79 free, high-quality past-paper-style questions on Simple harmonic motion for IB Physics SL, each with a full mark scheme. Covers: SHM: period & frequency; Energy in SHM; The defining equation a = −ω²x; and more. For when you have finished the past papers, or want to practise one topic at a time.C.1 at Higher Level →Read the C.1 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.411 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1C-1A-01SHM: period & frequencyPaper 1AEasy1 mark
An object oscillating with simple harmonic motion completes one full oscillation in 0.40 s.
What is its frequency?
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Step 1f = 1/T = 1/0.40 = 2.5 Hz.
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Answer C
Answer: C · 1 stage of work, one mark
Every option, and why
AThis is T/2π: the period has been treated as if it were ω and converted with f = ω/2π.
BThis is the period itself. The frequency is the reciprocal of the period, not the period.
CCorrect: f = 1/T = 1/0.40 = 2.5 Hz.
DThis is the angular frequency ω = 2π/T in rad s⁻¹, not the frequency in Hz.
The bob of a simple pendulum is released from rest at one end of its swing and moves to the other end. Air resistance is negligible.
Which row describes the changes in the kinetic energy and in the gravitational potential energy of the bob during this motion?
Kinetic energyGravitational potential energy
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Step 1The bob is momentarily at rest at the highest points (each end of the swing) and moves fastest at the lowest point, so the kinetic energy rises then falls while the gravitational potential energy falls then rises.
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Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: kinetic energy is greatest at the lowest point and gravitational potential energy is greatest at the ends, with a constant sum.
BBoth energies cannot peak together. With no air resistance their sum is constant, so one rises while the other falls.
CBoth energies cannot fall together either: the bob speeds up as it descends, because gravitational potential energy is converted to kinetic energy.
DThe two positions are the wrong way round: at the ends of the swing the bob is at rest, so its kinetic energy is zero there, not a maximum.
A small mass on a horizontal spring oscillates with simple harmonic motion of amplitude 3.0 cm and period 0.40 s.
Which statement about the motion is correct?
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Step 1In one period the mass moves from the equilibrium position to one extreme, back through equilibrium to the other extreme, and back again: four amplitudes, 4 × 3.0 cm = 12 cm.
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Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: one complete oscillation covers four amplitudes, so the distance travelled in one period is 4 × 3.0 cm = 12 cm.
BAfter one whole period the mass is back where it started, so its displacement has not changed at all; 6.0 cm is the distance between the two extremes.
CThe mass is at rest at the two extremes, which are 2 × 3.0 = 6.0 cm apart. The amplitude is measured from the equilibrium position, not from one extreme to the other.
D2.5 is the frequency in Hz, 1/T. The angular frequency is ω = 2π/T = 15.7 rad s−1.
4C-1A-61The defining equation a = −ω²xPaper 1AEasy1 mark
The defining equation of simple harmonic motion is a = −ω2x, where x is the displacement from the equilibrium position.
What is the significance of the minus sign in this equation?
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Step 1ω2 is always positive, so the minus sign means that a and x always have opposite signs: when x is positive a is negative and vice versa, so the acceleration always points back towards the equilibrium position.
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Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: ω2 > 0, so a and x always have opposite signs and the acceleration is always directed towards the equilibrium position.
BThis confuses displacement with velocity. While the oscillator moves towards equilibrium its velocity and acceleration are in the same direction, and it speeds up.
CThe minus sign does not describe damping. In a = −ω2x the amplitude, and so the largest acceleration, stays the same in every cycle.
DThe direction of a is not fixed: for negative x, −ω2x is positive. Only its direction relative to the displacement is fixed.
5C-1A-65The defining equation a = −ω²xPaper 1AEasy1 mark
The needle of a sewing machine moves up and down with simple harmonic motion. At one instant the displacement of the needle from its equilibrium position is +3.5 mm and its acceleration is −5.6 m s−2.
What is the angular frequency of the oscillation?
