion sensor records the displacement x of the trolley from its equilibrium position, and the phone's accelerometer records the acceleration a along the track at the same instants. The accelerometer also registers any component of the gravitational field along the track.
The graph shows a against x with the line of best fit.
x / m
−0.060
−0.040
−0.020
0.000
0.020
0.040
0.060
a / m s−2
3.35
2.30
1.21
0.11
−0.94
−2.02
−3.05
Acceleration a against displacement x with the line of best fit (drawn to scale).
(a)
(i)
Determine the gradient of the line, including its unit, and hence the period of oscillation.
(2)
(b)
(i)
Determine the combined spring constant of the two springs acting on the trolley.
(1)
(c)
(i)
In a static test, a horizontal force of 2.00 N displaces the trolley by 6.2 cm. Discuss whether this agrees with (b).
(1)
(d)
(i)
The line does not pass through the origin. Suggest a cause consistent with the stem, determine the size of the effect, and state its effect on the period found in (a).
(3)
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Notes
Part (a)(i)
Gradient = −54 s−2 from a large triangle
✓ 1
Accept −52 to −56. Unit s−2 required.
ω = √53.6 = 7.32 rad s−1, T = 2π/ω = 0.86 s
✓ 1
Accept 0.84–0.87 s. The minus sign is not included in the square root.
Part (b)(i)
ma = −kx, so k = mω² = 0.600 × 53.6 = 32.1 N m−1
✓ 1
Accept 31–34 N m−1. Allow ECF from (a).
Part (c)(i)
k = 2.00/0.062 = 32.3 N m−1; within ≈ 1 % of (b), so the two agree (within the uncertainty of the gradient)
✓ 1
Allow ECF from (b): the conclusion must match their numbers.
Part (d)(i)
The track is slightly tilted, so the accelerometer registers g sin α even at equilibrium; intercept = 0.14 m s−2
✓ 1
Accept intercept 0.10–0.16 m s−2.
sin α = 0.14/9.81, α = 0.8°
✓ 1
Accept 0.6°–0.9°.
A constant offset changes only the intercept; the gradient, and hence ω and T, are unaffected
✓ 1
Allow ECF from (a).
Answers: (a)(i) −54 s−2; 0.86 s · (b)(i) 32.1 N m−1 · (c)(i) 32.3 N m−1; agree · (d)(i) ≈ 0.8°; no effect on T(the remaining parts are explanations — see the table above)
72C-1B-13SHM energy and equationsPaper 1BMedium7 marks
A glider of mass 0.350 kg oscillates on a horizontal air track between two springs. A card of width 10.0 mm on the glider passes through a light gate at distance x from the equilibrium position, and the time t for which the gate is blocked gives the speed v at that displacement. The gate is moved to different positions and the glider is released from the same point each time.
The graph shows v² against x² with the line of best fit.
x / cm
1.0
2.0
3.0
4.0
5.0
6.0
t / ms
24.4
24.8
25.8
27.8
30.9
36.4
v² / m² s−2
0.168
0.163
0.150
0.129
0.105
v² against x² with the line of best fit (drawn to scale).
(a)
(i)
Calculate the missing value of v².
(1)
(b)
(i)
Determine the gradient of the line and its intercept on the v² axis.
(2)
(c)
(i)
Hence determine the angular frequency ω and the amplitude x0 of the oscillation.
(2)
(d)
(i)
The card is later found to be 9.5 mm wide, not 10.0 mm. Deduce the effect of this error on the values of ω and of x0 found in (c).
(2)
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Notes
Part (a)(i)
v = 0.0100/0.0364 = 0.275 m s−1, v² = 0.075 m² s−2
✓ 1
Accept 0.075–0.076.
Part (b)(i)
Gradient = (0.064 − 0.173)/(40 × 10−4) = −27 s−2
✓ 1
Accept −26 to −28 s−2. The point from (a) should lie on the line: allow ECF from (a).
Intercept = 0.173 m² s−2
✓ 1
Accept 0.170–0.175.
Part (c)(i)
Gradient = −ω², so ω = √27 = 5.2 rad s−1
✓ 1
Accept 5.1–5.3 rad s−1. Allow ECF from (b).
Intercept = ω²x0², so x0 = √(0.173/27.1) = 0.080 m
✓ 1
Accept 0.079–0.081 m. Allow ECF from (b).
Part (d)(i)
Every v was calculated with a width 10.0/9.5 = 1.05 times too large, so every v² (and both the gradient and the intercept) is 1.11 times too large
✓ 1
ω ∝ √|gradient| is about 5 % too large; x0 = √(intercept/|gradient|) is unaffected because the factor cancels
✓ 1
Both conclusions needed. Allow ECF from (c).
Answers: (a)(i) 0.075 m² s−2 · (b)(i) −27 s−2; 0.173 m² s−2 · (c)(i) 5.2 rad s−1; 0.080 m · (d)(i) ω ≈ 5 % too large; x0 unchanged (the remaining parts are explanations — see the table above)
A student investigates whether the period T of a simple pendulum of length 0.800 m depends on the amplitude θ0, the angle from the vertical at which the bob is released. An electronic timer started and stopped by a light gate measures the time for 20 oscillations to ±0.02 s.
The student claims that the period is independent of the amplitude for amplitudes up to 15°. A second student suggests that T = T0(1 + kθ0²), where T0 is the period for very small amplitudes, k is a constant and θ0 is in radians. The graph shows T against θ0² with error bars and the line of best fit.
θ0 / °
5
10
15
20
25
30
35
40
20T / s
35.90
35.96
36.04
36.17
36.32
36.52
36.73
37.01
T / s
1.7950
1.7980
1.8020
1.8085
1.8160
1.8260
1.8365
1.8505
θ0² / rad²
0.0076
0.0305
0.0685
0.1218
0.2742
0.3732
0.4874
T against θ0² with error bars of ±0.001 s and the line of best fit (drawn to scale; false origin on the T axis).
(a)
(i)
Calculate the missing value of θ0² for θ0 = 25°.
(1)
(b)
(i)
Deduce, using the data for three amplitudes, whether the first student's claim is supported.
(2)
(c)
(i)
Use the graph to determine T0 and hence a value for g.
Uncertainty in T = 0.02/20 = ±0.001 s; between 5°, 10° and 15° T increases by 0.0030 s and then 0.0040 s
✓ 1
Both the uncertainty in T and the use of three amplitudes are needed.
These changes are larger than the combined uncertainty (±0.002 s), so T increases with amplitude even below 15°: the claim is not supported at this precision
✓ 1
Accept the conclusion that the change is small (< 0.5 %) but still detectable. Conclusion must follow from a comparison with the uncertainty.
Part (c)(i)
T0 = intercept on the T axis = 1.794 s
✓ 1
Accept 1.793–1.796 s.
g = 4π²l/T0² = 4π² × 0.800/1.794² = 9.81 m s−2
✓ 1
Accept 9.79–9.82 m s−2. Allow ECF from their intercept.
Part (d)(i)
Gradient = kT0 = 0.115 s rad−2
✓ 1
Accept 0.110–0.120 s rad−2.
k = gradient/T0 = 0.115/1.794 = 0.064 rad−2
✓ 1
Accept 0.061–0.067. Accept with no unit. Allow ECF from (c).
Answers: (a)(i) 0.1904 rad² · (b)(i) not supported · (c)(i) 1.794 s; 9.81 m s−2 · (d)(i) 0.064 rad−2(the remaining parts are explanations — see the table above)
A small magnet hangs from a spring and oscillates vertically just above a thick copper plate. A sensor records the displacement x of the magnet from its equilibrium position, and the graph shows the trace. The amplitude of each successive positive peak is read from the data file and listed in the table.
The student suggests that the amplitude decreases by the same fraction in every cycle.
Peak number
0
1
2
3
4
5
6
Amplitude / mm
40.0
34.2
28.7
24.7
20.8
17.8
15.1
Displacement x of the magnet against time t (drawn to scale).
(a)
(i)
Use the trace to determine the period of oscillation.
(1)
(b)
(i)
Deduce whether the student's suggestion is supported by the data.
(2)
(c)
(i)
Determine the fraction of the oscillator's energy that is dissipated in one cycle.
(2)
(d)
(i)
Predict the time at which the amplitude of a positive peak is first less than 5.0 mm.
(1)
(e)
(i)
The copper plate is replaced by an identical plate with many narrow slots cut in it. State and explain the effect on the ratio found in (b).
(1)
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Notes
Part (a)(i)
6 complete cycles between the peaks at t = 0 and t = 2.40 s, so T = 2.40/6 = 0.400 s
✓ 1
Accept 0.39–0.41 s. Timing a single cycle: [0].
Part (b)(i)
Ratios of successive amplitudes for at least three pairs, e.g. 0.855, 0.861, 0.848
✓ 1
ALT: equal-interval check, e.g. A0/A2, A2/A4, A4/A6.
The ratios are constant (≈ 0.85) within about ±1.5 %, so the suggestion is supported
✓ 1
Conclusion must follow from their values. Two pairs only: [1 max].
Part (c)(i)
Energy ∝ (amplitude)², so the energy ratio per cycle = 0.85² = 0.72
✓ 1
Allow ECF from (b).
Fraction dissipated = 1 − 0.72 = 0.28 (≈ 28 %)
✓ 1
Answer 0.15 (amplitude fraction): [1 max].
Part (d)(i)
0.8513 × 40 = 4.8 mm < 5.0 mm (0.8512 × 40 = 5.7 mm), so after 13 cycles: t = 13 × 0.400 = 5.2 s
✓ 1
Accept 12–14 cycles (4.8–5.6 s) consistent with their ratio. Allow ECF from (a) and (b).
Part (e)(i)
The slots reduce the induced (eddy) currents in the plate, so less energy is dissipated per cycle: the ratio becomes closer to 1
✓ 1
Reference to smaller induced currents (Lenz/Faraday) is needed. "Less damping" alone: [0].
Answers: (a)(i) 0.400 s · (b)(i) supported; ratio ≈ 0.85 · (c)(i) 0.28 · (d)(i) 5.2 s · (e)(i) ratio increases towards 1 (the remaining parts are explanations — see the table above)
75IBP2H-019-01-09SHM energy and equationsPaper 2Easy4 marks
The brush head of an electric toothbrush oscillates from side to side with simple harmonic motion. The frequency is 260 Hz and the amplitude is 1.5 mm.
(a)
State the condition on the acceleration of the brush head for its motion to be simple harmonic.
(1)
(b)
Calculate the maximum speed of the brush head.
(2)
(c)
Calculate the maximum acceleration of the brush head.
(1)
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Mark
Notes
Part (a)
The acceleration is proportional to the displacement from equilibrium and is directed towards the equilibrium position, i.e. opposite to the displacement
✓ 1
Part (b)
ω = 2π × 260 = 1.63×103 rad s−1
✓ 1
vmax = ωx0 = 1634 × 1.5×10−3 = 2.5 m s−1 (2.45 m s−1; accept 2.4 m s−1 from rounded ω)
✓ 1
Part (c)
amax = ω2x0 = (1.63×103)2 × 1.5×10−3 = 4.0×103 m s−2. Allow ECF for ω from the maximum-speed part
✓ 1
Common error: Convert the amplitude from millimetres to metres before substituting. Leaving it in mm makes the answer 1000 times too large.
76IBP2H-019-00-09Mass–spring and pendulum periodsPaper 2Easy5 marks
A demolition crane swings a heavy steel ball on a cable of length 12 m. The ball is pulled slightly to one side and released, so that it swings as a simple pendulum.
(a)
Calculate the period of the swing.
(2)
(b)
State one assumption needed for your answer to (a) to be valid.
(1)
(c)
A ball of twice the mass is attached to the same cable. Explain why the period is unchanged.
(2)
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Notes
Part (a)
Uses T = 2π√(L/g) . T = 2π√(12/9.8) = 6.95 s (≈ 7.0 s) .
✓ 2
Part (b)
Any one of: the angle of swing is small (less than about 10°); the cable is light and inextensible; the ball acts as a point mass at the end of the cable; air resistance is negligible .
