Simple harmonic motion is defined by an acceleration proportional to displacement and directed towards equilibrium, a = −ω²x. Both levels use the periods of a mass–spring system and a simple pendulum and the exchange between kinetic and potential energy during an oscillation.
At HL you also describe SHM with equations that include a phase angle, find the velocity at any displacement, v = ±ω√(x₀² − x²), and calculate kinetic and total energy from the amplitude and angular frequency. Phase differences between oscillators are a frequent multiple-choice theme.
52 questions
265 marks
Paper 1A: 28
Paper 1B: 10
Paper 2: 14
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Showing 52 of 52 questions · 265 marks
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26 practice questions on C.1 Simple harmonic motion
1C-1A-01
SHM velocity and displacement·C.1 Simple harmonic motion
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A particle performs simple harmonic motion with amplitude A and maximum speed v0.
What is the speed of the particle when its displacement is A/2?
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Notes
Step 1The speed at displacement x is v = ω√(A² − x²), and the maximum speed (at x = 0) is v0 = ωA.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2At x = A/2: v = ω√(A² − A²/4) = ωA√(3/4).
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Step 3v = (√3/2)v0 ≈ 0.87v0.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis assumes the speed falls linearly with displacement. It follows a circle: x² + (v/ω)² = A².
BThis is the speed at x = A/√2, where kinetic and potential energies are equal.
CCorrect: v = ω√(A² − A²/4) = (√3/2)ωA.
DThis is the fraction of the kinetic energy remaining (3/4), not of the speed.
Syllabus understandingC.1 — that a particle undergoing simple harmonic motion has velocity v = ±ω√(x0² − x²) Command term: Determine
2C-1A-02
Mass–spring and pendulum periods·C.1 Simple harmonic motion
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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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Notes
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.
✓ 1
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.
CCorrect: m/k doubles, and T ∝ √(m/k).
DThis forgets the square root.
Syllabus understandingC.1 — the period of a mass–spring system as given by T = 2π√(m/k) Command term: Determine
3C-1A-03
Phase and timing in SHM·C.1 Simple harmonic motion
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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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Notes
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.
✓ 1
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.
DThis is the time to reach the full amplitude A.
Syllabus understandingC.1 — that the displacement of a particle in SHM is given by x = x0 sin ωt or x = x0 cos ωt; the phase of an oscillation Command term: Determine
4C-1A-15
Energy in SHM·C.1 Simple harmonic motion
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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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Notes
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.
✓ 1
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.
Syllabus understandingC.1 (HL) — that problems can be solved using the equations for simple harmonic motion as given by ET = ½mω²x0²; C.1 — the time period of a mass–spring system as given by T = 2π√(m/k) Command term: Determine
5C-1A-17
Mass–spring and pendulum periods·C.1 Simple harmonic motion
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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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Notes
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.
✓ 1
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.
Syllabus understandingC.1 — the time period of a simple pendulum as given by T = 2π√(l/g); Tools 3 — linearizing a relationship and interpreting the gradient Command term: Determine
6C-1A-22
Acceleration in SHM·C.1 Simple harmonic motion
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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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Notes
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.
✓ 1
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.
Syllabus understandingC.1 — the defining equation of simple harmonic motion as given by a = −ω²x; the time period in terms of frequency of oscillation and angular frequency as given by T = 1/f = 2π/ωCommand term: Determine
7C-1A-31
Conditions for SHM·C.1 Simple harmonic motion
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksIdentify
Four objects oscillate about an equilibrium position. Resistive forces are negligible.
Which object performs simple harmonic motion?
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Notes
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.
✓ 1
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.
Syllabus understandingC.1 — conditions that lead to simple harmonic motion Command term: Identify
8C-1A-32
Acceleration in SHM·C.1 Simple harmonic motion
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
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.
Velocity of the object against time (drawn to scale).Show mark scheme
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Notes
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.
✓ 1
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.
Syllabus understandingC.1 — the defining equation of simple harmonic motion as given by a = −ω²x; a qualitative approach to energy changes during one cycle of an oscillation Command term: Deduce
9C-1A-33
Phase and timing in SHM·C.1 Simple harmonic motion
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
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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Notes
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.
✓ 1
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.
Syllabus understandingC.1 (HL) — that a particle undergoing simple harmonic motion can be described using phase angle; x = x0 sin(ωt + φ), v = ωx0 cos(ωt + φ); a = −ω²xCommand term: Deduce
10C-1A-34
Phase difference between oscillators·C.1 Simple harmonic motion
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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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Notes
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.
✓ 1
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).
Syllabus understandingC.1 (HL) — that a particle undergoing simple harmonic motion can be described using phase angle (Guidance: radians are used for phase angle calculations) Command term: Determine
11C-1A-35
Energy in SHM·C.1 Simple harmonic motion
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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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Notes
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.
✓ 1
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.
Syllabus understandingC.1 — a qualitative approach to energy changes during one cycle of an oscillation; T = 2π/ω; C.1 (HL) — v = ±ω√(x0² − x²) Command term: Determine
12C-1A-36
Energy in SHM·C.1 Simple harmonic motion
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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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Notes
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).
✓ 1
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.
Syllabus understandingC.1 (HL) — that problems can be solved using the equations for simple harmonic motion as given by Ep = ½mω²x², ET = ½mω²x0² Command term: Determine
13C-1A-63
Damped oscillations·C.1 Simple harmonic motion
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
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?
PeriodMaximum speed
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Mark
Notes
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.
✓ 1
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.
Syllabus understandingC.1 — a particle undergoing simple harmonic motion can be described using time period, frequency, angular frequency, amplitude, equilibrium position and displacement; the time period of a mass–spring system T = 2π√(m/k); C.4 — the effects of light damping on the system Command term: Deduce
14C-1A-66
Mass–spring and pendulum periods·C.1 Simple harmonic motion
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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?
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Notes
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.
✓ 1
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.
Syllabus understandingC.1 — the time period of a simple pendulum as given by T = 2π√(l/g); A.2 — Newton’s three laws of motion; tension Command term: Determine
15C-1A-70
Phase and timing in SHM·C.1 Simple harmonic motion
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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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Notes
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.
✓ 1
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.
Syllabus understandingC.1 (HL) — that a particle undergoing SHM can be described using phase angle; x = x0 sin(ωt + φ); ET = ½mω²x0² and Ep = ½mω²x² Command term: Determine
16C-1A-73
Amplitude and equilibrium position·C.1 Simple harmonic motion
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A glider oscillates with simple harmonic motion on a horizontal air track. A motion sensor at one end of the track records the distance d between the sensor and the glider. The graph shows how d varies with time t.
What is the maximum speed of the glider?
Distance d between the sensor and the glider against time t (drawn to scale).Show mark scheme
Marking point
Mark
Notes
Step 1The glider oscillates between d = 0.30 m and d = 0.50 m, so the equilibrium position is at d = 0.40 m and the amplitude is x0 = 0.10 m.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Successive maxima are 0.80 s apart, so T = 0.80 s and ω = 2π/0.80 = 7.85 rad s−1.
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Step 3vmax = ωx0 = 7.85 × 0.10 = 0.79 m s−1.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis uses the frequency instead of the angular frequency: fx0 = 1.25 × 0.10 = 0.125 m s−1. The factor 2π is missing.