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Step 1a = −ω2x, so ω = √(−a/x) = √(5.6/(3.5 × 10−3)) = √1600 = 40 rad s−1.
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Answer C
Answer: C · 1 stage of work, one mark
Every option, and why
AThe displacement has been left in millimetres: √(5.6/3.5) = 1.26. x must be in metres, 3.5 × 10−3 m.
BThis is ω/2π = 6.37, the frequency f in Hz. The angular frequency itself is ω = 40 rad s−1.
CCorrect: ω2 = 5.6/(3.5 × 10−3) = 1600 s−2, so ω = 40 rad s−1.
DThis is ω2 = 1600 s−2: the square root has not been taken.
6IBP2S-018-00-01The mass–spring oscillatorPaper 1AEasy1 mark
A sensitive optical instrument rests on a set of springs and oscillates vertically with period T. It is replaced by an instrument with four times the mass, on the same springs. What is the new period?
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For a mass–spring system \(T = 2\pi\sqrt{m/k}\), so T ∝ √m. Multiplying the mass by 4 multiplies the period by √4 = 2, giving 2T. Choosing 4T treats T as proportional to m. Choosing T confuses this system with a pendulum, whose period does not depend on mass. Choosing T/2 inverts the relationship.
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Answer C
Answer: C
Common error: The period depends on the square root of the mass. Only a pendulum's period is independent of the mass.
7IBP2S-018-00-02Maximum speed & accelerationPaper 1AEasy1 mark
A truck driver's seat is mounted on a spring and oscillates vertically in simple harmonic motion. Which statement describes the seat at the instant it reaches its maximum displacement?
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In SHM a = −ω2x. At maximum displacement the magnitude of the acceleration is greatest, and the minus sign means it points back towards equilibrium. At this turning point the seat is momentarily at rest, so its speed is zero.
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Answer B
Answer: B
Exam tip: The minus sign in a = −ω²x tells you the acceleration always points towards the equilibrium position, opposite to the displacement.
8IBP2S-018-00-06SHM: period & frequencyPaper 1AEasy1 mark
The needle of a sewing machine moves up and down in approximately simple harmonic motion. It completes 90 oscillations in 6.0 s. What is its angular frequency?
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The frequency is f = 90/6.0 = 15 Hz, so ω = 2πf = 30π ≈ 94 rad s−1. The option 0.067 is the period in seconds. The option 15 is the ordinary frequency f. The option 47 uses πf instead of 2πf.
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Answer D
Answer: D
Exam tip: Angular frequency ω = 2πf is measured in rad s⁻¹ and is always about six times larger than f.
9IBP2S-018-02-01The simple pendulumPaper 1AEasy1 mark
A pendulum clock keeps correct time at 15 °C. Its pendulum rod is made of brass, which expands when heated. The clock is moved to a room at 30 °C. What happens to the clock?
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The period of a simple pendulum is T = 2π√(L/g). When the rod expands, L increases, so T increases. Each swing takes longer, so the clock counts fewer swings per real hour and runs slow. The period does not depend on the mass of the bob or on a small amplitude, but it does depend on length.
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Answer B
Answer: B
Common error: Mass and small amplitude do not affect a pendulum's period, but length does. A longer period means fewer ticks per hour, so the clock loses time.
A child on a playground swing behaves as a simple pendulum with a period of 2.0 s. What is the period of the variation of the child's kinetic energy?
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The kinetic energy is a maximum each time the child passes through the lowest point, which happens twice per oscillation. It is zero at each extreme, which also happens twice per oscillation. The kinetic energy therefore repeats every half period, which is 1.0 s. The answer 2.0 s confuses the energy cycle with the motion cycle.
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Answer B
Answer: B
Common error: Energy cannot be negative, so it repeats every half cycle. The energy oscillates at twice the frequency of the motion.