✓ 1
Part (c)
The restoring force (mg sin θ) and the inertia (mass) are both proportional to m . So the acceleration, and therefore the period, does not depend on m .
✓ 2
Exam tip: For 'explain' about mass independence, mention both the force and the inertia scaling with m. Saying only 'mass is not in the formula' is weak.
77IBP3H-C1-20r7-09Mass–spring and pendulum periodsPaper 2Easy6 marks
A beehive scale has a platform supported by springs. The platform and hive together have a mass of 45 kg and the springs together act as a single spring of spring constant 1.8 × 104 N m−1. The platform oscillates vertically when disturbed.
(a)
State the two conditions that must be met for the platform to move with simple harmonic motion.
(2)
(b)
Calculate the period of oscillation of the loaded platform.
(2)
(c)
Bees increase the mass of the hive so that the total mass becomes 60 kg. Determine the new period, in s, to 3 significant figures.
(2)
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Notes
Part (a)
The acceleration (or resultant force) is proportional to the displacement from the equilibrium position
✓ 1
The acceleration (or resultant force) is directed towards the equilibrium position (opposite to the displacement)
✓ 1
Part (b)
Use T = 2π√(m/k)
✓ 1
T = 2π√(45/18000) = 0.314 s
✓ 1
Part (c)
T = 2π√(60/18000)
✓ 1
T = 0.363 s
✓ 1
Exam tip: Check that mass is in kilograms and that k is in N m⁻¹ before substituting into the period formula.
79IBP3H-C1-20r10-09Mass–spring and pendulum periodsPaper 2Easy5 marks
A display in a library holds a 0.250 kg model hanging from a light vertical spring of force constant 18.0 N m−1. The model is pulled down 4.0 cm below its equilibrium position and released. The motion is simple harmonic.
Diagram not to scale
(a)
Calculate the period of oscillation. Give your answer to two significant figures.
(2)
(b)
Calculate the magnitude of the maximum acceleration of the model. Give your answer to three significant figures.
(2)
(c)
State the direction of the acceleration at the instant of release.
(1)
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Part (a)
uses T = 2π√(m/k)
✓ 1
T = 2π√(0.250/18.0) = 0.74 s
✓ 1
Part (b)
amax = ω2x0 with ω2 = k/m = 72.0 s−2
✓ 1
amax = 72.0 × 0.040 = 2.88 m s−2
✓ 1
Part (c)
upwards, towards the equilibrium position
✓ 1
Exam tip: Use x₀ in metres: 4.0 cm is 0.040 m. Acceleration always points towards equilibrium.
81IBP3H-C1-20r20-09SHM energy and equationsPaper 2Easy4 marks
A small sensor of mass 0.150 kg is attached to a spring of spring constant 60.0 N m−1 inside a beehive. It oscillates horizontally with simple harmonic motion of amplitude 8.0 mm.
(a)
Calculate the period of the oscillation.
(2)
(b)
Calculate the maximum speed of the sensor.
(2)
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Part (a)
M1 T = 2π√(m/k) = 2π√(0.150/60.0)
✓ 1
A1 T = 0.314 s
✓ 1
Part (b)
M1 ω = 2π/T = 20.0 rad s−1 and vmax = ωx0 (accept ½kx02 = ½mv2)
✓ 1
A1 vmax = 0.160 m s−1
✓ 1
Exam tip: Give the period to three significant figures and use the exact value of ω from the unrounded period.
82IBP3H-C1-20r20-10Energy in SHMPaper 2Easy6 marks
A test mass of 0.40 kg on a spring of spring constant 200 N m−1 is used in a Formula 1 telemetry rig. It oscillates horizontally with simple harmonic motion of amplitude 0.050 m. Ignore friction.
(a)
Calculate the total energy of the oscillation.
(2)
(b)
Calculate the kinetic energy of the mass when its displacement is 0.030 m.
(2)
(c)
Calculate the speed of the mass at this displacement.
(2)
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Part (a)
M1 E = ½kx02 = ½ × 200 × 0.0502
✓ 1
A1 E = 0.25 J
✓ 1
Part (b)
M1 KE = E − ½kx2 = 0.25 − ½ × 200 × 0.0302
✓ 1
A1 KE = 0.16 J
✓ 1
Part (c)
M1 ½mv2 = 0.16 (ECF from (b))
✓ 1
A1 v = 0.89 m s−1
✓ 1
Exam tip: In SHM, total energy is constant, so kinetic energy at any displacement is the total energy minus the potential energy.
83IBP3H-C1-20r39-09Acceleration in SHMPaper 2Easy6 marks
In a school library demonstration, a 0.40 kg trolley rests on a frictionless horizontal track. It is attached to a fixed support by a light spring of spring constant 25 N m−1. The trolley is pulled 0.060 m from equilibrium and released from rest.
Diagram not to scale
(a)
State the condition for the motion of the trolley to be simple harmonic.
(2)
(b)
Calculate the period of the oscillation.
(2)
(c)
Calculate the maximum speed of the trolley.
(2)
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Part (a)
the acceleration is proportional to the displacement from the equilibrium position
✓ 1
the acceleration is directed towards the equilibrium position (opposite to the displacement)
✓ 1
Part (b)
T = 2π√(m/k) = 2π√(0.40/25)
✓ 1
T = 0.795 s
✓ 1
Part (c)
vmax = ωx0 with ω = √(k/m) = 7.91 rad s−1 (or 2π/T using the unrounded period)
✓ 1
vmax = 0.474 m s−1
✓ 1
Exam tip: Use the unrounded value of the period or angular frequency when it feeds a later part; this avoids small rounding differences in the final answer.
84IBP3H-C1-20r39-10Mass–spring and pendulum periodsPaper 2Easy6 marks
A timer beside a swimming pool uses a simple pendulum: a small dense bob on a light string of length 0.64 m. The bob swings through small angles. Use g = 9.8 m s−2.
(a)
Calculate the period of the pendulum.
(2)
(b)
The pendulum is shortened so that its period is 1.20 s. Calculate the new length.
(2)
(c)
The bob is replaced by a heavier bob of the same size. State and explain the effect on the period.
(2)
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Part (a)
T = 2π√(l/g) = 2π√(0.64/9.8)
✓ 1
T = 1.61 s
✓ 1
Part (b)
l = gT2/(4π2) = 9.8 × 1.202 / (4π2)
✓ 1
l = 0.357 m
✓ 1
Part (c)
the period is unchanged
✓ 1
the period depends only on l and g; the restoring force and the inertia both increase in proportion to the mass
✓ 1
Common error: The period of a simple pendulum does not depend on the mass of the bob, in contrast to a mass–spring system.
85C-2-05Mass–spring and pendulum periodsPaper 2Medium9 marks
A lander on Mars carries a simple pendulum of length 0.500 m that is used to test its instruments. Mars has mass 6.42 × 1023 kg and radius 3.39 × 106 m.
(a)
(i)
Show that the gravitational field strength at the surface of Mars is about 3.7 N kg−1.
(2)
(ii)
Calculate the period of the pendulum on Mars.
(1)
(b)
(i)
Show that, for planets of the same mean density, the period of a given pendulum is proportional to R−½, where R is the radius of the planet.
(2)
(c)
(i)
The bob is released from rest at a horizontal displacement of 4.0 cm. Determine the speed of the bob when its displacement is 2.0 cm.
(3)
(ii)
Explain why a mass–spring oscillator would have the same period on Mars as on the Earth.
86C-2-11Conditions for simple harmonic motionPaper 2Medium12 marks
A wave-energy buoy is a vertical cylinder of uniform cross-sectional area A floating in seawater of density ρ. At equilibrium a length d of the cylinder is under water. When the buoy is pushed down a further small distance x and released, it bobs up and down.
Diagram not to scale
(a)
(i)
Show that the motion of the buoy is simple harmonic with ω² = g/d.
(2)
(ii)
The buoy floats with d = 2.20 m. Calculate the period of its oscillation.
(1)
(b)
Ocean swell of wavelength 13.9 m travels past the buoy at 4.67 m s−1.
(i)
Show that the swell drives the buoy close to resonance.
(1)
(ii)
A generator inside the buoy extracts energy from the motion and so acts as damping. Explain why the buoy is designed to operate near resonance, and why the damping must not be too heavy.
(2)
(c)
(i)
A heavier generator is fitted, so that d becomes 2.60 m. Determine the wavelength of swell, travelling at the same speed, that now drives the buoy at resonance.
(2)
(ii)
With d = 2.20 m, the cross-sectional area of the buoy is 7.0 m2 and the density of seawater is 1030 kg m−3. The buoy oscillates with amplitude 0.60 m, and the generator removes 15 % of the energy of the oscillation in each cycle. Determine the mean power of the generator.
(3)
(iii)
State one reason why the motion of a real buoy is simple harmonic only for small displacements.
(1)
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Part (a)(i)
The extra buoyancy ρAgx acts upwards when x is downwards, so the resultant force is −ρAgx: proportional to the displacement and directed towards equilibrium
✓ 1
At equilibrium mg = ρAdg, so a = −(ρAg/m)x = −(g/d)x, which is a = −ω²x with ω² = g/d
✓ 1
The hidden step is eliminating m using flotation.
Part (a)(ii)
T = 2π√(d/g) = 2π√(2.20/9.81) = 2.98 s
✓ 1
Allow ECF from (a)(i).
Part (b)(i)
Period of the swell = λ/v = 13.9/4.67 = 2.98 s, which is very close to the natural period 2.98 s
✓ 1
Allow ECF from (a)(ii).
Part (b)(ii)
Near resonance the amplitude of the buoy, and so the energy it can transfer from the waves, is greatest
✓ 1
Damping lowers (and broadens) the resonance peak: very heavy damping would reduce the amplitude so much that little energy is extracted
✓ 1
Also accept: with very light damping little energy is extracted per cycle.
Part (c)(i)
T = 2π√(2.60/9.81) = 3.23 s
✓ 1
Allow ECF from (a)(i).
λ = vT = 4.67 × 3.23 = 15.1 m
✓ 1
Part (c)(ii)
Flotation: m = ρAd = 1030 × 7.0 × 2.20 = 1.59 × 104 kg
✓ 1
The hidden step: the mass is not given.
ET = ½mω²x0² = ½ × 1.59 × 104 × (2.11)² × 0.60² = 1.3 × 104 J
✓ 1
Allow ECF from (a).
P = 0.15 × 1.3 × 104/2.98 = 6.4 × 102 W
✓ 1
Allow ECF from (a)(ii).
Part (c)(iii)
The restoring force is proportional to x only while the cross-section at the waterline stays the same: the top must not go under water and the bottom must not leave the water
✓ 1
Also accept: drag of the water becomes significant at large speeds.
Answers: (a)(ii) 2.98 s · (b)(i) 2.98 s · (c)(i) 15.1 m · (c)(ii) 6.4 × 102 W (the remaining parts are explanations — see the table above)
87C-2-20Phase difference between oscillatorsPaper 2Medium10 marks
In a three-cylinder engine the pistons move approximately with simple harmonic motion. The graph shows how the displacements x of pistons A and B from the centres of their cylinders vary with time t while the engine runs at a constant rate.
Displacement of pistons A (solid line) and B (dashed line) against time (drawn to scale).
(a)
(i)
Show that the angular frequency of the pistons is about 310 rad s−1.
(1)
(ii)
Determine the phase difference between the motions of A and B, in radians, and state which piston leads.
(2)
(b)
Each piston is moved by a crank pin that travels in a circle of radius equal to the amplitude of the piston, at constant angular speed. The piston's displacement equals the component of the pin's position along the cylinder.
(i)
Show that the maximum acceleration of a piston equals the centripetal acceleration of its crank pin.
(2)
(ii)
A piston has mass 0.35 kg. Determine the maximum resultant force on it.
(2)
(iii)
Determine the velocity of piston B at an instant when piston A is at its maximum positive displacement.
(3)
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Part (a)(i)
T = 20 ms from the graph, so ω = 2π/0.020 = 314 rad s−1
✓ 1
Period must be read from the graph.
Part (a)(ii)
B reaches each maximum 6.7 ms after A, so the phase difference = 2π × 6.7/20 = 2.1 rad (2π/3)
✓ 1
Accept 2.0–2.2 rad. Allow ECF from (a)(i).