BCorrect: x0 = 0.10 m, ω = 7.85 rad s−1, vmax = ωx0 = 0.79 m s−1.
CThis takes the amplitude as the full range of d (0.50 − 0.30 = 0.20 m): 7.85 × 0.20 = 1.57 m s−1. The amplitude is half of this.
DThis takes the largest distance from the sensor, 0.50 m, as the amplitude: 7.85 × 0.50 = 3.93 m s−1. The displacement is measured from the equilibrium position, d = 0.40 m.
Syllabus understandingC.1 — a particle undergoing simple harmonic motion can be described using time period T, frequency f, angular frequency ω, amplitude, equilibrium position, and displacement; C.1 — the time period in terms of frequency of oscillation and angular frequency as given by T = 1/f = 2π/ω; C.1 (HL) — that problems can be solved using the equations for simple harmonic motion as given by v = ±ω√(x0² − x²) (with x = 0) Command term: Determine
17C-1A-74
The minus sign in a = −ω²x·C.1 Simple harmonic motion
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
An object performs simple harmonic motion about an equilibrium position.
Which quantities are always directed towards the equilibrium position whenever the object is not at the equilibrium position?
I. The acceleration of the object
II. The resultant force on the object
III. The velocity of the object
Show mark scheme
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Mark
Notes
Step 1a = −ω²x: the minus sign means that the acceleration is always opposite to the displacement, i.e. towards the equilibrium position (I is correct).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The resultant force is ma, so it always has the same direction as the acceleration (II is correct).
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Step 3The velocity is away from the equilibrium position for half of each cycle (from equilibrium out to each extreme), so III is not correct.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the acceleration and the resultant force (F = −mω²x) always point towards equilibrium; the velocity does not.
BIII is not correct: as the object moves out towards an extreme its velocity points away from equilibrium. II is also correct, since F = ma.
CI is also correct (the acceleration is in the direction of the resultant force), and III is not: the velocity points away from equilibrium for half of every cycle.
DThis assumes that the object always moves towards equilibrium. While it moves out to an extreme it is slowing down: its velocity is away from equilibrium and its acceleration is towards it.
Syllabus understandingC.1 — the defining equation of simple harmonic motion as given by a = −ω²x (Guidance: the significance of the minus sign in the defining equation for simple harmonic motion); A.2 — Newton's second law Command term: Deduce
18C-1A-75
Pendulum length and period·C.1 Simple harmonic motion
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
Two simple pendulums P and Q are set swinging with small amplitudes at the same place. In the time that P completes 6 oscillations, Q completes 4 oscillations.
What is (length of Q)/(length of P)?
Show mark scheme
Marking point
Mark
Notes
Step 1Equal times: 6TP = 4TQ, so TQ/TP = 6/4 = 3/2. Q has the longer period.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2T = 2π√(l/g), so l ∝ T² at the same place.
—
Step 3lQ/lP = (3/2)² = 9/4.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis squares the ratio of the numbers of oscillations without inverting it: (4/6)² = 4/9. It makes Q, which oscillates less often, the shorter pendulum.
BThis takes the length as proportional to the frequency (4/6) instead of to the square of the period.
CThis is the ratio of the periods, 3/2; the length is proportional to the square of the period.
DCorrect: TQ/TP = 3/2 and l ∝ T², so the ratio is 9/4.
Syllabus understandingC.1 — the time period of a simple pendulum as given by T = 2π√(l/g); C.1 — the time period in terms of frequency of oscillation and angular frequency as given by T = 1/f = 2π/ωCommand term: Determine
19C-1A-76
Maximum speed and maximum acceleration·C.1 Simple harmonic motion
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
An object performs simple harmonic motion. Its maximum speed is vmax and the maximum magnitude of its acceleration is amax.
What is the period of the motion?
Show mark scheme
Marking point
Mark
Notes
Step 1vmax = ωx0 (at equilibrium) and amax = ω²x0 (at the extremes).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Dividing: amax/vmax = ω, so ω = amax/vmax.
—
Step 3T = 2π/ω = 2πvmax/amax.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis is 1/ω, not the period: the factor 2π in T = 2π/ω is missing.
BThis is half the period: it uses T = π/ω.
CCorrect: ω = amax/vmax and T = 2π/ω = 2πvmax/amax.
DThis inverts the ratio, taking ω = vmax/amax; this expression does not even have the unit of time (it has the unit s−1).
Syllabus understandingC.1 — the defining equation of simple harmonic motion as given by a = −ω²x; C.1 — the time period in terms of frequency of oscillation and angular frequency as given by T = 1/f = 2π/ω; C.1 (HL) — that problems can be solved using the equations for simple harmonic motion as given by v = ±ω√(x0² − x²) Command term: Determine
20C-1A-77
Equation of motion with a phase angle·C.1 Simple harmonic motion
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksIdentify
The graph shows how the displacement x of an oscillator varies with time t.
Which equation describes the motion, with x in cm and t in s?
Displacement x of the oscillator against time t (drawn to scale).Show mark scheme
Marking point
Mark
Notes
Step 1Amplitude x0 = 3.0 cm; period T = 0.40 s, so ω = 2π/0.40 = 5π rad s−1.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2At t = 0, x = 0 and x then becomes negative: x0 sin φ = 0 with the displacement decreasing requires φ = π.
—
Step 3x = 3.0 sin(5πt + π).
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis uses ω = 2πT instead of 2π/T: 2π × 0.40 = 0.8π rad s−1.
BThis omits the phase angle: 3.0 sin(5πt) starts at zero but moves to positive x first.
CThis takes the amplitude as the peak-to-peak range of the graph (6.0 cm); the amplitude is the maximum displacement from equilibrium.
DCorrect: x0 = 3.0 cm, ω = 5π rad s−1 and φ = π so that x starts at zero and decreases.
Syllabus understandingC.1 (HL) — that a particle undergoing simple harmonic motion can be described using phase angle; C.1 (HL) — that problems can be solved using the equations for simple harmonic motion as given by x = x0 sin(ωt + φ); C.1 — the time period in terms of frequency of oscillation and angular frequency as given by T = 1/f = 2π/ω (Guidance: radians are used for phase angle calculations) Command term: Identify
21C-1A-78
The minus sign in a = −ω²x·C.1 Simple harmonic motion
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
The graph shows how the displacement x of an object performing simple harmonic motion varies with time t. P is a point on the graph.
Which row gives the signs of the velocity and of the acceleration of the object at the instant P?
Displacement x against time t, with the point P marked (drawn to scale).
VelocityAcceleration
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Mark
Notes
Step 1At P the gradient of the x–t graph is negative, so the velocity is negative.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2At P the displacement is positive (+2.8 cm); a = −ω²x, so the acceleration is negative.
—
Step 3Velocity and acceleration are both negative: the object is moving towards equilibrium and speeding up.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis takes the sign of the velocity from the sign of the displacement. x is positive at P, but it is decreasing, so the velocity is negative.
BCorrect: the gradient of the graph is negative and a = −ω²x is negative because x is positive.
CThis assumes that the acceleration always opposes the velocity. Between an extreme and equilibrium the object speeds up, so its acceleration is in the same direction as its velocity.
DThis copies the sign of the displacement for both quantities; it ignores the minus sign in a = −ω²x and the negative gradient at P.