11IBP2S-018-02-06The simple pendulumPaper 1AEasy1 mark
Why is the equation T = 2π√(L/g) for a simple pendulum valid only for small angular amplitudes?
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The restoring force on the bob is mg sin θ. SHM requires the acceleration to be proportional to the displacement, and the displacement is proportional to θ. Only when θ is small (in radians) is sin θ ≈ θ, which gives a = −(g/L)x. The tension varies even at small amplitudes. Air resistance and mass are not the reason the formula fails.
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Answer C
Answer: C
Exam tip: Link the small-angle condition to sin θ ≈ θ (in radians). This approximation is what makes the restoring force proportional to displacement.
12IBP2S-018-01-01The defining equation a = −ω²xPaper 1AEasy1 mark
A data logger records the motion of a vibrating sensor. Four possible relationships between the acceleration a of the sensor (in m s−2) and its displacement x from equilibrium (in m) are given. Which relationship shows that the sensor performs simple harmonic motion?
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SHM requires the acceleration to be proportional to the displacement and directed opposite to it, a = −ω²x. Only a = −9x has this form, with ω² = 9 s−2. a = −9x² is not linear in x and does not reverse sign when x does. a = 9x points away from equilibrium, so the motion would run away rather than oscillate. a = −9 is a constant acceleration that does not depend on x.
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Answer A
Answer: A
Exam tip: Check two things: the acceleration must be linear in x, and it must carry a minus sign. Both are needed for SHM.
13IBP2S-018-01-02Conditions for SHMPaper 1AEasy1 mark
Which statement gives the conditions for an object to perform simple harmonic motion?
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The defining condition of SHM is a ∝ −x. The acceleration (and so the resultant force) is proportional to the displacement from a fixed equilibrium point and always points back towards that point. Acceleration proportional to velocity describes a damping force, not SHM. Acceleration directed away from equilibrium gives unstable motion. A constant-magnitude acceleration describes motion such as a bouncing ball, which is periodic but not simple harmonic.
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Answer C
Answer: C
Common error: Stating only 'acceleration proportional to displacement' is incomplete. The direction (towards equilibrium, opposite to displacement) is part of the condition.
14IBP2S-018-01-10SHM: period & frequencyPaper 1AEasy1 mark
The cantilever tip of an atomic force microscope performs simple harmonic motion. When its displacement is +3.0 nm, its acceleration is −12 m s−2. What is its acceleration when its displacement is −1.5 nm?
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In SHM, a = −ω²x, so acceleration is proportional to displacement and opposite in direction. ω² = 12 / (3.0×10−9) = 4.0×109 s−2. At x = −1.5×10−9 m: a = −4.0×109 × (−1.5×10−9) = +6.0 m s−2. Halving the displacement halves the acceleration. Reversing the displacement reverses the acceleration. Option −6.0 keeps the original sign and ignores the reversal of displacement. Options ±24 treat acceleration as inversely proportional to displacement.
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Answer A
Answer: A
Exam tip: In SHM, acceleration and displacement always have opposite signs. Use this as a quick check on every answer.
15C-1A-02The defining equation a = −ω²xPaper 1AMedium1 mark
A body oscillates with simple harmonic motion about its equilibrium position. Its displacement from the equilibrium position is x and its acceleration is a.
Which graph shows the variation of a with x during one complete oscillation?
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Step 1a = −ω²x, so a is directly proportional to x: the graph is a straight line through the origin.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The minus sign means a always points opposite to x (towards equilibrium), so the gradient, −ω², is negative: graph A.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: a straight line through the origin with negative gradient −ω², as a = −ω²x requires.
BThe minus sign has been ignored: an acceleration in the same direction as the displacement would drive the body further from equilibrium.
CThis acceleration is always directed towards equilibrium but has a constant size. SHM also needs a to be proportional to x.
DThis confuses acceleration with speed: the speed is greatest at equilibrium, but the acceleration there is zero and is greatest at the extremes.