A leads B
✓ 1
Part (b)(i)
The pin has speed v = ωr and centripetal acceleration v²/r = ω²r
✓ 1
The piston has amax = ω²x0 with x0 = r, so they are equal
✓ 1
Part (b)(ii)
amax = ω²x0 = 314² × 0.040 = 3.9 × 103 m s−2
✓ 1
Allow ECF from (a)(i).
F = ma = 0.35 × 3.9 × 103 = 1.4 × 103 N
✓ 1
Part (b)(iii)
A is at its positive maximum when its phase is π/2
✓ 1
B lags by 2π/3, so its phase is π/2 − 2π/3 = −π/6
✓ 1
Allow ECF from (a)(ii).
vB = ωx0 cos(−π/6) = 314 × 0.040 × 0.866 = +10.9 m s−1 (in the positive direction)
✓ 1
Sign or direction required. Allow ECF from (a)(i).
Answers: (a)(i) 314 rad s−1 · (a)(ii) 2.1 rad; A leads · (b)(ii) 1.4 × 103 N · (b)(iii) +10.9 m s−1(the remaining parts are explanations — see the table above)
88IBP2S-018-00-14SHM energy and equationsPaper 2Medium7 marks
A diver of mass 65 kg stands at the end of a springboard. She plans to set the board oscillating vertically in simple harmonic motion with a frequency of 1.8 Hz and an amplitude of 0.12 m. Treat the diver as the only oscillating mass and take g = 9.81 m s−2.
(a)
Calculate the maximum speed of the diver in the planned motion.
(2)
(b)
By calculating the maximum acceleration, determine whether the diver can stay in contact with the board throughout the planned motion.
(3)
(c)
Assuming the planned motion could take place, calculate its total energy.
(2)
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Part (a)
ω = 2π × 1.8 = 11.3 rad s−1 . vmax = ωx0 = 11.3 × 0.12 = 1.36 m s−1 (accept 1.4 m s−1) .
✓ 2
Part (b)
amax = ω2x0 = 11.32 × 0.12 = 15 m s−2 . At the top of the motion the diver would need a downward acceleration of 15 m s−2, which is greater than g . The board can only push up on the diver and cannot pull her down, so she cannot stay in contact and would leave the board . Allow error carried forward from the value of ω in the first part.
✓ 3
Part (c)
Uses \(E_T = \frac{1}{2}m\omega^2x_0^2\) or \(E_T = \frac{1}{2}mv_{max}^2\) . ET = 0.5 × 65 × 11.32 × 0.122 = 60 J (accept answers from 60 J to 64 J, depending on rounding of vmax) . Allow error carried forward from the first part.
✓ 2
Exam tip: If ω²x₀ > g, an object resting on a vertically oscillating support loses contact at the top of the motion. State both values and compare them explicitly. Carry unrounded values forward to avoid rounding drift.
89IBP2H-019-01-10Mass–spring and pendulum periodsPaper 2Medium7 marks
A student investigates a light spring for use in a vehicle crash-test sensor. Different masses m are hung from the spring, and the period T of the vertical oscillations is measured for each mass. The results are shown below.
m / kg: 0.10, 0.20, 0.30, 0.40, 0.50 T / s: 0.50, 0.70, 0.86, 0.99, 1.11
The mass of the spring may be assumed negligible.
(a)
Explain why a graph of T2 against m is expected to be a straight line through the origin.
(2)
(b)
Determine the spring constant k of the spring.
(3)
(c)
The experiment is repeated with a spring whose own mass is not negligible. Outline how the graph of T2 against m would differ, and whether k could still be found.
(2)
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Part (a)
T = 2π√(m/k), so T2 = (4π2/k)m
✓ 1
4π2/k is constant, so T2 ∝ m, giving a straight line through the origin with gradient 4π2/k
k = 4π2/gradient = 39.5/2.45 ≈ 16 N m−1; ECF for gradient
✓ 1
Part (c)
Part of the spring's mass also oscillates, so T2 is larger for every load. The line is shifted up and has a positive intercept on the T2 axis
✓ 1
The gradient is still 4π2/k, so k can still be found from the gradient
✓ 1
Exam tip: When finding a gradient from tabulated data, use two points that are widely separated. Show both the T² values and the gradient calculation, because each step can earn credit.
90IBP2H-019-01-12SHM with a phase anglePaper 2Medium6 marks
A skyscraper has a tuned mass damper of mass 3.0×105 kg. The damper can be modelled as a mass on a spring that oscillates horizontally with SHM. During a storm its period is 6.8 s and its amplitude is 0.80 m.
(a)
Show that the effective spring constant of the damper system is about 2.6×105 N m−1.
(2)
(b)
Calculate the total energy of the oscillating damper.
(2)
(c)
Calculate the speed of the damper when its displacement from equilibrium is 0.50 m.
(2)
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Part (a)
ω = 2π/T = 2π/6.8 = 0.924 rad s−1
✓ 1
k = mω2 = 3.0×105 × 0.9242 = 2.56×105 N m−1
✓ 1
Equivalent: k = 4π2m/T2.
—
Part (b)
E = ½kx02 (or ½mω2x02) = ½ × 2.56×105 × 0.802
✓ 1
E = 8.2×104 J. Accept 8.3×104 J if the quoted value 2.6×105 N m−1 is used. Allow error carried forward from the spring constant.
✓ 1
Part (c)
v = ω√(x02 − x2) = 0.924 × √(0.802 − 0.502). Alternatively use energy conservation: ½mv2 = ½k(x02 − x2).
✓ 1
v = 0.58 m s−1. Accept 0.57–0.59 m s−1. Allow error carried forward from earlier parts.
✓ 1
Exam tip: In a 'show that' question, give your answer to at least one more significant figure than the value quoted. This shows you calculated it rather than copied it.
91IBP2H-019-01-16Phase and timing in SHMPaper 2Medium8 marks
The cantilever tip of an atomic force microscope oscillates with SHM at a frequency of 75 kHz and an amplitude of 10.0 nm. Its motion is described by x = x0 sin(ωt + φ). At t = 0 the tip is at x = +6.0 nm and is moving towards the equilibrium position. The effective mass of the oscillating tip is 1.2×10−11 kg.
(a)
Calculate the speed of the tip at t = 0.
(2)
(b)
Determine the phase angle φ.
(2)
(c)
Determine the time after t = 0 at which the tip first passes through the equilibrium position.
(2)
(d)
Calculate the total energy of the oscillating tip.
(2)
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Part (a)
ω = 2π × 75×103 = 4.71×105 rad s−1
✓ 1
v = ω√(x02 − x2) = 4.71×105 × √((10.0×10−9)2 − (6.0×10−9)2) = 4.71×105 × 8.0×10−9 = 3.8×10−3 m s−1
✓ 1
Part (b)
sin φ = 6.0/10.0 = 0.60
✓ 1
The tip moves towards equilibrium, so the velocity is negative and cos φ < 0. Therefore φ = π − 0.644 = 2.50 rad
✓ 1
Part (c)
The tip first reaches equilibrium when ωt + φ = π
✓ 1
t = (π − 2.50)/4.71×105 = 0.644/4.71×105 = 1.4×10−6 s (allow error carried forward from the value of φ)
✓ 1
Part (d)
E = ½mω2x02 = ½ × 1.2×10−11 × (4.71×105)2 × (1.0×10−8)2
✓ 1
E = 1.3×10−16 J (allow error carried forward from the value of ω)
✓ 1
Common error: A positive displacement does not mean a positive velocity. Read the direction of motion carefully, because it decides which phase angle is correct.
92IBP2H-019-00-10Mass–spring and pendulum periodsPaper 2Medium6 marks
A car of mass 1200 kg has a suspension that may be modelled as a single vertical spring. Four passengers of total mass 300 kg get in, and the car body sinks by 2.5 cm.
(a)
Show that the spring constant of the suspension is about 1.2 × 105 N m−1.
(2)
(b)
Calculate the period of vertical oscillation of the loaded car, assuming no damping.
(2)
(c)
The loaded car drives over a bump and oscillates with amplitude 4.0 cm. Calculate the maximum vertical acceleration of the car body.
(2)
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Part (a)
The extra weight is 300 × 9.8 = 2940 N . k = 2940 / 0.025 = 1.18 × 105 N m−1 .
✓ 2
Part (b)
Uses T = 2π√(m/k) with m = 1500 kg . T = 2π√(1500 / 1.18 × 105) = 0.71 s; accept 0.70 s if k = 1.2 × 105 N m−1 is used . Allow ECF from (a).
✓ 2
Part (c)
Uses amax = ω²x0 with ω² = k/m = 1.18 × 105 / 1500 = 78 s−2, or ω = 2π/T . amax = 78 × 0.040 = 3.1 m s−2; accept 3.2 m s−2 if k = 1.2 × 105 N m−1 is used . Allow ECF from (a) or (b).
✓ 2
Common error: Use the total oscillating mass (car plus passengers) in the period formula. The passengers' mass alone only gives you k.
93IBP2H-019-00-11Mass–spring and pendulum periodsPaper 2Medium7 marks
An engineering student characterises a spring for a vibration damper. Different masses m are hung on the spring and the period T of vertical oscillation is measured each time. m / kg: 0.10, 0.20, 0.30, 0.40, 0.50 T / s: 0.40, 0.56, 0.69, 0.80, 0.89
(a)
Outline why a graph of T2 against m is expected to be a straight line.
(2)
(b)
Suggest how the graph of T2 against m would differ if the mass of the spring were not negligible.
(2)
(c)
Determine the spring constant k using the first and last data points.
(3)
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Part (a)
T = 2π√(m/k), so T² = (4π²/k)m . k is constant, so T² ∝ m: the graph is a straight line through the origin with gradient 4π²/k .
✓ 2
Part (b)
Part of the spring would also oscillate, so the effective oscillating mass would be greater than m . The line would still be straight with the same gradient, but it would have a positive intercept on the T² axis and would not pass through the origin .
✓ 2
Part (c)
T² = 0.160 s² and 0.792 s² . Gradient = (0.792 − 0.160)/(0.50 − 0.10) = 1.58 s² kg−1 . k = 4π²/gradient = 25 N m−1 .
✓ 3
Exam tip: Find k from the gradient rather than from a single (m, T) pair. A gradient is unaffected by a constant offset such as the spring's own mass.
94IBP2H-019-00-12Energy in SHMPaper 2Medium8 marks
The tip of an atomic force microscope cantilever oscillates with SHM at a frequency of 300 kHz and an amplitude of 20 nm. The effective mass of the tip is 1.0 × 10−11 kg.
(a)
Calculate the maximum speed of the tip.
(2)
(b)
Calculate the speed of the tip when its displacement is 10 nm.
(2)
(c)
Calculate the total energy of the oscillation.
(2)
(d)
Determine the displacement at which the kinetic energy equals the potential energy.
95IBP2H-019-00-16SHM energy and equationsPaper 2Medium7 marks
The brush head of an electric toothbrush oscillates with SHM at a frequency of 260 Hz and an amplitude of 1.5 mm.
(a)
Calculate the maximum acceleration of the brush head.
(2)
(b)
Calculate the maximum speed of the brush head.
(2)
(c)
Determine the shortest time for the brush head to move from its equilibrium position to a displacement of 1.0 mm.
(3)
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Part (a)
ω = 2π × 260 = 1.63 × 103 rad s−1 and amax = ω2x0 . amax = (1634)2 × 1.5 × 10−3 = 4.0 × 103 m s−2 .
✓ 2
Part (b)
vmax = ωx0 . vmax = 1634 × 1.5 × 10−3 = 2.45 m s−1; accept answers in the range 2.4 to 2.5 m s−1 .
✓ 2
Part (c)
Uses x = x0 sin ωt with x = 0 at t = 0, or any equivalent valid method . sin ωt = 1.0/1.5, so ωt = 0.730 rad . t = 0.730/(1.63 × 103) = 4.5 × 10−4 s . Allow ECF for the value of ω used in (a).