Syllabus understandingC.1 — the defining equation of simple harmonic motion as given by a = −ω²x (Guidance: the significance of the minus sign in the defining equation for simple harmonic motion); C.1 — a particle undergoing simple harmonic motion can be described using time period T, frequency f, angular frequency ω, amplitude, equilibrium position, and displacement Command term: Deduce
22C-1A-79
Energy changes during one cycle·C.1 Simple harmonic motion
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksIdentify
An object performs simple harmonic motion with period T. At time t = 0 the object passes through its equilibrium position.
Which graph shows how the potential energy Ep of the object varies with t from t = 0 to t = T?
Four possible graphs of the potential energy Ep of the object against time t for one period (not to scale).Show mark scheme
Marking point
Mark
Notes
Step 1Ep = 0 at equilibrium, so Ep = 0 at t = 0, T/2 and T (the object passes through equilibrium twice per cycle); Ep is greatest at each extreme, at T/4 and 3T/4.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Ep ∝ x² and x ∝ sin(2πt/T), so Ep ∝ sin²(2πt/T): two identical humps per period.
—
Step 3Near equilibrium Ep ∝ x² changes slowly, so the graph touches zero with zero gradient (rounded minima) — graph D.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis shows Ep ∝ |x|: two humps, but with sharp V-shaped minima at equilibrium. Ep depends on the square of the displacement, so the graph is flat at its minima.
BThis is the kinetic energy (or the potential energy for an oscillation that starts at an extreme): it is a maximum at t = 0, when the object is at equilibrium, where Ep = 0.
CThis shows only one hump per period, as if Ep had the same period as the displacement. Ep depends on x², which repeats every half period.
DCorrect: Ep ∝ sin²(2πt/T), zero at 0, T/2 and T with rounded minima and maxima at T/4 and 3T/4.
Syllabus understandingC.1 — a qualitative approach to energy changes during one cycle of an oscillation; C.1 (HL) — that problems can be solved using the equations for simple harmonic motion as given by Ep = ½mω²x² Command term: Identify
23C-1A-80
Mass–spring period and static extension·C.1 Simple harmonic motion
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A load hangs at rest from a vertical spring, which is extended by e. The load is pulled down a little and released, and it oscillates with period Ts. A simple pendulum of length 4e, at the same place, oscillates with small amplitude with period Tp.
What is Tp/Ts?
Show mark scheme
Marking point
Mark
Notes
Step 1At rest ke = mg, so m/k = e/g.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Ts = 2π√(m/k) = 2π√(e/g); Tp = 2π√(4e/g).
—
Step 3Tp/Ts = √4 = 2.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis inverts the ratio and omits the square root.
BThis inverts the ratio: the pendulum is four times longer than the "equivalent length" e of the spring, so it has the longer period.
CCorrect: the load on the spring has the same period as a pendulum of length e, and the period is proportional to the square root of the length.
DThis omits the square root, taking the period to be proportional to the length.
Syllabus understandingC.1 — the time period of a mass–spring system as given by T = 2π√(m/k); C.1 — the time period of a simple pendulum as given by T = 2π√(l/g); A.2 — Hooke's law and translational equilibrium Command term: Determine
24C-1A-81
Pendulum length and period·C.1 Simple harmonic motion
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A pendulum clock keeps correct time when the length of its pendulum is l. On a hot day the length of the pendulum increases by 0.20 %. The clock counts the oscillations of the pendulum and assumes each one has its original period.
Which row describes the behaviour of the clock over one day (86 400 s)?
The clockTime gained or lost
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Marking point
Mark
Notes
Step 1T ∝ √l, so a 0.20 % increase in l increases T by about ½ × 0.20 % = 0.10 % (√1.0020 = 1.0010).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Each oscillation now takes longer, so fewer oscillations are counted in a day: the clock loses time.
—
Step 3Time lost ≈ 0.10 % of 86 400 s = 86 s.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThe size is right, but a longer period means fewer oscillations counted per day, so the clock runs slow, not fast.
BCorrect: the period increases by 0.10 %, so the clock loses about 86 s per day.
CThe clock does lose time, but this takes the period to be proportional to the length: 0.20 % of 86 400 s = 173 s.
DThis omits the square root (T ∝ l) and also reverses the effect of a longer period.
Syllabus understandingC.1 — the time period of a simple pendulum as given by T = 2π√(l/g); C.1 — the time period in terms of frequency of oscillation and angular frequency as given by T = 1/f = 2π/ωCommand term: Deduce
Multiple choice · 1 mark3 steps to full marksDetermine
The graph shows how the velocity v of an object performing simple harmonic motion varies with its displacement x during one cycle.
What is the period of the motion?
Velocity v against displacement x for one cycle (drawn to scale).Show mark scheme
Marking point
Mark
Notes
Step 1From the graph: amplitude x0 = 5.0 cm = 0.050 m (where v = 0) and vmax = 0.40 m s−1 (where x = 0).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2ω = vmax/x0 = 0.40/0.050 = 8.0 rad s−1.
—
Step 3T = 2π/ω = 0.79 s.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis is x0/vmax = 1/ω = 0.125 s; the factor 2π is missing.
BThis uses T = π/ω, which is half the period: 0.39 s.
CCorrect: ω = 8.0 rad s−1 and T = 2π/8.0 = 0.79 s.
DThis leaves the amplitude in centimetres: 2π × 5.0/0.40 = 79.
Syllabus understandingC.1 (HL) — that problems can be solved using the equations for simple harmonic motion as given by v = ±ω√(x0² − x²); C.1 — the time period in terms of frequency of oscillation and angular frequency as given by T = 1/f = 2π/ωCommand term: Determine
26C-1A-83
Maximum acceleration·C.1 Simple harmonic motion
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A horizontal platform oscillates vertically with simple harmonic motion of frequency 2.5 Hz. A small block rests on the platform.
What is the largest amplitude of the platform for which the block stays in contact with it throughout the motion?
Show mark scheme
Marking point
Mark
Notes
Step 1The block is most likely to leave the platform at the top of the motion, where the acceleration is downwards with magnitude ω²x0. Only gravity can give the block a downward acceleration, so contact is lost if ω²x0 > g.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2ω = 2π × 2.5 = 15.7 rad s−1.
—
Step 3x0 = g/ω² = 9.81/15.7² = 0.0398 m ≈ 4.0 cm.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the block stays in contact while amax ≤ g, so x0 = g/ω² = 4.0 cm.
BThis uses ω = πf instead of 2πf: 9.81/(π × 2.5)² = 0.16 m.
CThis does not square ω: 9.81/15.7 = 0.62 m.
DThis uses the frequency in place of the angular frequency: 9.81/2.5² = 1.57 m.
Syllabus understandingC.1 — the defining equation of simple harmonic motion as given by a = −ω²x; A.2 — Newton's second law; normal force Command term: Determine
ACorrect: eliminating ω between the two speed equations gives x0²(v1² − v2²) = v1²x2² − v2²x1². Check: if v2 = 0 the expression gives x0 = x2, as it must (the particle is momentarily at rest at x2).
BThis pairs each speed with its own displacement instead of with the other one. Check: with v2 = 0 it gives x1, not x2.