20C-1A-47The defining equation a = −ω²xPaper 1AMedium1 mark
The first graph shows how the displacement x of a body moving with simple harmonic motion varies with time t. The body is released from rest at x = x0 at t = 0.
Which graph shows how the acceleration a of the body varies with t over the same time?
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Step 1For SHM, a = −ω2x: at every instant the acceleration is proportional to the displacement and opposite in sign.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2So the a–t graph has the same shape as the x–t graph, inverted: a is most negative at t = 0 and T, and most positive at T/2 — graph B.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThe minus sign in a = −ω2x has been ignored. An acceleration in the same direction as the displacement would drive the body further from equilibrium.
BCorrect: a = −ω2x, so a is greatest in the negative direction when x = +x0 and zero whenever x = 0.
CThis is the velocity–time graph: the velocity is zero at the extremes and the body first moves in the negative direction. The acceleration is zero where the velocity is greatest.
DThis acceleration is always towards equilibrium but constant in size. In SHM the size of a is proportional to the displacement, so it falls to zero at x = 0.
21C-1A-60SHM: period & frequencyPaper 1AMedium1 mark
The head of an electric toothbrush vibrates with simple harmonic motion. The graph shows how the displacement x of the tip of the head from its equilibrium position varies with time t.
Which row gives the amplitude and the angular frequency of the oscillation?
Graph drawn to scale
AmplitudeAngular frequency
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Step 1The amplitude is the maximum displacement from the equilibrium position (x = 0): 1.5 mm.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Successive crests are at 1.0 ms and 5.0 ms (and 9.0 ms), so T = 4.0 ms.
A block attached to a horizontal spring oscillates with simple harmonic motion of frequency 2.0 Hz on a frictionless surface.
Which row describes how the kinetic energy of the block and the total energy of the block–spring system vary with time?
Kinetic energy of the blockTotal energy of the block–spring system
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Step 1The kinetic energy depends on the speed, not on the direction of motion. It is greatest each time the block passes through the equilibrium position and zero at each extreme, so it goes through two complete cycles in every oscillation: its frequency is 2 × 2.0 Hz = 4.0 Hz.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2With no friction no energy leaves the system: energy is transferred back and forth between kinetic energy and elastic potential energy, so the total energy is constant.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis treats the kinetic energy as if it varied like the velocity. The velocity goes through one cycle per oscillation, but the kinetic energy is greatest at both passes through the equilibrium position, in either direction, so it varies at twice the frequency.
BCorrect: the kinetic energy is greatest twice and zero twice in every oscillation, so it varies at 4.0 Hz, while the total energy stays constant.
CThe kinetic energy is right, but with no friction no energy leaves the system. The elastic potential energy rises as the kinetic energy falls, so their sum is constant.
DConservation of energy applies to the total energy, not to the kinetic energy alone. The block is momentarily at rest at each extreme, so its kinetic energy falls to zero twice in every oscillation.
24IBP2S-018-00-03The mass–spring oscillatorPaper 1AMedium1 mark
A prototype energy-harvesting floor tile of mass m is supported by two identical springs, each of spring constant k. In design X the springs are connected in series. In design Y they are connected side by side (in parallel). What is \(\frac{T_X}{T_Y}\), the ratio of the period of vertical oscillation in design X to that in design Y?
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In series the effective spring constant is k/2. In parallel it is 2k. Since T ∝ 1/√k, \(\frac{T_X}{T_Y} = \sqrt{\frac{2k}{k/2}} = \sqrt{4} = 2\). The series arrangement is softer, so it oscillates more slowly. Choosing 4 forgets the square root. Choosing 1/2 swaps the effective stiffnesses.
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Answer A
Answer: A
Common error: Springs in series are less stiff than one spring alone, and springs in parallel are stiffer. Find the effective k first, then take the square root.