✓ 3
Exam tip: Put your calculator in radian mode when solving for ωt. An angle in degrees divided by ω gives a meaningless time. Keep unrounded values of ω in your calculator to avoid rounding errors in later parts.
96IBP2S-018-02-14Mass–spring and pendulum periodsPaper 2Medium8 marks
A tall building contains a tuned mass damper: a mass of 4.0 × 105 kg hanging from cables, which behaves as a simple pendulum. During a storm the damper swings with a period of 6.5 s and a horizontal amplitude of 0.80 m.
(a)
Calculate the effective length of the pendulum.
(2)
(b)
Calculate the maximum speed of the damper mass.
(2)
(c)
Determine the total energy of the oscillation.
(2)
(d)
Calculate the speed of the damper mass when its horizontal displacement is 0.40 m.
(2)
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Part (a)
L = gT²/(4π²) = 9.8 × 6.5²/(4π²)
✓ 1
L = 10.5 m (accept 10 m)
✓ 1
Part (b)
ω = 2π/6.5 = 0.967 rad s−1
✓ 1
vmax = ωx0 = 0.967 × 0.80 = 0.77 m s−1
✓ 1
Part (c)
E = ½mvmax² = ½ × 4.0 × 105 × 0.773² (or E = ½mω²x0²)
✓ 1
E = 1.2 × 105 J (allow ECF from the maximum speed calculated earlier)
✓ 1
Part (d)
v = ω√(x0² − x²) = 0.967 × √(0.80² − 0.40²)
✓ 1
v = 0.67 m s−1 (allow ECF for ω from the maximum speed calculation)
✓ 1
Exam tip: Calculate ω once, from ω = 2π/T, and reuse it. It links the period to the maximum speed, the energy and the speed at any displacement.
97IBP3H-C1-20r7-12Energy in SHMPaper 2Medium8 marks
A vibration monitor in a hospital consists of a block of mass 0.25 kg on a horizontal spring. When it oscillates, the period is 0.50 s and the amplitude is 0.080 m. Friction is negligible.
(a)
Calculate the angular frequency of the oscillation.
(2)
(b)
Describe, in words, a sketch graph of the kinetic energy of the block against time for two complete oscillations, starting when the block is at its maximum displacement.
(3)
(c)
Determine the total energy of the oscillation, in J, to 3 significant figures.
(3)
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Part (a)
ω = 2π/T
✓ 1
ω = 12.6 rad s−1
✓ 1
Part (b)
Starts at zero kinetic energy
✓ 1
Rises to a maximum then returns to zero, repeating with a period of 0.25 s (twice the frequency of the displacement)
✓ 1
Maximum value equals the total energy and the graph never goes below zero
✓ 1
Part (c)
k = mω² = 0.25 × 12.57² = 39.5 N m−1
✓ 1
E = ½kA² = ½ × 39.5 × 0.080²
✓ 1
E = 0.126 J
✓ 1
Common error: The kinetic energy oscillates at twice the frequency of the displacement because it peaks at both the equilibrium position in both directions of travel.
98IBP3H-C1-20r7-13Conditions for SHMPaper 2Medium7 marks
A mass of 0.35 kg hangs at rest from a vertical spring and stretches it by 0.070 m. The mass is pulled down a small distance and released. Use g = 9.8 m s−2.
Diagram not to scale
(a)
Show that the mass moves with simple harmonic motion. Take displacement x as positive downwards from the equilibrium position.
(3)
(b)
Calculate the spring constant.
(2)
(c)
Determine the period of the oscillation, in s, to 3 significant figures.
(2)
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Part (a)
At equilibrium ke = mg
✓ 1
At displacement x the extension is e + x so the resultant force is mg − k(e + x)
✓ 1
Resultant force F = −kx, which is proportional to and opposite to displacement, so SHM
✓ 1
Part (b)
k = mg/e
✓ 1
k = 0.35 × 9.8/0.070 = 49 N m−1
✓ 1
Part (c)
T = 2π√(m/k)
✓ 1
T = 2π√(0.35/49) = 0.531 s
✓ 1
Exam tip: In a show-that question, state clearly the force at equilibrium before adding the displacement.
100IBP3H-C1-20r10-11Energy in SHMPaper 2Medium7 marks
A tuned-damper block of mass 3.5 kg in a wind-turbine nacelle is attached to a horizontal spring of force constant 2.80 × 103 N m−1. The block slides on a frictionless track and oscillates with amplitude 0.060 m.
(a)
Calculate the total energy of the oscillation.
(2)
(b)
Calculate the kinetic energy of the block when its displacement is 0.036 m.
(3)
(c)
Calculate the maximum speed of the block. Give your answer to three significant figures.
(2)
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Part (a)
E = ½kx02
✓ 1
E = ½ × 2.80 × 103 × 0.0602 = 5.04 J
✓ 1
Part (b)
elastic potential energy = ½kx2 = 1.81 J
✓ 1
kinetic energy = total energy − potential energy (ECF from (a))
✓ 1
KE = 3.23 J
✓ 1
Part (c)
maximum KE equals the total energy, so ½mvmax2 = 5.04 J (ECF from (a))
✓ 1
vmax = √(2 × 5.04 / 3.5) = 1.70 m s−1
✓ 1
Exam tip: At the equilibrium position all of the energy is kinetic, which gives the maximum speed without needing ω.
101IBP3H-C1-20r10-12Phase and timing in SHMPaper 2Medium9 marks
A pendulum clock in a library has a bob on a light rod of effective length 0.850 m. The bob swings through a small angle with amplitude 6.0°. Treat the bob's horizontal displacement x from the lowest point as the arc length. At t = 0 the bob passes through its lowest point moving to the right (positive x), so that x = x0 sin ωt. Use g = 9.8 m s−2 and unrounded values in later parts.
Diagram not to scale
(a)
Calculate the period of the pendulum. Give your answer to three significant figures.
(2)
(b)
Determine the angular frequency ω and the amplitude x0 of the displacement.
(3)
(c)
Calculate the displacement of the bob at t = 0.50 s. Give your answer to two significant figures.
(2)
(d)
State whether the bob is moving towards or away from the lowest point at t = 0.50 s, and justify your answer.
(2)
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Part (a)
uses T = 2π√(L/g)
✓ 1
T = 1.85 s
✓ 1
Part (b)
ω = 2π/T = √(g/L) = 3.40 rad s−1
✓ 1
x0 = Lθ with θ = 6.0° converted to 0.105 rad
✓ 1
x0 = 0.0890 m
✓ 1
Part (c)
x = 0.0890 sin(3.40 × 0.50) with the calculator in radian mode
✓ 1
x = 0.088 m
✓ 1
Part (d)
towards the lowest point
✓ 1
ωt = 1.70 rad lies between π/2 and π, so x is positive but decreasing (velocity ∝ cos ωt is negative)
✓ 1
Common error: Evaluating sin(ωt) with the calculator in degree mode gives a completely different displacement; ωt is in radians.
102IBP3H-C1-20r10-13Phase and timing in SHMPaper 2Medium8 marks
A vibration sensor on a wind-turbine blade moves with simple harmonic motion described by x = 0.15 cos(2.8t), where x is in metres and t is in seconds.
(a)
State the amplitude and calculate the period of the motion. Give the period to two significant figures.
(2)
(b)
Determine the first time at which the displacement of the sensor is 0.075 m. Give your answer to two significant figures.
(2)
(c)
Calculate the speed of the sensor at this time. Give your answer to two significant figures.
(2)
(d)
Calculate the magnitude of the acceleration at this time and state its direction. Give the magnitude to two significant figures.
(2)
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Part (a)
amplitude = 0.15 m
✓ 1
T = 2π/2.8 = 2.2 s
✓ 1
Part (b)
cos(2.8t) = 0.5, so 2.8t = π/3
✓ 1
t = 0.37 s
✓ 1
Part (c)
uses v = ω√(x02 − x2) (or v = ωx0 sin ωt)
✓ 1
v = 2.8 × √(0.152 − 0.0752) = 0.36 m s−1
✓ 1
Part (d)
a = ω2x = 2.82 × 0.075 = 0.59 m s−2
✓ 1
directed towards the equilibrium position (negative x)
✓ 1
Exam tip: With x = x₀ cos ωt the motion starts at maximum displacement; solve cos(ωt) = 0.5 using radians and take the smallest positive angle for the first time.
103IBP3H-C1-20r10-14Conditions for simple harmonic motionPaper 2Medium7 marks
A uniform solid cylinder of mass 0.300 kg, length 0.400 m and cross-sectional area 2.00 × 10−3 m2 floats upright in a swimming pool. The density of the water is 1.00 × 103 kg m−3. At equilibrium the upthrust equals the weight. The cylinder is pushed down a small distance x and released. It stays partly submerged and water resistance is negligible.
Diagram not to scale
(a)
Show that the resultant force on the cylinder when it is displaced a distance x below equilibrium is F = −ρAgx, where A is the cross-sectional area and ρ the density of the water.
(3)
(b)
Calculate the period of the oscillation. Give your answer to three significant figures.
(3)
(c)
Suggest why the motion would not be simple harmonic if the cylinder were pushed down until it was fully submerged.
(1)
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Part (a)
extra submerged volume is Ax, so the upthrust increases by ρAgx
✓ 1
weight is unchanged and was equal to the original upthrust, so the resultant force equals the extra upthrust, directed upwards
✓ 1
F = −ρAgx is proportional to the displacement and opposite to it, so the motion is simple harmonic
✓ 1
Part (b)
effective force constant k = ρAg = 1.00 × 103 × 2.00 × 10−3 × 9.8 = 19.6 N m−1
✓ 1
uses T = 2π√(m/k)
✓ 1
T = 0.777 s
✓ 1
Part (c)
once fully submerged the upthrust no longer increases with displacement, so the resultant force is not proportional to x
✓ 1
Exam tip: To show SHM, find an expression for the resultant force in terms of displacement and compare it with F = −kx; the coefficient gives the effective k.
104IBP3H-C1-20r20-11Phase difference between oscillatorsPaper 2Medium7 marks
The piston of a hospital ventilator moves with simple harmonic motion of amplitude 6.0 cm and period 4.0 s. At t = 0 the piston is at its equilibrium position, moving in the positive direction, so that its displacement is x = x0 sin ωt.
(a)
Calculate the angular frequency ω.
(1)
(b)
Calculate the displacement of the piston at t = 0.50 s.
(2)
(c)
Calculate the speed of the piston at t = 0.50 s.
(2)
(d)
Calculate the maximum acceleration of the piston in m s−2. Give your answer to three significant figures.
(2)
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Part (a)
A1 ω = 2π/4.0 = 1.57 rad s−1
✓ 1
Part (b)
M1 x = 6.0 sin(1.571 × 0.50), with the calculator in radian mode
✓ 1
A1 x = 4.2 cm
✓ 1
Part (c)
M1 v = ωx0 cos ωt = 1.571 × 6.0 × cos(0.785)
✓ 1
A1 v = 6.7 cm s−1
✓ 1
Part (d)
M1 amax = ω2x0 = 1.5712 × 0.060
✓ 1
A1 amax = 0.148 m s−2
✓ 1
Common error: Using degrees instead of radians in sin ωt. Check the calculator mode before evaluating.
105IBP3H-C1-20r20-13Damped oscillationsPaper 2Medium7 marks
A lightly damped sensor in a beehive monitor oscillates with an initial amplitude of 8.0 mm. After 5 complete oscillations the amplitude is 5.2 mm.
(a)
Calculate the percentage of the oscillation energy lost during the 5 oscillations. Give your answer to two significant figures.
(3)
(b)
Assume the amplitude falls by the same factor in each further set of 5 oscillations. Calculate the amplitude after 10 oscillations. Give your answer to two significant figures.
(2)
(c)
State what is meant by critical damping and give one example of its use.
(2)
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Part (a)
M1 energy is proportional to amplitude squared
✓ 1
A1 energy remaining = (5.2/8.0)2 = 0.4225
✓ 1
A1 energy lost = 58%
✓ 1
Part (b)
M1 amplitude = 8.0 × (5.2/8.0)2
✓ 1
A1 3.4 mm
✓ 1
Part (c)
The system returns to equilibrium in the shortest possible time without oscillating
✓ 1
Any suitable use, e.g. a vehicle suspension or a door closer
✓ 1
Common error: Percentage energy lost uses the square of the amplitude ratio. Using 5.2/8.0 directly gives the amplitude loss, not the energy loss.