CThis reverses the signs when the equations are combined (adding instead of subtracting); with numbers, e.g. x0 = 7.0 cm, x1 = 2.0 cm and x2 = 5.0 cm, it gives 4.2 cm instead of 7.0 cm.
DThis is 1/ω, found from ω² = (v1² − v2²)/(x2² − x1²); it is only the first step and has the unit of time, not of length.
Syllabus understandingC.1 (HL) — that problems can be solved using the equations for simple harmonic motion as given by v = ±ω√(x0² − x²) Command term: Determine
28C-1A-85
Period of a modified pendulum·C.1 Simple harmonic motion
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A simple pendulum of length l has period T0 for small oscillations. A thin peg P is fixed a distance 3l/4 vertically below the point of suspension S. Each time the bob swings to the right of the vertical, the string catches on the peg and the bob moves in an arc about P. The amplitude stays small.
What is the new period of the motion?
A simple pendulum whose string catches on a peg P on one side of each swing (not to scale).Show mark scheme
Marking point
Mark
Notes
Step 1To the left of the vertical the bob swings on length l: it spends half of a period T0 there, T0/2.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2To the right it swings on length l − 3l/4 = l/4, whose period is T0√(1/4) = T0/2; it spends half of this, T0/4, on that side.
—
Step 3New period = T0/2 + T0/4 = 3T0/4.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis uses the shortened length l/4 for the whole cycle: T0√(1/4). The bob swings on the full length for half of each cycle.
BThis combines the two halves with the period proportional to the length: ½T0 + ½(¼T0) = 5T0/8. The period is proportional to √l.
CCorrect: ½T0 + ½(T0/2) = 3T0/4.
DThis takes the length on the right as the distance from S to the peg, 3l/4: ½T0 + ½√(3/4)T0 = (2 + √3)T0/4. The bob swings about P, on the part of the string below the peg.
Syllabus understandingC.1 — the time period of a simple pendulum as given by T = 2π√(l/g); C.1 — the time period in terms of frequency of oscillation and angular frequency as given by T = 1/f = 2π/ωCommand term: Determine
29C-1B-01
Mass–spring and pendulum periods·C.1 Simple harmonic motion
Paper 1BMedium7 marks
Data-based question5 steps to full marksDetermine
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)
Show mark scheme
Marking point
Mark
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)
Syllabus understandingC.1 — the period of a mass–spring system as given by T = 2π√(m/k); A.2 — Hooke's law, the elastic restoring force F = −kx; Tools — propagating uncertainty through a power, error bars, uncertainty in a gradient from maximum and minimum gradient lines Command term: Determine
30C-1B-07
Mass–spring and pendulum periods·C.1 Simple harmonic motion
Paper 1BMedium7 marks
Data-based question5 steps to full marksDetermine
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)
Show mark scheme
Marking point
Mark
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)
Syllabus understandingC.1 — the period of a simple pendulum T = 2π√(l/g); D.1 — gravitational field strength g = GM/r²; Tools — gradient with its unit, intercept and systematic error, comparing a difference with the uncertainty Command term: Determine
31C-1B-09
Damped oscillations·C.1 Simple harmonic motion
Paper 1BHard7 marks
Data-based question6 steps to full marksDeduce
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)
Show mark scheme
Marking point
Mark
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)
Syllabus understandingC.4 — the effect of damping on oscillations (qualitative); C.1 (HL) — the total energy of an oscillator, ET = ½mω²x0²; D.4 (HL) — Lenz's law and induced currents; Tools — timing from a trace by counting cycles, testing a hypothesis with several points Command term: Deduce
32C-1B-11
SHM with a phase angle·C.1 Simple harmonic motion
Paper 1BMedium7 marks
Data-based question5 steps to full marksDetermine
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)
Show mark scheme
Marking point
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)
Syllabus understandingC.1 (HL) — a particle in SHM described using phase angle; x = x0 sin(ωt + φ), v = ωx0 cos(ωt + φ) and ET = ½mω²x0²; Tools — using the full range of data to reduce uncertainty Command term: Determine
33C-1B-12
Acceleration in SHM·C.1 Simple harmonic motion
Paper 1BMedium7 marks
Data-based question5 steps to full marksDetermine
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)
Show mark scheme
Marking point
Mark
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)
Syllabus understandingC.1 — the defining equation of SHM a = −ω²x; T = 2π/ω; A.2 — Newton's second law and Hooke's law, components of weight on an inclined plane; Tools — gradient with its unit, a systematic offset read from an intercept Command term: Determine
34C-1B-13
SHM energy and equations·C.1 Simple harmonic motion
Paper 1BMedium7 marks
Data-based question5 steps to full marksDetermine
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)
Show mark scheme
Marking point
Mark
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)
Syllabus understandingC.1 (HL) — the equations for SHM, v = ±ω√(x0² − x²) and ET = ½mω²x0²; Tools — linearising, gradient and intercept with units, systematic error Command term: Determine
35C-1B-24
Conditions for SHM·C.1 Simple harmonic motion
Paper 1BMedium7 marks
Data-based question5 steps to full marksDeduce
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)
Syllabus understandingC.1 — conditions that lead to simple harmonic motion; the time period of a simple pendulum T = 2π√(l/g); Tools — absolute uncertainty, testing a hypothesis with data, linearising a relationship, intercept and gradient of a graph Command term: Deduce
36C-1B-26
Maximum speed and amplitude·C.1 Simple harmonic motion
Paper 1BEasy7 marks
Data-based question6 steps to full marksDetermine
A load of mass 0.250 kg hangs from a vertical spring. A light gate is placed level with a flag on the load when the load is at rest in its equilibrium position. The flag is 1.00 ± 0.01 cm wide. The load is pulled down a distance x0 and released from rest, so x0 is the amplitude of the oscillation. An electronic timer measures the time t for which the flag blocks the light gate as the load first passes through equilibrium; its resolution is ±0.1 ms. The flag is narrow, so the speed calculated from t is taken to be the maximum speed vmax.
The graph shows vmax against x0 with the line of best fit.
x0 / cm
t / ms
vmax / m s−1
2.0
62.5
0.160
4.0
30.7
0.326
6.0
20.5
0.488
8.0
15.3
0.654
10.0
12.5
0.800
12.0
10.2
vmax against x0 with the line of best fit (drawn to scale).
(a)
(i)
Calculate the missing value of vmax for x0 = 12.0 cm.
(1)
(b)
(i)
Determine the absolute uncertainty in your answer to (a).
(2)
(c)
(i)
Determine the gradient of the line of best fit, including its unit.
(1)
(ii)
Explain how the graph supports the hypothesis that vmax is proportional to x0.
Both contributions are needed. Accept percentages (2.0 %).
Δvmax = 0.020 × 0.980 = ±0.02 m s−1
✓ 1
Accept ±0.019 to ±0.02 m s−1. Allow ECF from (a).
Part (c)(i)
Gradient from a large triangle, e.g. (0, 0) to (14.0 cm, 1.14): 8.13 s−1
✓ 1
Accept 7.9–8.3. Unit s−1 (or rad s−1) required; a gradient in m s−1 cm−1 must be converted.
Part (c)(ii)
The points lie on (or very close to) a straight line that passes through the origin, within the uncertainties
✓ 1
Both "straight line" and "through the origin" are needed.