25IBP2S-018-00-04Maximum speed & accelerationPaper 1AMedium1 mark
The proof mass in a MEMS gyroscope chip is driven in simple harmonic motion. The engineers double the amplitude and halve the frequency. What happens to the maximum acceleration of the proof mass?
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The maximum acceleration is amax = ω2x0. Halving the frequency halves ω, which multiplies ω2 by 1/4. Doubling x0 multiplies by 2. Overall the factor is 2 × 1/4 = 1/2. 'Unchanged' would be correct for the maximum speed ωx0, not the acceleration. 'Doubled' and 'quartered' each take account of only one of the two changes.
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Answer B
Answer: B
Exam tip: Maximum speed scales with ωx₀ but maximum acceleration scales with ω²x₀. Check which one the question asks for.
26IBP2S-018-00-05The simple pendulumPaper 1AMedium1 mark
A robotic lander carries a simple pendulum and a separate mass–spring oscillator from Earth to the surface of Mars, where the gravitational field strength is smaller. Both are set oscillating with small amplitude. Compared with their periods on Earth, which describes the periods on Mars?
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For a pendulum \(T = 2\pi\sqrt{l/g}\), so a smaller g gives a longer period. For a mass–spring system \(T = 2\pi\sqrt{m/k}\), which contains neither g nor weight. Gravity only shifts the equilibrium position of the spring, so its period is unchanged. Only the pendulum's period increases.
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Answer C
Answer: C
Common error: A vertical spring stretches less in weaker gravity, but its period does not depend on g. Only its equilibrium position moves.
27IBP2S-018-00-09SHM: period & frequencyPaper 1AMedium1 mark
A baby is placed in a bouncer seat of negligible mass that hangs from a vertical spring. The spring extends by 9.8 cm as the seat settles at its new equilibrium position. What is the period of small vertical oscillations of the baby? (g = 9.8 m s−2)
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At equilibrium kΔx = mg, so m/k = Δx/g. Then \(T = 2\pi\sqrt{\Delta x/g} = 2\pi\sqrt{0.098/9.8} = 2\pi \times 0.10 \approx 0.63\) s, so the answer is 0.63 s. The option 6.3 s comes from using 9.8 cm without converting it to metres. The option 0.10 s omits the factor 2π. The option 0.063 s omits the square root.
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Answer B
Answer: B
Exam tip: A static extension gives you m/k without needing m or k separately. Convert the extension to metres before substituting.
28IBP2S-018-02-02The mass–spring oscillatorPaper 1AMedium1 mark
A simple pendulum has period T when it hangs in a stationary lift. The lift then accelerates upwards at a constant rate of g/3. What is the new period of small oscillations?
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In a lift accelerating upwards, the effective gravitational field strength is g + g/3 = 4g/3. Since T = 2π√(L/geff), the new period is T√(3/4) = 0.87T. The answer 0.75T forgets the square root. The answer 1.15T uses a reduced effective g. The answer T wrongly assumes the period is fixed by length alone.
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Answer B
Answer: B
Exam tip: In an accelerating frame, replace g with the effective field. Upward acceleration increases it and downward acceleration decreases it.
29IBP2S-018-02-03Energy in SHMPaper 1AMedium1 mark
A body oscillates with simple harmonic motion of amplitude x0. What fraction of the total energy is kinetic energy when the displacement is x0/2?
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Potential energy is proportional to x². At x = x0/2 the potential energy is (1/2)² = 1/4 of the total energy. The kinetic energy is therefore 1 − 1/4 = 3/4 of the total. The answer 1/2 assumes energy varies linearly with displacement. The answer 1/4 is the potential energy fraction. The answer √3/2 is the fraction of the maximum speed, not of the energy.
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Answer C
Answer: C
Common error: Halfway out in displacement is not halfway in energy. Energy depends on x², so square the fraction first.
30IBP2S-018-02-04Maximum speed & accelerationPaper 1AMedium1 mark
A vibration sensor contains a small mass on a light spring that performs simple harmonic motion. The amplitude of its oscillation is doubled. Which row gives the effect on the period, the maximum speed and the total energy?