107IBP3H-C1-20r39-12Phase and timing in SHMPaper 2Medium9 marks
The top of a wind-turbine tower oscillates horizontally with simple harmonic motion of amplitude 0.30 m and frequency 0.50 Hz. The displacement x is measured from the equilibrium position, and at t = 0 the top is at its maximum positive displacement.
(a)
Determine the angular frequency and write down an expression for x in terms of t.
(2)
(b)
Calculate the displacement at t = 0.70 s.
(2)
(c)
Calculate the velocity at t = 0.70 s. Take velocity to be positive in the direction of positive x.
(3)
(d)
Calculate the acceleration at t = 0.70 s. Use the unrounded value of x from part (b).
(2)
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Part (a)
ω = 2πf = π rad s−1 (3.14 rad s−1)
✓ 1
x = 0.30 cos(πt) (with x in m and t in s)
✓ 1
Part (b)
substitutes t = 0.70 s into x = 0.30 cos(0.70π), calculator in radian mode
✓ 1
x = −0.176 m
✓ 1
Part (c)
v = −x0ω sin(ωt)
✓ 1
substitutes: −0.30 × π × sin(0.70π)
✓ 1
v = −0.762 m s−1
✓ 1
Part (d)
a = −ω2x = −π2 × (−0.1763)
✓ 1
a = +1.74 m s−2
✓ 1
Common error: Set the calculator to radians when evaluating cos(ωt) and sin(ωt); degrees give a completely different value.
108IBP3H-C1-20r39-13Mass–spring and pendulum periodsPaper 2Medium8 marks
A sensor of mass 0.50 kg hangs at rest from a light vertical spring inside the nacelle of a wind turbine. The spring extends by 0.049 m from its natural length. Use g = 9.8 m s−2.
Diagram not to scale
(a)
Calculate the spring constant of the spring.
(2)
(b)
The sensor is pulled down a small distance and released. Calculate the period of its oscillation.
(2)
(c)
The same sensor and spring are taken to a planet where the gravitational field strength is 4.0 N kg−1. State the period of small oscillations there and explain your answer.
(2)
(d)
The sensor is pulled 0.020 m below its equilibrium position and released. Calculate its maximum speed.
(2)
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Part (a)
kΔx = mg at equilibrium
✓ 1
k = 0.50 × 9.8 / 0.049 = 100 N m−1
✓ 1
Part (b)
T = 2π√(m/k) = 2π√(0.50/100)
✓ 1
T = 0.444 s
✓ 1
Part (c)
the period is unchanged, 0.444 s
✓ 1
the period depends on m and k only, which are unchanged; it does not depend on g
✓ 1
Part (d)
vmax = ωx0 with ω = √(k/m) = 14.1 rad s−1
✓ 1
vmax = 0.283 m s−1
✓ 1
Common error: Do not use the equilibrium extension as the amplitude; the amplitude is the extra displacement from the equilibrium position.
109IBP3H-C1-20r39-14Acceleration in SHMPaper 2Medium9 marks
In a library demonstration, a 0.20 kg trolley is attached to a light horizontal spring of spring constant 30 N m−1. The trolley moves on a frictionless track with simple harmonic motion of amplitude 0.10 m.
(a)
Calculate the total energy of the oscillation.
(2)
(b)
Calculate the kinetic energy of the trolley when its displacement is 0.060 m.
(2)
(c)
Determine the speed of the trolley at this displacement.
(2)
(d)
Describe the shape of a graph of the kinetic energy of the trolley against its displacement, for displacements from −0.10 m to +0.10 m.
(3)
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Part (a)
E = ½kx02 = ½ × 30 × 0.102
✓ 1
E = 0.15 J
✓ 1
Part (b)
Ek = ½k(x02 − x2) = ½ × 30 × (0.102 − 0.0602) or 0.15 − ½ × 30 × 0.0602
✓ 1
Ek = 0.096 J
✓ 1
Part (c)
½mv2 = 0.096 so v = √(2 × 0.096 / 0.20)
✓ 1
v = 0.980 m s−1
✓ 1
Part (d)
an inverted parabola, symmetrical about zero displacement
✓ 1
maximum of 0.15 J at zero displacement
✓ 1
falls to zero at displacements of ±0.10 m
✓ 1
Exam tip: Kinetic energy plus potential energy equals the constant total energy; find one from the total and the other to avoid long calculations.
110IBP3H-C1-20r39-16Conditions for SHMPaper 2Medium7 marks
A pendulum-type damper in a wind-turbine tower consists of a heavy bob of mass m on a light cable of length L = 2.5 m. Treat it as a simple pendulum swinging through small angles. Use g = 9.8 m s−2.
Diagram not to scale
(a)
Show that, for small angles, the motion of the bob is simple harmonic with period T = 2π√(L/g).
(3)
(b)
Calculate the period of the damper.
(2)
(c)
The bob is released from rest with a horizontal displacement of 0.20 m. Determine its maximum speed.
(2)
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Notes
Part (a)
the restoring force along the arc is F = −mg sin θ
✓ 1
for small θ, sin θ ≈ θ = x/L, so F = −(mg/L)x
✓ 1
a = −(g/L)x, so a is proportional to −x with ω2 = g/L, giving T = 2π/ω = 2π√(L/g)
✓ 1
Part (b)
T = 2π√(2.5/9.8)
✓ 1
T = 3.17 s
✓ 1
Part (c)
vmax = ωx0 with ω = √(g/L) = 1.98 rad s−1
✓ 1
vmax = 0.396 m s−1
✓ 1
Common error: The small-angle approximation sin θ ≈ θ needs θ in radians; a displacement of 0.20 m on a 2.5 m cable is about 0.08 rad, so it holds.
111C-2-01SHM energy and equationsPaper 2Hard14 marks
The accelerometer in a smartphone contains a proof mass of 4.0 × 10−9 kg held between tiny springs inside a frame that is fixed to the phone. The proof mass can move only along one axis, and its natural frequency of oscillation along this axis is 2.50 kHz.
When the phone has a constant acceleration a along the axis, the proof mass comes to rest relative to the frame at a displacement x from the position where the springs are unstretched. A circuit measures x.
(a)
(i)
Show that x = a/ω², where ω is the angular frequency of the natural oscillation of the proof mass.
(2)
(ii)
The phone is held at rest with the sensing axis vertical. Show that x is about 40 nm.
(2)
(iii)
Suggest why the designer chooses a high natural frequency even though it makes x very small.
(1)
(b)
The phone, held at rest with the sensing axis vertical, is dropped and falls freely.
(i)
Explain why the proof mass oscillates relative to the frame as the phone falls, and determine its maximum speed relative to the frame.
(3)
(ii)
Calculate the total energy of this oscillation.
(1)
(iii)
The phone falls 1.20 m from rest and hits a hard floor, which stops it uniformly in 2.0 ms. Determine the displacement of the proof mass relative to the frame during the impact. Assume that the proof mass stays at rest relative to the frame while the phone is being stopped.
(3)
(c)
(i)
Gas inside the sensor damps the motion of the proof mass. Explain why the designer chooses damping close to critical.
(2)
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Notes
Part (a)(i)
The only horizontal force on the proof mass is the spring force, and it must give the proof mass the same acceleration as the phone: kx = ma
✓ 1
Newton's second law applied to the proof mass at rest relative to the frame.
ω² = k/m for a mass–spring system, so x = ma/k = a/ω²
✓ 1
Symbolic answer required; no numerical substitution is expected.
Part (a)(ii)
The springs now support the weight, so g plays the part of a: ω = 2π × 2500 = 1.57 × 104 rad s−1
✓ 1
x = 9.81/(1.57 × 104)² = 3.98 × 10−8 m
✓ 1
Answer to at least 2 s.f. (3.98 or 4.0 × 10−8 m) or full substitution required. Allow ECF from (a)(i).
Part (a)(iii)
The natural frequency is then far above the frequencies at which a phone is normally shaken, so the proof mass follows changes of acceleration quickly and is never driven at resonance
✓ 1
Accept either idea: fast response or no resonance with ordinary motion.
Part (b)(i)
In free fall the frame and the proof mass have the same acceleration g, so the equilibrium position becomes the unstretched position, x = 0
✓ 1
The proof mass starts at rest relative to the frame 4.0 × 10−8 m from this new equilibrium, so it oscillates with amplitude 4.0 × 10−8 m
✓ 1
This is the hidden step: the old equilibrium becomes the turning point.
Speed at impact v = √(2gh) = √(2 × 9.81 × 1.20) = 4.85 m s−1
✓ 1
Deceleration a = v/Δt = 4.85/2.0 × 10−3 = 2.4 × 103 m s−2
✓ 1
x = a/ω² = 2.4 × 103/(1.57 × 104)² = 9.8 × 10−6 m
✓ 1
Allow ECF from (a)(i) and (a)(ii). The weight of the proof mass (adding g/ω²) is negligible.
Part (c)(i)
With critical damping the proof mass returns to its equilibrium position in the shortest time without oscillating, so the reading settles quickly
✓ 1
Light damping would let the proof mass keep oscillating and give false readings; heavy damping would make the reading respond too slowly
✓ 1
Both alternatives needed for this mark.
Answers: (a)(ii) 3.98 × 10−8 m · (b)(i) 6.25 × 10−4 m s−1 · (b)(ii) 7.8 × 10−16 J · (b)(iii) 9.8 × 10−6 m (the remaining parts are explanations — see the table above)
112C-2-18Mass–spring and pendulum periodsPaper 2Hard12 marks
A bungee jumper of mass 70.0 kg steps off a high platform. She is attached to an elastic cord of unstretched length 20.0 m that obeys Hooke's law with spring constant 140 N m−1. Air resistance and the mass of the cord can be ignored.
Diagram not to scale
(a)
(i)
Calculate her speed when the cord first becomes taut.
(1)
(ii)
Show that the extension of the cord when she would hang at rest is about 4.9 m.
(1)
(b)
(i)
Explain why, while the cord is taut, her motion is simple harmonic about the position 4.9 m below the point where the cord becomes taut.
(2)
(ii)
Determine the amplitude of the simple harmonic motion.
(3)
(iii)
Determine her maximum acceleration and compare it with g.
(2)
(iv)
Use the conservation of energy to confirm your answer to (b)(ii).
(2)
(c)
(i)
Explain why her motion as a whole is not simple harmonic.
(1)
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Part (a)(i)
v = √(2gh) = √(2 × 9.81 × 20.0) = 19.8 m s−1
✓ 1
Part (a)(ii)
e = mg/k = 70.0 × 9.81/140 = 4.905 m
✓ 1
Answer to at least 3 s.f. required.
Part (b)(i)
At a distance x below that position the resultant force is k(e + x) − mg = kx upwards, since ke = mg
✓ 1
The resultant force (and acceleration) is proportional to x and directed towards this position, which is the condition for SHM
✓ 1
Part (b)(ii)
ω = √(k/m) = √(140/70.0) = 1.414 rad s−1
✓ 1
When the cord becomes taut, x = 4.91 m and v = 19.8 m s−1: 19.8 = 1.414√(x0² − 4.91²)
✓ 1
Or energy: ½mv² = ½k(x0² − x²). Allow ECF from (a).
x0 = 14.8 m
✓ 1
Accept 14.8–14.9 m. Taking x = 0 at the taut point (x0 = 14.0 m): [2 max].
Part (b)(iii)
amax = ω²x0 = 2.00 × 14.8 = 30 m s−2
✓ 1
Allow ECF from (b)(ii).
This is about 3g, directed upwards, at the lowest point
✓ 1
Direction or position required.
Part (b)(iv)
Greatest extension = 4.91 + 14.8 = 19.75 m, so she falls 20.0 + 19.75 = 39.75 m in all
✓ 1
Allow ECF from (b)(ii).
Loss of gravitational potential energy = 70.0 × 9.81 × 39.75 = 2.73 × 104 J; elastic energy stored = ½ × 140 × 19.75² = 2.73 × 104 J: equal, which confirms the amplitude
✓ 1
Comparison and conclusion required.