Part (d)(i)
vmax = ωx0, so ω = gradient = 8.13 rad s−1
✓ 1
Allow ECF from (c)(i).
k = mω² = 0.250 × 8.13² = 16.5 N m−1
✓ 1
Accept 15.6–17.2 N m−1. Allow ECF from (c)(i).
Answers: (a)(i) 0.980 m s−1 · (b)(i) ±0.02 m s−1 · (c)(i) 8.13 s−1 · (d)(i) 16.5 N m−1(the remaining parts are explanations — see the table above)
Syllabus understandingC.1 (HL) — that problems can be solved using the equations for simple harmonic motion as given by v = ±ω√(x0² − x²) (at x = 0, vmax = ωx0); C.1 — the time period of a mass–spring system as given by T = 2π√(m/k); Tools — absolute and fractional uncertainty, gradient with its unit, proportionality Command term: Determine
37C-1B-27
Mass–spring period: power law·C.1 Simple harmonic motion
Paper 1BMedium7 marks
Data-based question6 steps to full marksDetermine
A student hangs a load of mass 0.200 kg from a chain of n identical light springs joined end to end, for n = 1 to 6. For each chain she times 10 vertical oscillations with a stopwatch; the uncertainty in each timing is ±0.3 s. She suggests that the period is T = Cnp, where C and p are constants, and plots lg(T / s) against lg n. The error bars show the uncertainty in lg(T / s).
A chain of n identical springs, each of spring constant k1, behaves as a single spring of spring constant k1/n.
n
10T / s
T / s
lg n
lg(T / s)
1
8.10
0.810
0.000
−0.092
2
11.51
1.151
0.301
0.061
3
14.03
1.403
0.477
0.147
4
16.19
1.619
0.602
5
18.04
1.804
0.699
0.256
6
19.84
1.984
0.778
0.298
lg T against lg n with error bars, the line of best fit (solid) and the lines of maximum and minimum gradient (dashed), drawn to scale.
(a)
(i)
Calculate the missing value of lg(T / s) for n = 4.
(1)
(b)
(i)
Use the graph to determine p with its absolute uncertainty.
(2)
(c)
(i)
Deduce whether your answer to (b) is consistent with the theory for a mass–spring system.
(1)
(d)
(i)
Use the intercept of the line on the lg(T / s) axis to determine k1.
(2)
(e)
(i)
In fact each spring has a small mass. Suggest the effect this has on the value of p obtained in (b).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
T = 16.19/10 = 1.619 s, lg T = 0.209
✓ 1
Accept 0.21.
Part (b)(i)
p = gradient of the line of best fit = 0.50
✓ 1
Accept 0.48–0.52. Large triangle expected.
Maximum and minimum gradients 0.53 and 0.47, so Δp = ±0.03
✓ 1
Accept ±0.02 to ±0.04. Accept (max − best) or (best − min).
Part (c)(i)
Theory: T = 2π√(nm/k1) ∝ n½, so p = 0.5, which lies within 0.50 ± 0.03: consistent
✓ 1
The theoretical value 0.5 and a comparison with the range are both needed. Allow ECF from (b).
Part (d)(i)
Intercept = −0.091, so for n = 1: T1 = 10−0.091 = 0.811 s
✓ 1
Accept intercept −0.081 to −0.101. ALT: use any point on the line with T = 2π√(nm/k1).
k1 = 4π²m/T1² = 4π² × 0.200/0.811² = 12.0 N m−1
✓ 1
Accept 11.5–12.6 N m−1 (from an intercept of −0.081 to −0.101). Allow ECF from their intercept.
Part (e)(i)
Adding springs adds mass that oscillates, so the oscillating mass increases with n and T increases faster than n½: the measured p is (slightly) larger than 0.5
✓ 1
Direction of the effect with a reason required. Allow ECF from (b).
Answers: (a)(i) 0.209 · (b)(i) 0.50 ± 0.03 · (c)(i) consistent · (d)(i) 12.0 N m−1(the remaining parts are explanations — see the table above)
Syllabus understandingC.1 — the time period of a mass–spring system as given by T = 2π√(m/k); A.2 — Hooke's law; Tools — linearising a power law with logarithms, error bars, gradient uncertainty from maximum and minimum gradient lines, intercept, systematic error Command term: Determine
38C-1B-28
Pendulum in a reduced field·C.1 Simple harmonic motion
Paper 1BHard7 marks
Data-based question5 steps to full marksDetermine
A small puck floats on a cushion of air on a flat air table, so friction is negligible. The puck is tied by a light string of length l = 0.500 m to a pin fixed in the table. The table is tilted at an angle α to the horizontal, and the puck hangs at rest directly downhill from the pin. The puck is displaced sideways across the table by a small amount and released, and it oscillates on the surface of the table.
For several angles α a student times 10 oscillations; the uncertainty in each timing is ±0.2 s. The graph shows T−2 against sin α with the line of best fit.
α / °
sin α
10T / s
T−2 / s−2
6
0.105
43.6
0.0526
10
0.174
34.1
0.0860
15
0.259
27.9
0.1285
20
0.342
24.2
30
0.500
20.1
0.2475
45
0.707
16.9
0.3501
60
0.866
15.2
0.4328
T−2 against sin α with the line of best fit (drawn to scale).
(a)
(i)
Show that the theory predicts T−2 = (g/4π²l) sin α.
(2)
(b)
(i)
Calculate the missing value of T−2 for α = 20°, together with its absolute uncertainty.
(2)
(c)
(i)
Determine g from the graph.
(2)
(d)
(i)
Suggest why the student did not use angles smaller than 5°.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
On the table only the component of the gravitational force along the surface, mg sin α, acts in the plane of the motion: the puck behaves as a pendulum in a field of strength g sin α
✓ 1
The idea of the component along the slope is needed.
T = 2π√(l/(g sin α)), so T−2 = g sin α/(4π²l)
✓ 1
Part (b)(i)
T = 24.2/10 = 2.42 s, T−2 = 0.1708 s−2
✓ 1
Accept 0.171 s−2.
ΔT = 0.02 s; fractional uncertainty in T−2 = 2 × 0.02/2.42 = 0.017, so Δ = ±0.003 s−2
✓ 1
Accept ±0.0028 s−2. Using ΔT = 0.2 s (not divided by 10), or omitting the factor 2: [0] for this mark.
Part (c)(i)
Gradient from a large triangle = 0.498 s−2
✓ 1
Accept 0.48–0.52 s−2.
g = 4π²l × gradient = 4π² × 0.500 × 0.498 = 9.83 m s−2
✓ 1
Accept 9.5–10.3 m s−2. Allow ECF from (a) and their gradient.
Part (d)(i)
At small α the restoring force (∝ sin α) is very small, so small effects such as residual friction, air currents or a slight tilt of the table across its width become comparable with it
✓ 1
Accept: the period becomes very long, so few oscillations occur before the motion is disturbed or dies away.