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For SHM the period is independent of amplitude. The maximum speed is ωx0, so it doubles when the amplitude doubles. The total energy is proportional to x0², so it increases by a factor of 4. This is consistent with E = ½mvmax², since doubling vmax also gives 4 times the energy.
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Answer A
Answer: A
Exam tip: Isochronous means the period does not depend on amplitude. Energy scales with amplitude squared, and maximum speed scales with amplitude.
31IBP2S-018-02-08Energy in SHMPaper 1AMedium1 mark
Two simple pendulums have the same length. Bob P has mass 0.10 kg and bob Q has mass 0.40 kg. Both are released from rest at the same small angle. Which row gives the ratio of maximum speeds (Q : P) and the ratio of total energies (Q : P)?
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The maximum speed is v = √(2gh), and h is the same for both pendulums because the length and angle are the same. The mass cancels, so the speeds are equal. The total energy is mgh, so it is proportional to mass and the ratio is 4 : 1. The answer 1 : 1 for energy assumes equal energy. The answer 1 : 2 for speed assumes equal energy shared between different masses.
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Answer A
Answer: A
Common error: Mass cancels in the speed calculation but not in the energy calculation. Check whether the mass appears on only one side of your equation before cancelling it.
32IBP2S-018-02-09The simple pendulumPaper 1AMedium1 mark
A pendulum on a Mars lander has a period of 2.0 s when tested on Earth (g = 9.8 m s−2). The gravitational field strength at the surface of Mars is 3.7 N kg−1. What is its period on Mars?
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Since T ∝ 1/√g, TMars = 2.0 × √(9.8/3.7) = 2.0 × 1.63 = 3.3 s. The answer 5.3 s omits the square root. The answer 1.2 s inverts the ratio. The answer 0.75 s inverts the ratio and also omits the square root.
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Answer C
Answer: C
Exam tip: Check the direction of the change first. Weaker gravity gives a smaller restoring force, so the swing is slower and the period is longer.
33IBP2S-018-01-03SHM: period & frequencyPaper 1AMedium1 mark
The proof mass inside a smartphone accelerometer chip oscillates with acceleration a = −(4.0 × 106 s−2)x, where x is its displacement from equilibrium. What is the approximate frequency of the oscillation?
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Comparing with a = −ω²x gives ω² = 4.0 × 106 s−2, so ω = 2.0 × 103 rad s−1. Then f = ω/2π = 2000/6.28 ≈ 320 Hz. The value 2000 Hz takes ω to be the frequency. The value 1.3 × 104 Hz multiplies ω by 2π instead of dividing. The value 6.4 × 105 Hz divides ω² by 2π without taking the square root first.
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Answer A
Answer: A
Exam tip: The constant in a = −(constant)x is ω², not ω. Take the square root first, then divide by 2π to get the frequency.
34IBP2S-018-01-04The defining equation a = −ω²xPaper 1AMedium1 mark
An accelerometer is fixed to the tip of a diving board after a diver leaves it. A graph of the tip's acceleration against its displacement from equilibrium is a straight line through the origin with gradient −16π2 s−2. What is the period of oscillation of the board?
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The gradient of an a–x graph for SHM equals −ω², so ω² = 16π² s−2 and ω = 4π rad s−1. Then T = 2π/ω = 2π/4π = 0.50 s. The value 2.0 s is the frequency in Hz, not the period. The value 0.25 s comes from T = π/ω, which drops the factor of 2. The value 0.040 s comes from using ω² in place of ω in T = 2π/ω.
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Answer C
Answer: C
Common error: The gradient of the a–x graph is −ω², not −ω. Square-root its magnitude before using T = 2π/ω.
35IBP2S-018-01-05Conditions for SHMPaper 1AMedium1 mark
Which of the following motions is periodic but is not simple harmonic?