Part (c)(i)
While the cord is slack the only force is her weight, so her acceleration is constant (g) and not proportional to her displacement from equilibrium
✓ 1
Answers: (a)(i) 19.8 m s−1 · (a)(ii) 4.91 m · (b)(ii) 14.8 m · (b)(iii) 30 m s−2 ≈ 3g(the remaining parts are explanations — see the table above)
113C-2-19SHM energy and equationsPaper 2Hard10 marks
A singly charged calcium ion of mass 6.64 × 10−26 kg is held in an ion trap. Along the axis of the trap the electric potential V varies with position x as shown, so that V = Cx², where C is a constant.
Electric potential against position along the axis of the trap (drawn to scale).
(a)
(i)
By comparing the electric potential energy of the ion with the potential energy of a simple harmonic oscillator, show that the ion oscillates with ω = √(2eC/m), where e is the elementary charge.
(2)
(ii)
Use the graph to determine C, and hence determine the frequency of oscillation of the ion.
(3)
(b)
The total energy of the oscillation of the ion is 3.0 eV.
(i)
Use the graph to determine the amplitude of the oscillation.
(2)
(ii)
Determine the maximum speed of the ion.
(2)
(iii)
Express the unit of C in fundamental SI units.
(1)
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Part (a)(i)
Ep = eV = eCx²
✓ 1
This has the form ½mω²x², so ½mω² = eC and ω = √(2eC/m)
✓ 1
Part (a)(ii)
At x = 1.00 mm, V = 5.24 V, so C = 5.24 V/(1.00 × 10−3 m)² = 5.2 × 106 V m−2
✓ 1
Accept 5.0–5.5 × 106 V m−2; any point read correctly.
At the turning points all the energy is potential: eV = 3.0 eV, so V = 3.0 V
✓ 1
The hidden step: an energy of 3.0 eV for a charge e corresponds to 3.0 V.
Reading the graph at 3.0 V: x0 = 0.76 mm
✓ 1
Accept 0.72–0.80 mm. Allow ECF from (a)(ii) if calculated from C.
Part (b)(ii)
½mvmax² = 3.0 × 1.60 × 10−19 J
✓ 1
Or vmax = ωx0. Allow ECF from (a)(ii) and (b)(i).
vmax = 3.8 × 103 m s−1
✓ 1
Accept 3.6–4.0 × 103 m s−1.
Part (b)(iii)
V m−2 = (J C−1) m−2 = kg s−3 A−1
✓ 1
Answer must be in kg, s and A only.
Answers: (a)(ii) 5.2 × 106 V m−2; 8.0 × 105 Hz · (b)(i) 0.76 mm · (b)(ii) 3.8 × 103 m s−1 · (b)(iii) kg s−3 A−1(the remaining parts are explanations — see the table above)
114C-2-21SHM energy and equationsPaper 2Hard11 marks
Carbon dioxide strongly absorbs infrared radiation of wavelength 15.0 μm. A simple model of the vibration responsible treats an oxygen atom of mass 2.66 × 10−26 kg as a mass on a spring of spring constant k, with the rest of the molecule fixed.
(a)
(i)
Calculate the frequency of the radiation.
(1)
(ii)
Assuming that the natural frequency of the vibration equals this frequency, determine k.
(2)
(iii)
Explain why radiation of this frequency is absorbed strongly while nearby frequencies are absorbed much less.
(2)
(b)
(i)
Calculate the energy of one photon of this radiation, in eV.
(1)
(ii)
The atom vibrates with amplitude 3.0 × 10−12 m. Determine its amplitude after the molecule absorbs one photon of this radiation.
(3)
(iii)
The mean temperature of the Earth's surface is 288 K. Explain how the absorption described in this question contributes to the greenhouse effect.
(2)
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Part (a)(i)
f = c/λ = 3.00 × 108/15.0 × 10−6 = 2.0 × 1013 Hz
✓ 1
Part (a)(ii)
k = m(2πf)² = 2.66 × 10−26 × (2π × 2.0 × 1013)²
✓ 1
k = 420 N m−1
✓ 1
Accept 420 N m−1. Allow ECF from (a)(i).
Part (a)(iii)
The oscillating electric field of the radiation drives the vibration at its own frequency; when this equals the natural frequency there is resonance
✓ 1
At resonance energy is transferred from the radiation to the vibration most efficiently, so the amplitude (and absorption) is greatest
✓ 1
Part (b)(i)
E = hf = 1.33 × 10−20 J = 0.083 eV
✓ 1
Unit eV required. Allow ECF from (a)(i).
Part (b)(ii)
Initial energy = ½kx0² = ½ × 420 × (3.0 × 10−12)² = 1.9 × 10−21 J
✓ 1
Allow ECF from (a)(ii).
Final energy = 1.9 × 10−21 + 1.33 × 10−20 = 1.52 × 10−20 J
✓ 1
Allow ECF from (b)(i).
x0 = √(2E/k) = 8.5 × 10−12 m
✓ 1
Accept 8.4–8.6 × 10−12 m.
Part (b)(iii)
By Wien's law the Earth's surface radiates most strongly at about 2.9 × 10−3/288 = 1 × 10−5 m, close to 15 μm, so CO2 absorbs a large part of this radiation
✓ 1
The absorbed energy is re-emitted in all directions, so part of it returns to the surface, which is warmer than it would otherwise be
✓ 1
Answers: (a)(i) 2.0 × 1013 Hz · (a)(ii) 420 N m−1 · (b)(i) 0.083 eV · (b)(ii) 8.5 × 10−12 m (the remaining parts are explanations — see the table above)
115C-2-39Conditions for simple harmonic motionPaper 2Hard14 marks
A U-tube of uniform internal cross-sectional area A = 1.5 × 10−4 m2 contains water of density ρ = 1000 kg m−3. The total length of the water column, measured along the tube, is L = 0.600 m. Both ends of the tube are open to the atmosphere. The water is displaced and released, and it oscillates. Viscous forces can be ignored.
Diagram not to scale
(a)
(i)
At one instant the water surface in the left arm is a distance y above its equilibrium level, and the surface in the right arm is y below its equilibrium level. Show that the resultant force on the water is 2ρAgy, directed towards the equilibrium position.
(2)
(ii)
Hence show that the water performs simple harmonic motion with angular frequency ω given by ω² = 2g/L.
(2)
(iii)
Calculate the period of the oscillation.
(1)
(b)
The right-hand arm is now sealed while the water is at rest in equilibrium, trapping a column of air of length H = 0.200 m at atmospheric pressure p0 = 1.01 × 105 Pa. When the water rises a small distance y in the sealed arm, the pressure of the trapped air increases by γp0y/H, where γ = 1.40 for air.
(i)
Determine the ratio (restoring force due to the trapped air)/(restoring force due to gravity) for the same displacement y.
(2)
(ii)
Determine the period of oscillation with the arm sealed.
(3)
(iii)
The compression of the trapped air is adiabatic. Explain why, and state the effect on the temperature of the trapped air as the water rises.
(2)
(c)
(i)
With both ends open again, the water oscillates with amplitude 0.020 m. Determine the total energy of the oscillation.
(2)
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Part (a)(i)
The levels differ by 2y, so the unbalanced part is the weight of a column of water of length 2y: mass = ρA × 2y
✓ 1
ALT: pressure difference ρg(2y) acting on area A.
F = 2ρAgy, acting to lower the higher surface, i.e. towards equilibrium
✓ 1
Direction needed.
Part (a)(ii)
The whole column, of mass ρAL, has acceleration a: ρALa = −2ρAgy
✓ 1
Allow ECF from (a)(i).
a = −(2g/L)y, which has the form a = −ω²x, so the motion is SHM with ω² = 2g/L
✓ 1
Reference to the defining equation (acceleration proportional to displacement and opposite to it) is needed.
Each compression lasts only a fraction of a second, so there is too little time for significant thermal energy transfer to the tube or the water: Q ≈ 0
✓ 1
Work is done on the air, so its internal energy, and hence its temperature, increases
✓ 1
Reference to work done on the gas or ΔU = −W is needed.
Part (c)(i)
ET = ½mω²x0² with m = ρAL = 1000 × 1.5 × 10−4 × 0.600 = 0.090 kg
✓ 1
Allow ECF from (a)(ii). ALT: ET = ρAgy0² from the work done against the force in (a)(i).
ET = ½ × 0.090 × 5.72² × 0.020² = 5.9 × 10−4 J
✓ 1
Answers: (a)(iii) 1.10 s · (b)(i) 36 · (b)(ii) 0.18 s · (c)(i) 5.9 × 10−4 J (the remaining parts are explanations — see the table above)
116C-2-41Mass–spring and pendulum periodsPaper 2Hard19 marks
Astronauts on a space station orbiting 410 km above the Earth's surface cannot weigh themselves with bathroom scales. Instead, an astronaut sits strapped into a chair of mass 12.0 kg that is attached to springs and can oscillate along one axis. With the chair empty, its period of oscillation is 0.900 s.
The graph shows the displacement x of the chair, with the astronaut in it, against time t after it is pulled to one side and released.
Displacement of the chair carrying the astronaut against time (drawn to scale).
(a)
(i)
Explain why a bathroom scale on the space station reads zero.
(2)
(ii)
Show that the gravitational field strength at the height of the station is about 8.7 N kg−1.
(2)
(b)
(i)
The period of the empty chair is T1 and the period with the astronaut is T2. Show that the mass of the astronaut is m = mc(T2²/T1² − 1), where mc is the mass of the chair.
(2)
(ii)
State the unit of the spring constant k in fundamental SI units.
(1)
(c)
(i)
Use the graph to determine T2.
(2)
(ii)
Determine the mass of the astronaut.
(2)
(iii)
Determine the maximum speed of the astronaut during the first oscillation.
(2)
(iv)
Calculate the total energy of the oscillation at the start.
(1)
(d)
(i)
The oscillation is lightly damped. Use the graph to estimate the percentage of the energy of the oscillation that is dissipated in each cycle.
(2)
(ii)
Explain why this damping does not significantly affect the measured mass.
(1)
(e)
(i)
The springs are attached to the station, which has a mass of 4.2 × 105 kg. Estimate the amplitude with which the station itself oscillates, and state one assumption you make.
(2)
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Part (a)(i)
The station, the scale and the astronaut are all in free fall, with the same acceleration towards the Earth (the gravitational field strength)
✓ 1
so no contact force between the astronaut and the scale is needed to give the astronaut this acceleration, and the scale reads zero
✓ 1
Answers stating that there is no gravity in orbit: [0] for the question part.
Dividing: T2²/T1² = (mc + m)/mc, so m = mc(T2²/T1² − 1)
✓ 1
Part (b)(ii)
kg s−2
✓ 1
Only fundamental units accepted; N m−1: [0].
Part (c)(i)
Five complete oscillations take 11.6 s
✓ 1
Measurement over several oscillations expected; a single period read: [1 max].
T2 = 2.32 s
✓ 1
Accept 2.28–2.36 s.
Part (c)(ii)
m = 12.0 × ((2.32/0.900)² − 1)
✓ 1
Allow ECF from (b)(i) and (c)(i).
m = 68 kg
✓ 1
Accept 66–70 kg.
Part (c)(iii)
ω = 2π/2.32 = 2.71 rad s−1, and x0 = 0.100 m from the graph
✓ 1
Allow ECF from (c)(i).
vmax = ωx0 = 0.27 m s−1
✓ 1
Part (c)(iv)
ET = ½(mc + m)ω²x0² = ½ × 80 × 2.71² × 0.100² = 2.9 J
✓ 1
The mass of the chair must be included; using the astronaut alone (2.5 J): [0]. Allow ECF from (c)(ii) and (c)(iii).
Part (d)(i)
After five cycles the amplitude is 8.6 cm, so the energy has fallen to (8.6/10.0)² = 0.74 of its initial value
✓ 1
Accept 8.4–8.8 cm. The hidden step is energy ∝ amplitude².
Per cycle: 0.741/5 = 0.941, so about 6 % is dissipated each cycle
✓ 1
Accept 5–7 %. Dividing 26 % by 5 is accepted. Allow ECF.