Answers: (b)(i) 0.171 ± 0.003 s−2 · (c)(i) 9.83 m s−2(the remaining parts are explanations — see the table above)
Syllabus understandingC.1 — the time period of a simple pendulum as given by T = 2π√(l/g); C.1 — conditions that lead to simple harmonic motion; A.2 — components of the weight on an inclined plane; Tools — propagating uncertainty through a power, linearising, gradient Command term: Determine
39C-2-01
SHM energy and equations·C.1 Simple harmonic motion
Paper 2Hard14 marks
Short answer & extended response8 steps to full marksExplain
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)
Show mark scheme
Marking point
Mark
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)
Syllabus understandingC.1 — the defining equation of SHM a = −ω²x; the time period of a mass–spring system; C.1 (HL) — v = ±ω√(x0² − x²) and ET = ½mω²x0²; C.4 — the effects of light, critical and heavy damping on the system; A.1 — motion with uniform acceleration; A.2 — Newton's second law Command term: Explain
40C-2-05
Mass–spring and pendulum periods·C.1 Simple harmonic motion
Paper 2Medium9 marks
Short answer & extended response6 steps to full marksDetermine
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.
T = 2π√(m/k) does not depend on g: the weight only changes the equilibrium extension, not the restoring force per unit displacement
✓ 1
Answers: (a)(i) 3.73 N kg−1 · (a)(ii) 2.30 s · (c)(i) 0.095 m s−1(the remaining parts are explanations — see the table above)
Syllabus understandingC.1 — the time period of a simple pendulum as given by T = 2π√(l/g); C.1 (HL) — v = ±ω√(x0² − x²); D.1 — gravitational field strength g = GM/r² Command term: Determine
41C-2-11
Conditions for simple harmonic motion·C.1 Simple harmonic motion
Paper 2Medium12 marks
Short answer & extended response8 steps to full marksShow that
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.
(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)
Show mark scheme
Marking point
Mark
Notes
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)
Syllabus understandingC.1 — conditions that lead to simple harmonic motion; the defining equation a = −ω²x; the time period T = 2π/ω; A.2 — buoyancy Fb = ρVg; C.4 — the nature of resonance; the effect of damping on the maximum amplitude; C.1 (HL) — ET = ½mω²x0²; C.2 — v = λ/T; A.3 — power Command term: Show that
42C-2-18
Mass–spring and pendulum periods·C.1 Simple harmonic motion
Paper 2Hard12 marks
Short answer & extended response8 steps to full marksDetermine
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.
(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)
Show mark scheme
Marking point
Mark
Notes
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)
Syllabus understandingC.1 — conditions that lead to simple harmonic motion; the defining equation a = −ω²x; C.1 (HL) — v = ±ω√(x0² − x²); A.1 — motion with uniform acceleration; A.3 — conservation of energy Command term: Determine
43C-2-19
SHM energy and equations·C.1 Simple harmonic motion
Paper 2Hard10 marks
Short answer & extended response6 steps to full marksDetermine
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)
Show mark scheme
Marking point
Mark
Notes
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)
Syllabus understandingC.1 (HL) — Ep = ½mω²x² and ET = ½mω²x0²; C.1 — the defining equation of SHM; D.2 — electric potential energy Ep = qV; Tools — fundamental SI units Command term: Determine
44C-2-20
Phase difference between oscillators·C.1 Simple harmonic motion
Paper 2Medium10 marks
Short answer & extended response6 steps to full marksDetermine
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)
Show mark scheme
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Notes
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)
Syllabus understandingC.1 (HL) — that a particle undergoing SHM can be described using phase angle; x = x0 sin(ωt + φ) and v = ωx0 cos(ωt + φ); radians are used for phase angle; A.2 — centripetal acceleration a = v²/r and Newton's second law Command term: Determine
45C-2-21
SHM energy and equations·C.1 Simple harmonic motion
Paper 2Hard11 marks
Short answer & extended response7 steps to full marksDetermine
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)
Show mark scheme
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Notes
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)
Syllabus understandingC.1 — the defining equation of SHM; C.1 (HL) — ET = ½mω²x0²; C.4 — the nature of resonance (linking question: modelling greenhouse gases as oscillators); E.1 — photon energy E = hf; B.1 — Wien's displacement law λmaxT = 2.9 × 10−3 m K; B.2 — the greenhouse effect Command term: Determine
46C-2-39
Conditions for simple harmonic motion·C.1 Simple harmonic motion
Paper 2Hard14 marks
Short answer & extended response8 steps to full marksShow that
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.
(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)
Show mark scheme
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Notes
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)
Syllabus understandingC.1 — conditions that lead to simple harmonic motion; the defining equation of SHM a = −ω²x; T = 2π/ω; C.1 (HL) — ET = ½mω²x0²; A.2 — Newton's second law; B.3 — pressure; B.4 (HL) — adiabatic processes and the first law of thermodynamics Command term: Show that
47C-2-40
Pendulum model of a swing·C.1 Simple harmonic motion
Paper 2Easy10 marks
Short answer & extended response7 steps to full marksDetermine
A child sits on a playground swing. The swing is modelled as a simple pendulum of length 2.40 m, measured from the pivot to the centre of mass of the child. Air resistance and friction at the pivot are negligible.
(a)
(i)
Calculate the period of small oscillations of the swing.
(1)
(ii)
State the condition that the motion must satisfy for this model to give the period correctly.
(1)
(b)
The child is pulled back until her centre of mass is 0.15 m above its lowest position and is then released from rest.
(i)
Show that her speed at the lowest point is about 1.7 m s−1.
(1)
(ii)
Determine the amplitude of the oscillation, treating the motion as simple harmonic.
(3)
(iii)
Describe the energy changes during one complete oscillation, starting from the moment of release.
(2)
(c)
(i)
An adult then gives the swing a small push each time it reaches its highest point on one side. Explain why the amplitude increases steadily.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
T = 2π√(2.40/9.81) = 3.11 s
✓ 1
Part (a)(ii)
The amplitude (angle) must be small, so that the restoring force is proportional to the displacement (sin θ ≈ θ)
✓ 1
Accept "small angle" with a reference to proportionality. "Small swing" alone: [0].
Part (b)(i)
½mv² = mgh, so v = √(2 × 9.81 × 0.15) = 1.72 m s−1
✓ 1
Full substitution or an answer to at least 3 s.f. required.
Part (b)(ii)
ω = 2π/T = 2π/3.11 = 2.02 rad s−1
✓ 1
Or ω = √(g/l). Allow ECF from (a)(i).
vmax = ωx0 at the lowest point (equilibrium)
✓ 1
x0 = 1.72/2.02 = 0.85 m
✓ 1
Accept 0.84–0.86 m. Allow ECF from (b)(i).
Part (b)(iii)
From release to the lowest point gravitational potential energy is transferred to kinetic energy; from the lowest point to the opposite extreme kinetic energy is transferred back to gravitational potential energy
✓ 1
Both halves of the swing needed.
This happens twice per cycle (kinetic energy is greatest twice and zero twice per oscillation), and the total energy stays constant
✓ 1
Accept a correct statement that the kinetic energy has period T/2.
Part (c)(i)
The pushes are applied at the natural frequency of the swing (once per period, at the same point of every cycle): this is resonance
✓ 1
Reference to the natural frequency or resonance needed.
Each push is applied just as the swing starts to move away from the adult, in the direction in which it then moves, so positive work is done and energy is added in every cycle, faster than it is lost
✓ 1
Accept: the force is in the direction of the subsequent velocity / in phase with the velocity. "Pushing makes it go higher" alone: [0].