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Between bounces the ball has a constant downward acceleration g, whatever its height. Its acceleration is therefore not proportional to its displacement from any equilibrium point, so the motion repeats but is not SHM. The other three systems all have a restoring force proportional to displacement: two springs give F = −2kx; a pendulum at small angles gives a ≈ −(g/L)x; and the liquid column gives a = −(2g/L)x.
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Answer B
Answer: B
Exam tip: A motion that repeats is not automatically SHM. Ask whether the acceleration is proportional to displacement and directed towards a fixed point.
36IBP2S-018-01-06Maximum speed & accelerationPaper 1AMedium1 mark
A rider on a fairground ride is carried up and down in simple harmonic motion. At the instant the rider moves with maximum speed, the rider's acceleration is
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In SHM the speed is greatest as the object passes through the equilibrium position, where x = 0. Since a = −ω²x, the acceleration there is zero. The acceleration is greatest at the extremes of the motion, where the speed is zero. The option 'half of its maximum value' wrongly assumes acceleration and speed share their peak values.
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Answer A
Answer: A
Common error: Maximum speed does not mean maximum acceleration. In SHM the acceleration depends on displacement, not on speed.
37IBP2S-018-01-08The defining equation a = −ω²xPaper 1AMedium1 mark
A hydrometer is a uniform floating tube used to test liquids. When pushed down slightly and released, it bobs vertically in simple harmonic motion. The extra buoyancy force when it is pushed down a distance y is ρgSy, where ρ is the liquid density and S is the tube's cross-sectional area. The same hydrometer is moved into a liquid of greater density. What happens to the period of its vertical oscillations?
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The resultant force is F = −ρgSy, so a = −(ρgS/m)y. This gives ω² = ρgS/m. A larger ρ increases ω, so T = 2π/ω decreases. Floating higher does not change the stiffness of the restoring force. The period is independent of amplitude, but it does depend on ρ. The mass is unchanged, but ω also depends on ρ.
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Answer B
Answer: B
Exam tip: For any new system, write the resultant force as −(constant)×displacement, divide by mass to get ω², and then judge how T changes.
On the Earth, a mass hanging from a spring and a simple pendulum both oscillate with period T. Both are taken to a planet where the gravitational field strength is one quarter of that on the Earth.
Which row gives the periods on the planet?
Mass–spring systemSimple pendulum
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Step 1T = 2π√(m/k) contains no g: a weaker field only moves the equilibrium position of the hanging mass, so its period stays T.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2T = 2π√(l/g) ∝ 1/√g: dividing g by 4 multiplies the pendulum's period by √4 = 2, giving 2T.
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThe pendulum is right, but the mass–spring period does not depend on g; the field only changes the extension at equilibrium.
BThe mass–spring period is right, but the pendulum's dependence has been inverted: T ∝ 1/√g, so a weaker field gives a longer period.
CBoth errors: the mass–spring period is wrongly made to depend on g, and the pendulum's dependence on g is inverted.
DCorrect: the mass–spring period is independent of g, and the pendulum period doubles because T ∝ 1/√g.
39C-1A-48The mass–spring oscillatorPaper 1AHard1 mark
A small mass hangs from a light spring and oscillates vertically with period T. The spring is then cut into two equal halves, and the same mass is hung from one of the halves and set oscillating.
What is (new period)/T?
Show mark scheme
Marking point
Mark
Notes
Step 1With the same tension, each half stretches by half as much as the whole spring did, so the spring constant of one half is 2k.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2T = 2π√(m/k), so T ∝ 1/√k: doubling k multiplies the period by 1/√2.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis assumes that half a spring has half the spring constant and also leaves out the square root (T ∝ 1/k). A shorter spring is stiffer, not weaker.
BThe square root is right, but the spring constant has been halved. Each half carries the full tension with half the extension, so its spring constant is doubled.
CCorrect: the half spring has spring constant 2k, and T ∝ 1/√k gives a period of T/√2.