Part (d)(ii)
For light damping the period is almost the same as the undamped period, so T2 (and hence m) is essentially unchanged
✓ 1
Allow ECF from (d)(i).
Part (e)(i)
The centre of mass of the station + chair + astronaut stays fixed, so xstation = (mc + m)x0/Mstation = 80 × 0.100/4.2 × 105 = 2 × 10−5 m
✓ 1
Allow ECF from (c)(ii). Momentum argument (equal and opposite momenta) accepted.
Assumption: the station is rigid and no other external or internal forces act on it during the measurement (it is an isolated system)
✓ 1
Answers: (a)(ii) 8.66 N kg−1 · (b)(ii) kg s−2 · (c)(i) 2.32 s · (c)(ii) 68 kg · (c)(iii) 0.27 m s−1 · (c)(iv) 2.9 J · (d)(i) about 6 % · (e)(i) 2 × 10−5 m (the remaining parts are explanations — see the table above)
117IBP2S-018-00-15Mass–spring and pendulum periodsPaper 2Hard9 marks
On an orbiting space station, an astronaut's mass is measured by strapping them into a chair attached to a spring. The chair is set oscillating horizontally in simple harmonic motion. Mass of empty chair = 12 kg Period with chair empty = 0.90 s Period with astronaut strapped in = 2.10 s
(a)
Explain why bathroom scales cannot measure the astronaut's mass on the station, whereas this oscillating device can.
(2)
(b)
Show that the spring constant of the spring is about 580 N m−1.
(2)
(c)
Determine the mass of the astronaut.
(2)
(d)
With the astronaut strapped in, the chair is released from rest at a displacement of 0.050 m at time t = 0. Calculate the speed of the chair at t = 0.40 s.
(3)
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Part (a)
The astronaut and the scales are both in free fall, so there is no normal contact force for the scales to register . The period of a mass–spring system depends only on m and k and not on g, so it still gives the mass .
✓ 2
Part (b)
Rearranges to k = 4π2m/T2 = 4π2 × 12/0.902 . k = 585 N m−1, which is about 580 N m−1 .
✓ 2
Part (c)
Total mass = kT2/4π2 = 585 × 2.102/4π2 = 65 kg, or 12 × (2.10/0.90)2 = 65 kg . Astronaut mass = 65 − 12 = 53 kg . ECF from (b).
✓ 2
Part (d)
ω = 2π/2.10 = 2.99 rad s−1 . Uses v = −ωx0 sin ωt, with ωt = 1.20 rad and the calculator in radian mode . Speed = 2.99 × 0.050 × sin(1.20) = 0.14 m s−1 .
✓ 3
Common error: Remember to subtract the chair's mass in (c). In (d), using degree mode gives a wrong value for sin ωt.
118IBP2H-019-01-11Mass–spring and pendulum periodsPaper 2Hard8 marks
A whole-body vibration platform used in physiotherapy oscillates vertically with simple harmonic motion. The frequency is 30 Hz and the amplitude is 2.0 mm. A small loose weight rests on the platform.
Diagram not to scale
(a)
Calculate the maximum acceleration of the platform.
(2)
(b)
Explain why the loose weight does not stay in contact with the platform throughout each oscillation.
(2)
(c)
Determine the maximum amplitude at 30 Hz for which the weight stays in contact with the platform at all times.
(2)
(d)
The platform returns to an amplitude of 2.0 mm. At t = 0 it passes through its equilibrium position moving upwards. Upwards is taken as positive. Calculate the displacement of the platform at t = 4.0 ms.
(2)
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Part (a)
ω = 2π × 30 = 188 rad s−1
✓ 1
amax = ω2x0 = 188.52 × 2.0×10−3 ≈ 71 m s−2
✓ 1
Part (b)
Near the top of the motion, the platform's downward acceleration is larger than g (71 m s−2 > 9.8 m s−2)
✓ 1
The weight can accelerate downwards at no more than g, because the normal force can only push and cannot pull it down. The platform therefore moves away from the weight and contact is lost
✓ 1
Part (c)
Contact is just maintained when ω2x0 = g
✓ 1
x0 = 9.8/188.52 = 2.8×10−4 m (0.28 mm); allow error carried forward for ω from the first part
✓ 1
Part (d)
x = x0 sin(ωt), with ωt = 188.5 × 4.0×10−3 = 0.754 rad
✓ 1
x = 2.0 × sin(0.754) = +1.4 mm (1.37 mm); allow error carried forward for ω from the first part
✓ 1
Common error: Set your calculator to radians when evaluating sin(ωt). Using degrees here gives about 0.026 mm, which is clearly wrong.
119IBP2H-019-01-13Damped oscillationsPaper 2Hard8 marks
After a diver leaves a springboard, the end of the board oscillates vertically with a frequency of 1.2 Hz. The initial amplitude is 0.15 m. The oscillations are lightly damped, and the amplitude decreases by 8.0% in each complete cycle.
(a)
State what is meant by light damping.
(1)
(b)
Show that about 15% of the energy of the oscillation is lost in each cycle.
(2)
(c)
The amplitude is measured at the end of each complete cycle. Determine the time, from when the diver leaves, at which the measured amplitude is first below 0.050 m.
(3)
(d)
Explain why the frequency of the board stays approximately 1.2 Hz even though the amplitude decreases.
(2)
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Part (a)
The resistive force is small, so the system still oscillates, and the amplitude decreases gradually over many cycles
✓ 1
Part (b)
Energy is proportional to (amplitude)2, so each cycle the energy becomes 0.922 = 0.846 of its previous value
✓ 1
Fraction lost = 1 − 0.846 = 0.154 ≈ 15%
✓ 1
Part (c)
0.15 × 0.92n < 0.050, so 0.92n < 1/3
✓ 1
n > ln 3 / ln(1/0.92) = 13.2, so 14 complete cycles are needed (after 13 cycles the amplitude is 0.0507 m and after 14 cycles it is 0.0467 m)
✓ 1
t = 14/1.2 = 11.7 s ≈ 12 s
✓ 1
Part (d)
For SHM the period and frequency do not depend on amplitude
✓ 1
The damping force is small compared with the restoring force, so it lowers the frequency only very slightly
✓ 1
Common error: A fixed percentage loss per cycle is not a linear decrease. Use repeated multiplication (0.92ⁿ), not subtraction of 8% of the original amplitude each time. Round the number of cycles up to a whole number, because the amplitude is only checked once per complete cycle.
120IBP2H-019-01-15Conditions for simple harmonic motionPaper 2Hard9 marks
A cylindrical wave-energy buoy of mass 120 kg floats upright in seawater of density 1030 kg m−3. The buoy has a uniform cross-sectional area of 0.25 m2 and is tall enough that it never becomes fully submerged or leaves the water. The buoy is pushed down a distance y below its equilibrium floating position and then released. Take g = 9.81 m s−2.
Diagram not to scale
(a)
Show that the buoy performs simple harmonic motion, and deduce an expression for its angular frequency ω.
(3)
(b)
Calculate the period of oscillation of the buoy.
(2)
(c)
The buoy is released from rest 0.12 m below its equilibrium position. Calculate its maximum speed.
(2)
(d)
Outline why the oscillations of a real buoy are damped.
(2)
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Part (a)
The extra submerged volume is Ay, so there is an extra upthrust of ρgAy acting upwards
✓ 1
At equilibrium the weight balances the original upthrust, so the net force is F = −ρgAy. This force is proportional to the displacement and directed opposite to it
✓ 1
a = −(ρgA/m)y, which has the form a = −ω2x, so the motion is SHM with ω = √(ρgA/m)
✓ 1
Part (b)
ω2 = 1030 × 9.81 × 0.25/120 = 21.1 s−2, so ω = 4.59 rad s−1
✓ 1
T = 2π/ω = 1.37 s
✓ 1
Part (c)
vmax = ωx0 = 4.59 × 0.12
✓ 1
vmax = 0.55 m s−1. Allow an error carried forward from the value of ω calculated in the previous part
✓ 1
Part (d)
The water exerts a drag force that opposes the motion of the buoy, and the moving buoy also generates surface waves
✓ 1
Energy is transferred from the oscillation to the water, so the amplitude decreases with time
✓ 1
Exam tip: To prove SHM, show that the net force (or acceleration) is a negative constant multiplied by the displacement. Then identify that constant as ω², or as mω² if you are working with force.
121IBP2H-019-00-13SHM with a phase anglePaper 2Hard8 marks
A tuned mass damper near the top of a skyscraper oscillates horizontally with SHM of period 8.0 s and amplitude 0.60 m. Its displacement is modelled as x = x0 sin(ωt + φ). At t = 0 the damper is at x = +0.30 m, moving in the positive direction.
(a)
Show that ω ≈ 0.785 rad s−1.
(1)
(b)
Determine the phase angle φ.
(2)
(c)
Calculate the velocity of the damper at t = 0.
(2)
(d)
Determine the earliest time after t = 0 at which the damper is at x = −0.30 m moving in the positive direction.
(3)
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Part (a)
ω = 2π/T = 2π/8.0 = 0.785 rad s−1 .
✓ 1
Part (b)
0.30 = 0.60 sin φ, so sin φ = 0.5 . The velocity is positive, so cos φ > 0, giving φ = π/6 (0.52 rad) .
✓ 2
Part (c)
v = ωx0 cos φ, or v = ω√(x0² − x²) . v = 0.785 × 0.60 × cos(π/6) = 0.41 m s−1 . ECF from (b).
✓ 2
Part (d)
This needs sin(ωt + φ) = −0.5 with cos(ωt + φ) > 0, so ωt + φ = 11π/6 . ωt = 11π/6 − π/6 = 5π/3 . t = 5.236/0.785 = 6.7 s . ECF from (b).
✓ 3
Exam tip: List all the phase values that give the required displacement. Then use the sign of the velocity to select the correct one before converting to time.
122IBP2H-019-00-14Conditions for simple harmonic motionPaper 2Hard9 marks
An ocean wave-monitoring buoy is a vertical cylinder of mass 400 kg and uniform cross-sectional area 0.50 m2. It floats upright in calm seawater of density 1030 kg m−3. The buoy is pushed down slightly and released.
Diagram not to scale
(a)
State the conditions required for an object to move with simple harmonic motion.
(2)
(b)
The buoy is displaced a further distance x below its equilibrium position. Show that its motion satisfies the defining equation of SHM.
(3)
(c)
Calculate the period of oscillation.
(2)
(d)
Outline why the motion would not be simple harmonic if the buoy were pushed down far enough to be fully submerged.
(2)
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Part (a)
The acceleration (or restoring force) is proportional to the displacement from equilibrium . It is directed towards the equilibrium position, opposite to the displacement .
✓ 2
Part (b)
The extra submerged volume is Ax, so the buoyancy increases by ρgAx . At equilibrium weight and buoyancy cancel, so the net force is ρgAx upward, opposite to x: F = −ρgAx . Then a = −(ρgA/m)x, which has the form a = −ω²x with ω² = ρgA/m constant .
✓ 3
Part (c)
T = 2π√(m/(ρgA)) . T = 2π√(400/(1030 × 9.8 × 0.50)) = 1.8 s (1.77 s) . ECF from (b).
✓ 2
Part (d)
Once the buoy is fully submerged, the buoyancy no longer increases with further depth . So the restoring force is no longer proportional to displacement, and the motion is not SHM .
✓ 2
Exam tip: In a 'show that it is SHM' derivation, always end by matching your result to a = −ω²x. Then identify ω² as a positive constant.
123IBP2H-019-00-15Phase difference between oscillatorsPaper 2Hard8 marks
A kinetic sculpture has two simple pendulums of lengths 1.00 m and 0.81 m. They are released together, in phase, with small amplitudes.
Diagram not to scale
(a)
Calculate the period of each pendulum.
(2)
(b)
Explain why the phase difference between the pendulums increases with time.
(2)
(c)
Determine the phase difference between the pendulums 5.0 s after release.
(2)
(d)
Determine the shortest time after release at which the pendulums are next in phase.