Answers: (a)(i) 3.11 s · (b)(i) 1.72 m s−1 · (b)(ii) 0.85 m (the remaining parts are explanations — see the table above)
Syllabus understandingC.1 — the time period of a simple pendulum as given by T = 2π√(l/g); C.1 — conditions that lead to simple harmonic motion; C.1 — a qualitative approach to energy changes during one cycle of an oscillation; C.1 (HL) — that problems can be solved using the equations for simple harmonic motion as given by v = ±ω√(x0² − x²); A.3 — conservation of energy; C.4 — resonance Command term: Determine
48C-2-41
Velocity–time graph in SHM·C.1 Simple harmonic motion
Paper 2Medium10 marks
Short answer & extended response7 steps to full marksDetermine
The cone of a loudspeaker has a mass of 12 g and moves with simple harmonic motion along a horizontal line. The graph shows how the velocity v of the cone varies with time t. Positive velocity is away from the magnet at the back of the loudspeaker.
Velocity v of the loudspeaker cone against time t (drawn to scale).
(a)
(i)
Use the graph to show that the frequency of the oscillation is 250 Hz.
(1)
(ii)
Determine the amplitude of the oscillation of the cone.
(2)
(b)
(i)
Determine the maximum resultant force on the cone.
(2)
(c)
(i)
Determine the displacement of the cone at t = 0. Your answer should include the direction of the displacement.
(3)
(d)
The cone emits sound into air in which the speed of sound is 340 m s−1.
(i)
Calculate the wavelength of the sound.
(1)
(ii)
Describe the motion of a layer of air molecules directly in front of the cone as the sound passes.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Two cycles between the maxima at 1.33 ms and 9.33 ms (or 2.5 cycles in 10 ms): T = 4.0 ms, f = 1/(4.0 × 10−3) = 250 Hz
✓ 1
Reading over more than one cycle expected.
Part (a)(ii)
ω = 2π × 250 = 1571 rad s−1; vmax = 0.80 m s−1 from the graph
The gradient of the v–t graph is positive at t = 0, so the acceleration is positive; a = −ω²x, so the displacement is negative (towards the magnet)
✓ 1
The direction must be justified from the graph and the minus sign in a = −ω²x.
Part (d)(i)
λ = v/f = 340/250 = 1.36 m
✓ 1
Allow ECF from (a)(i).
Part (d)(ii)
The layer oscillates about a fixed equilibrium position, parallel to the direction in which the sound travels (along the axis of the cone), at 250 Hz
✓ 1
Reference to oscillation (no net movement) parallel to the direction of travel needed. Allow ECF from (a)(i) for the frequency.
Answers: (a)(i) 250 Hz · (a)(ii) 5.1 × 10−4 m · (b)(i) 15 N · (c)(i) −4.4 × 10−4 m (towards the magnet) · (d)(i) 1.36 m (the remaining parts are explanations — see the table above)
Syllabus understandingC.1 (HL) — that problems can be solved using the equations for simple harmonic motion as given by v = ±ω√(x0² − x²), v = ωx0 cos(ωt + φ); C.1 — the defining equation of simple harmonic motion as given by a = −ω²x (Guidance: the significance of the minus sign in the defining equation for simple harmonic motion); C.1 — the time period in terms of frequency of oscillation and angular frequency as given by T = 1/f = 2π/ω; A.2 — Newton's second law; C.2 — the nature of sound waves, v = fλCommand term: Determine
49C-2-42
Mass–spring model of a car suspension·C.1 Simple harmonic motion
Paper 2Medium10 marks
Short answer & extended response7 steps to full marksDetermine
The body of a car of mass 1100 kg is supported by springs, which are modelled as a single vertical spring that obeys Hooke's law. When four passengers of total mass 320 kg get into the car, the body moves down by 4.0 cm.
(a)
(i)
Show that the spring constant of the suspension is about 7.8 × 104 N m−1.
(1)
(ii)
Calculate the frequency of vertical oscillations of the body of the loaded car.
(2)
(b)
The loaded car drives at constant speed over a series of speed bumps 9.0 m apart.
(i)
Determine the speed of the car at which the vertical oscillations of the body are largest.
(2)
(ii)
Outline why the suspension is fitted with dampers (shock absorbers).
(1)
(c)
At one speed the body of the loaded car oscillates vertically with simple harmonic motion of amplitude 6.0 cm at the frequency found in (a)(ii).
(i)
Calculate the maximum acceleration of the body.
(1)
(ii)
A passenger of mass 80 kg moves with the body. Determine the maximum and the minimum force exerted on the passenger by the seat, and state where in the oscillation each occurs.
(3)
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Mark
Notes
Part (a)(i)
k = ΔF/Δx = 320 × 9.81/0.040 = 7.85 × 104 N m−1
✓ 1
Full substitution or an answer to at least 3 s.f. required.
Part (a)(ii)
Oscillating mass = 1100 + 320 = 1420 kg
✓ 1
Using 1100 kg (1.34 Hz): [1 max].
f = (1/2π)√(7.85 × 104/1420) = 1.18 Hz
✓ 1
Allow ECF from (a)(i).
Part (b)(i)
The largest oscillations occur at resonance, when the time between bumps equals the natural period: T = 1/1.18 = 0.845 s
✓ 1
Allow ECF from (a)(ii).
v = 9.0/0.845 = 10.6 m s−1
✓ 1
Accept 10.6–10.7 m s−1 (≈ 38 km h−1).
Part (b)(ii)
Damping removes energy from the oscillation, so the amplitude at resonance is much smaller and the oscillation after each bump dies away quickly
✓ 1
Reference to energy removal or reduced amplitude needed.
Part (c)(i)
amax = (2π × 1.18)² × 0.060 = 3.32 m s−2
✓ 1
Allow ECF from (a)(ii).
Part (c)(ii)
At the lowest point the acceleration is upwards: N − mg = mamax, N = 80(9.81 + 3.32) = 1.05 × 103 N
✓ 1
Allow ECF from (c)(i).
At the highest point the acceleration is downwards: N = 80(9.81 − 3.32) = 520 N
✓ 1
Allow ECF from (c)(i).
Maximum at the lowest point, minimum at the highest point
✓ 1
Both positions needed; this mark can be awarded without the calculations.
Answers: (a)(i) 7.85 × 104 N m−1 · (a)(ii) 1.18 Hz · (b)(i) 10.6 m s−1 · (c)(i) 3.32 m s−2 · (c)(ii) 1050 N (lowest point); 520 N (highest point) (the remaining parts are explanations — see the table above)
Syllabus understandingC.1 — the time period of a mass–spring system as given by T = 2π√(m/k); C.1 — the defining equation of simple harmonic motion as given by a = −ω²x; C.1 — the time period in terms of frequency of oscillation and angular frequency as given by T = 1/f = 2π/ω; A.2 — Hooke's law and Newton's second law; A.1 — speed; C.4 — resonance and damping Command term: Determine
Short answer & extended response8 steps to full marksDetermine
A tall building sways in strong winds with a natural period of 6.80 s. To reduce the sway, a steel mass of 6.0 × 105 kg is hung from long cables near the top of the building to form a simple pendulum, and hydraulic dampers connect the mass to the building. The pendulum is designed to have the same natural period as the building.
(a)
(i)
Show that the length of the pendulum is about 11 m.
(2)
(ii)
Explain why the pendulum reduces the sway of the building most effectively when its natural period equals that of the building.