(2)
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Part (a)
T1 = 2π√(1.00/9.8) = 2.01 s . T2 = 2π√(0.81/9.8) = 1.81 s .
✓ 2
Part (b)
The pendulums have different periods, so they have different angular frequencies . The phase of each increases at a rate equal to its ω, so the phase difference grows at the rate (ω2 − ω1) and increases steadily with time .
✓ 2
Part (c)
Δφ = 2πt(1/T2 − 1/T1) = 2π × 5.0 × (0.5536 − 0.4982) . Δφ = 1.7 rad (accept 1.7–1.8 rad from rounded periods) . ECF from (a).
✓ 2
Part (d)
They are next in phase when the shorter pendulum has completed one more oscillation than the longer one, so Δφ = 2π, i.e. t(1/T2 − 1/T1) = 1 . t = 18 s (equal to 9T1 = 10T2) . ECF from (a).
✓ 2
Common error: The pendulums are back in phase when one has completed exactly one more full cycle than the other, not one more half-cycle.
124IBP2S-018-02-16Mass–spring and pendulum periodsPaper 2Hard9 marks
A rope swing is used for training at a sports centre. The distance from the pivot to the centre of mass of the athlete is 5.0 m. The athlete is released from rest with the rope at 10° to the vertical. Air resistance can be ignored.
Diagram not to scale
(a)
Show that the athlete's centre of mass falls through about 0.076 m between the release point and the lowest point.
(2)
(b)
Use energy conservation to calculate the speed of the athlete at the lowest point.
(2)
(c)
Treating the motion as simple harmonic, calculate the maximum speed of the athlete.
(3)
(d)
The athlete is instead released at 50° to the vertical. State and explain which method, energy or SHM, now gives the larger maximum speed.
(2)
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Part (a)
h = L − L cos 10°
✓ 1
h = 5.0(1 − 0.9848) = 0.0760 m
✓ 1
Part (b)
mgh = ½mv², so v = √(2gh)
✓ 1
v = √(2 × 9.8 × 0.0760) = 1.22 m s−1
✓ 1
Part (c)
ω = √(g/L) = √(9.8/5.0) = 1.40 rad s−1
✓ 1
arc amplitude x0 = Lθ = 5.0 × 0.1745 = 0.873 m
✓ 1
vmax = ωx0 = 1.22 m s−1
✓ 1
Part (d)
The SHM model gives the larger (overestimated) value
✓ 1
because the true restoring force is proportional to sin θ, which is less than θ at large angles, so the real acceleration is smaller than the SHM model assumes
✓ 1
Exam tip: For SHM calculations the amplitude of a pendulum is the arc length Lθ, with θ in radians. Converting degrees to radians is an easy mark to lose.
125IBP3H-C1-20r7-15Energy in SHMPaper 2Hard11 marks
In a model of a honeybee wing in flight, the wing tip moves with simple harmonic motion of frequency 230 Hz and amplitude 1.1 mm. The effective mass of the wing is 1.0 × 10−6 kg. A second wing has the same frequency and amplitude but its maximum displacements occur 0.50 ms after those of the first wing.
(a)
Calculate the maximum speed of the wing tip.
(2)
(b)
Calculate the maximum acceleration of the wing tip.
(2)
(c)
Calculate the total energy of the oscillation of the wing.
(2)
(d)
Show that the kinetic energy equals the potential energy at a displacement of 0.78 mm.
(3)
(e)
Determine the phase difference between the two wings, in rad, to 3 significant figures.
(2)
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Part (a)
ω = 2π × 230 = 1445 rad s−1
✓ 1
vmax = ωA = 1.59 m s−1
✓ 1
Part (b)
amax = ω²A
✓ 1
amax = 1445² × 1.1 × 10−3 = 2.3 × 103 m s−2
✓ 1
Part (c)
E = maximum kinetic energy = ½mvmax²
✓ 1
E = ½ × 1.0 × 10−6 × 1.59² = 1.3 × 10−6 J (1.26 × 10−6 J)
✓ 1
Part (d)
Kinetic energy = potential energy means each is E/2
✓ 1
½mω²x² = ½ × ½mω²A² so x = A/√2
✓ 1
x = 1.1/√2 = 0.78 mm
✓ 1
Part (e)
φ = 2π × Δt/T = 2πf Δt
✓ 1
φ = 2π × 230 × 0.50 × 10−3 = 0.723 rad
✓ 1
Exam tip: Kinetic and potential energies are equal at x = A/√2, not at A/2.
126IBP3H-C1-20r7-16Energy in SHMPaper 2Hard10 marks
A tuned block of mass 0.60 kg oscillates on a horizontal spring in a test for a Formula 1 car. Its displacement in metres is given by x = 0.080 sin(ωt + φ), where the period is 0.90 s. At t = 0 the displacement is +0.040 m and the block is moving in the positive direction.
(a)
Calculate the angular frequency ω.
(2)
(b)
Determine the phase constant φ.
(3)
(c)
Calculate the time at which the block first reaches its maximum displacement.
(2)
(d)
Determine the kinetic energy of the block at t = 0, in J, to 3 significant figures.
(3)
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Part (a)
ω = 2π/T
✓ 1
ω = 7.0 rad s−1 (6.98 rad s−1)
✓ 1
Part (b)
0.040 = 0.080 sin φ so sin φ = 0.50
✓ 1
φ = π/6 (0.524 rad) or 5π/6
✓ 1
Velocity positive requires cos φ > 0, so φ = π/6 (0.524 rad)
✓ 1
Part (c)
ωt + φ = π/2
✓ 1
t = (π/2 − π/6)/ω = 0.15 s (ECF from (a) and (b))
✓ 1
Part (d)
Speed v = ωA cos φ = 6.98 × 0.080 × cos(π/6) = 0.484 m s−1
✓ 1
Ek = ½mv² = ½ × 0.60 × 0.484²
✓ 1
Ek = 0.0702 J
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Common error: When finding a phase constant from a displacement, use the direction of motion to choose between the two possible angles.
127IBP3H-C1-20r10-16Energy in SHMPaper 2Hard9 marks
An automated library shelf carriage of mass 0.40 kg is attached to a spring of force constant 64 N m−1 on a frictionless horizontal rail. The carriage is pulled 0.050 m from its equilibrium position and released from rest at t = 0, so that x = x0 cos ωt.
(a)
Calculate the angular frequency of the oscillation. Give your answer to three significant figures.
(2)
(b)
Determine the first time at which the kinetic energy of the carriage equals its elastic potential energy. Give your answer to three significant figures.
(3)
(c)
Calculate the speed of the carriage at this time. Give your answer to three significant figures.
(2)
(d)
Describe how the kinetic energy of the carriage varies with time over one period of the motion.
(2)
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Mark
Notes
Part (a)
uses ω = √(k/m)
✓ 1
ω = 12.6 rad s−1
✓ 1
Part (b)
½kx2 = ½ × ½kx02, so x = x0/√2
✓ 1
cos ωt = 1/√2, so ωt = π/4
✓ 1
t = 0.0621 s
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Part (c)
total energy ½kx02 = 0.080 J so KE = 0.040 J (or v = ωx0 sin ωt)
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v = √(2 × 0.040/0.40) = 0.447 m s−1
✓ 1
Part (d)
KE is zero at t = 0, rises to a maximum of 0.080 J at T/4 = 0.124 s and returns to zero at T/2, never negative
✓ 1
the KE graph repeats every T/2 = 0.248 s, so twice per displacement cycle
✓ 1
Exam tip: Equal kinetic and potential energy occurs at x = x₀/√2, not at half the amplitude.
128IBP3H-C1-20r20-15Energy in SHMPaper 2Hard9 marks
A block of mass 0.60 kg is attached to a horizontal spring of spring constant 150 N m−1 and oscillates with simple harmonic motion on a frictionless surface. When the displacement from equilibrium is 0.040 m, the speed of the block is 0.50 m s−1. Use unrounded values from earlier parts in later parts.
(a)
Calculate the angular frequency of the oscillation.
(1)
(b)
Determine the amplitude of the oscillation.
(3)
(c)
Calculate the total energy of the oscillation.
(2)
(d)
The block is released from its maximum positive displacement at t = 0, so that x = x0 cos ωt. Determine the first time at which the displacement is +0.040 m. Give your answer to three significant figures.
(3)
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Notes
Part (a)
A1 ω = √(k/m) = √(150/0.60) = 15.8 rad s−1
✓ 1
Part (b)
M1 v = ω√(x02 − x2)
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M1 x02 = x2 + (v/ω)2 = 0.0016 + 0.0010 = 0.0026
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A1 x0 = 0.0510 m
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Part (c)
M1 E = ½kx02 = ½ × 150 × 0.0026 (ECF from (b))
✓ 1
A1 E = 0.195 J (accept 0.20 J)
✓ 1
Part (d)
M1 cos ωt = 0.040/x0 = 0.784
✓ 1
M1 ωt = cos−1(0.784) = 0.669 rad (radian mode)
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A1 t = 0.0423 s
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Common error: Mixing up cos and sin forms. If the motion starts at maximum displacement, use x = x₀ cos ωt and work in radians.
129IBP3H-C1-20r20-16Damped oscillationsPaper 2Hard10 marks
A sensor of mass 0.250 kg hangs from a vertical spring in a beehive. At equilibrium the spring is extended by 0.049 m. The sensor is pulled down a further 0.030 m and released, and it then oscillates vertically with simple harmonic motion. Damping causes the total energy of the oscillation to decrease by 4.0% in each complete cycle. Use g = 9.8 m s−2.
Diagram not to scale
(a)
Show that the spring constant is 50 N m−1.
(2)
(b)
Calculate the period of the undamped oscillation.
(2)
(c)
Calculate the amplitude after 5 complete cycles. Give your answer to three significant figures.
(3)
(d)
Determine the smallest number of complete cycles after which the total energy is less than half of its initial value.
(3)
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Notes
Part (a)
M1 mg = ke at equilibrium
✓ 1
A1 k = 0.250 × 9.8/0.049 = 50 N m−1
✓ 1
Part (b)
M1 T = 2π√(m/k) = 2π√(0.250/50)
✓ 1
A1 T = 0.444 s
✓ 1
Part (c)
M1 energy is proportional to amplitude squared, so A5 = A0√(0.965)
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M1 0.965 = 0.815
✓ 1
A1 A5 = 0.0271 m
✓ 1
Part (d)
M1 0.96n < 0.50
✓ 1
M1 n > ln 0.50/ln 0.96 = 16.98
✓ 1
A1 17 complete cycles
✓ 1
Common error: A 4.0% energy loss per cycle is not a 4.0% amplitude loss. Energy is proportional to the amplitude squared.
130IBP3H-C1-20r39-15Phase difference between oscillatorsPaper 2Hard10 marks
Two identical buoys A and B in a swimming lane oscillate vertically with simple harmonic motion of the same amplitude 0.15 m and the same period 2.4 s. Their displacements from equilibrium, in metres, at time t in seconds are xA = 0.15 sin(ωt) and xB = 0.15 sin(ωt + π/4).
(a)
Determine the time by which the motion of B is ahead of the motion of A.
(2)
(b)
Calculate the displacement of B at t = 0.
(2)
(c)
Determine the first time, after t = 0, at which the two buoys have the same displacement.
(4)
(d)
Calculate the common displacement of the buoys at the time found in part (c). Use ωt = 3π/8.
(2)
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Notes
Part (a)
phase difference π/4 corresponds to a fraction 1/8 of a period
✓ 1
time = 2.4/8 = 0.30 s
✓ 1
Part (b)
xB = 0.15 sin(π/4)
✓ 1
xB = 0.106 m
✓ 1
Part (c)
sets sin(ωt) = sin(ωt + π/4)
✓ 1
uses sin θ = sin(π − θ), so ωt + π/4 = π − ωt
✓ 1
ωt = 3π/8
✓ 1
t = (3π/8) / (2π/2.4) = 0.45 s
✓ 1
Part (d)
x = 0.15 sin(3π/8)
✓ 1
x = 0.139 m
✓ 1
Exam tip: To find when two sinusoids are equal, use the fact that sin θ takes the same value at θ and π − θ, and choose the smallest positive time that fits.
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