(3)
(b)
During a storm the steel mass oscillates with amplitude 0.80 m.
(i)
Calculate the maximum speed of the steel mass.
(1)
(ii)
Show that the maximum resultant force on the steel mass is mgx0/l, where m is the mass and l the length of the pendulum. Ignore the force exerted by the dampers.
(2)
(iii)
Calculate this force.
(1)
(c)
(i)
Calculate the total energy of the oscillation of the steel mass.
(1)
(ii)
In one cycle the hydraulic dampers dissipate 25 % of this energy. Determine the amplitude of the steel mass at the end of the cycle, assuming no further energy is supplied.
(2)
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Mark
Notes
Part (a)(i)
l = gT²/4π²
✓ 1
l = 9.81 × 6.80²/4π² = 11.5 m
✓ 1
Answer to at least 3 s.f. or full substitution required.
Part (a)(ii)
The swaying building drives the pendulum at the building's frequency, which then equals the natural frequency of the pendulum: resonance
✓ 1
So the pendulum oscillates with a large amplitude
✓ 1
Large amounts of energy are transferred from the building to the pendulum and dissipated in the hydraulic dampers, so the amplitude of the building's sway decreases
✓ 1
Energy transfer away from the building is needed for this mark.
Part (b)(i)
vmax = (2π/6.80) × 0.80 = 0.74 m s−1
✓ 1
Allow ECF from (a)(i) if ω = √(g/l) is used.
Part (b)(ii)
The maximum acceleration is ω²x0, at the extremes of the motion
✓ 1
For a pendulum ω² = g/l, so F = mω²x0 = mgx0/l
✓ 1
Symbolic answer required.
Part (b)(iii)
F = 6.0 × 105 × 9.81 × 0.80/11.5 = 4.1 × 105 N
✓ 1
Allow ECF from (a)(i).
Part (c)(i)
ET = ½mvmax² = ½ × 6.0 × 105 × 0.739² = 1.6 × 105 J
✓ 1
Or ½mω²x0². Allow ECF from (b)(i).
Part (c)(ii)
ET ∝ x0², so the amplitude falls by the factor √0.75
✓ 1
Allow ECF from (c)(i).
x0 = 0.80 × √0.75 = 0.69 m
✓ 1
Taking the amplitude as falling by 25 % (0.60 m): [0] for this mark.
Answers: (a)(i) 11.5 m · (b)(i) 0.74 m s−1 · (b)(iii) 4.1 × 105 N · (c)(i) 1.6 × 105 J · (c)(ii) 0.69 m (the remaining parts are explanations — see the table above)
Syllabus understandingC.1 — the time period of a simple pendulum as given by T = 2π√(l/g); C.1 — the defining equation of simple harmonic motion as given by a = −ω²x; C.1 (HL) — that problems can be solved using the equations for simple harmonic motion as given by ET = ½mω²x0²; C.4 — resonance and damping; A.2 — Newton's second law Command term: Determine
51C-2-44
Resonant mass sensor·C.1 Simple harmonic motion
Paper 2Hard14 marks
Short answer & extended response8 steps to full marksDetermine
A biosensor uses a microscopic cantilever whose tip vibrates perpendicular to its length. The vibration is modelled as a mass m on a spring of spring constant k = 0.80 N m−1. The natural frequency of the vibration is f = 150 kHz. When protein molecules stick to the tip, the vibrating mass increases and the natural frequency falls; the spring constant is unchanged.
(a)
(i)
Show that m is about 9 × 10−13 kg.
(2)
(ii)
The tip vibrates with amplitude 15 nm. Determine the maximum acceleration of the tip, as a multiple of g.
(2)
(b)
(i)
Show that, when a small mass Δm is added, the fractional decrease in the natural frequency is Δf/f ≈ Δm/2m.
(2)
(ii)
After exposure to a protein solution, the natural frequency is 18 Hz lower. Determine the mass of protein on the tip.
(2)
(iii)
The molar mass of the protein is 66 kg mol−1. Estimate the number of protein molecules on the tip.
(3)
(c)
(i)
Calculate the total energy of the vibration of the tip when its amplitude is 15 nm.
(1)
(ii)
The sensor works best in a vacuum. Explain why air surrounding the cantilever makes the 18 Hz shift harder to detect.
For a small change the fractional change in a power is the power times the fractional change: Δf/f = ½ × Δm/m (in magnitude), with f decreasing as m increases
✓ 1
ALT: (f − Δf)/f = (1 + Δm/m)−½ ≈ 1 − Δm/2m.
Part (b)(ii)
Δm = 2m Δf/f = 2 × 9.0 × 10−13 × 18/1.50 × 105
✓ 1
Allow ECF from (a)(i) and (b)(i). ALT: k/(2π × 149 982)² − m.
Δm = 2.2 × 10−16 kg
✓ 1
Accept 2.1–2.2 × 10−16 kg. Using Δf/f = Δm/m (1.1 × 10−16 kg): [1 max].
Part (b)(iii)
Mass of one molecule = 66/(6.02 × 1023) = 1.1 × 10−22 kg
✓ 1
Treating the molar mass as 66 g mol−1 (0.066 kg mol−1) gives about 2 × 109 molecules: [0] for this mark.
N = Δm/(mass of one molecule)
✓ 1
Allow ECF from (b)(ii).
N = 2.2 × 10−16/1.1 × 10−22 = 2 × 106
✓ 1
Accept 1.9–2.0 × 106.
Part (c)(i)
ET = ½kx0² = ½ × 0.80 × (1.5 × 10−8)² = 9.0 × 10−17 J
✓ 1
Accept ½mω²x0² with their values.
Part (c)(ii)
Air resistance damps the vibration: the resonance peak (amplitude against driving frequency) becomes lower and broader
✓ 1
The frequency of the peak is then less sharply defined, so a shift of only 18 Hz in 150 kHz (about 1 part in 104) becomes difficult to resolve
✓ 1
Reference to the small fractional shift or to the broad peak hiding the shift needed.
Answers: (a)(i) 9.0 × 10−13 kg · (a)(ii) 1.4 × 103g · (b)(ii) 2.2 × 10−16 kg · (b)(iii) 2 × 106 · (c)(i) 9.0 × 10−17 J (the remaining parts are explanations — see the table above)
Syllabus understandingC.1 — the time period of a mass–spring system as given by T = 2π√(m/k); C.1 — the defining equation of simple harmonic motion as given by a = −ω²x; C.1 (HL) — that problems can be solved using the equations for simple harmonic motion as given by ET = ½mω²x0²; B.3 — the mole, molar mass and the Avogadro constant; C.4 — damping and resonance; Tools — fractional changes for a power law Command term: Determine
52C-2-66
Mass–spring and pendulum periods·C.1 Simple harmonic motion
Paper 2Hard19 marks
Short answer & extended response12 steps to full marksDetermine
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)
Show mark scheme
Marking point
Mark
Notes
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)
Syllabus understandingC.1 — the time period of a mass–spring system T = 2π√(m/k); C.1 (HL) — v = ωx0 cos(ωt + φ), ET = ½mω²x0²; C.4 — the effect of light damping; D.1 — gravitational field strength and orbital motion; A.2 — free fall and conservation of momentum; Tools — fundamental SI units and estimation Command term: Determine
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