DThe spring constant of the half spring is right (2k), but the square root in T = 2π√(m/k) has been left out.
A mass hangs from a vertical spring and oscillates vertically with simple harmonic motion. The spring remains stretched throughout the motion. Air resistance is negligible.
Where is the elastic potential energy of the spring least, and where is the kinetic energy of the mass greatest?
Elastic potential energy leastKinetic energy greatest
Show mark scheme
Marking point
Mark
Notes
Step 1The elastic potential energy depends on the extension of the spring. The spring is still stretched at the highest point, but its extension is smallest there, so the elastic potential energy is least at the highest point.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2The mass moves with SHM about the equilibrium position, so its speed, and its kinetic energy, are greatest as it passes through the equilibrium position.
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThe kinetic energy is right, but the elastic potential energy is not least at equilibrium: there the spring is stretched enough to support the weight. That is true only for a horizontal oscillator.
BCorrect: the extension, and so the elastic potential energy, is smallest at the highest point; the speed is greatest at the equilibrium position.
CThis confuses elastic with gravitational potential energy. At the lowest point the gravitational potential energy is least, but the extension and the elastic potential energy are greatest.
DThis assumes that kinetic plus elastic potential energy is constant. The gravitational potential energy also changes, and the mass is momentarily at rest at the highest point.
41IBP2S-018-00-10The defining equation a = −ω²xPaper 1AHard1 mark
A laboratory sieve shaker moves a tray of sand vertically in simple harmonic motion at a frequency of 5.0 Hz. What is the largest amplitude for which the sand stays in contact with the tray throughout the motion? (g = 9.8 m s−2)
Show mark scheme
Marking point
Mark
Notes
At the top of the motion the tray accelerates downwards at ω2x0. The tray can push the sand but cannot pull it down, so the sand can only follow if this acceleration is no more than g. So x0 = g/ω2 = 9.8/(2π × 5.0)2 = 9.9 × 10−3 m ≈ 9.9 mm. The option 0.39 m uses f2 instead of ω2. The option 0.31 m uses g/ω. The option 50 mm uses 4π2f instead of 4π2f2.
✓ 1
Answer A
Answer: A
Exam tip: Contact is lost at the highest point if the required downward acceleration ω²x₀ exceeds g.
42IBP2S-018-01-07Maximum speed & accelerationPaper 1AHard1 mark
Two sections of a footbridge, P and Q, oscillate in simple harmonic motion. P has amplitude A and period T. Q has amplitude 2A and period 2T. What is the ratio \(\frac{\text{maximum acceleration of P}}{\text{maximum acceleration of Q}}\)?
Show mark scheme
Marking point
Mark
Notes
Maximum acceleration = ω²A = 4π²A/T². For P: 4π²A/T². For Q: 4π²(2A)/(2T)² = 2π²A/T². The ratio P : Q = 4/2 = 2. The answer 1/2 considers only the amplitudes. The answer 1 assumes the doubled amplitude and doubled period cancel, ignoring that amax depends on the square of the period. The answer 1/8 treats amax as proportional to AT² instead of A/T².
✓ 1
Answer D
Answer: D
Exam tip: Write amax = 4π²A/T² and substitute ratios directly. The period appears squared and in the denominator.
43IBP2S-018-01-09The simple pendulumPaper 1AHard1 mark
An imaginary straight, frictionless tunnel is drilled through the centre of the Earth. A capsule released from rest at one end has acceleration a = −(g/R)x, where x is its displacement from the centre of the Earth along the tunnel, g = 9.8 m s−2 and R = 6.4 × 106 m is the Earth's radius. What is the period of the capsule's oscillation?
Show mark scheme
Marking point
Mark
Notes
Comparing with a = −ω2x gives ω2 = g/R. So T = 2π√(R/g) = 2π√(6.4 × 106/9.8) = 2π × 808 s ≈ 5.1 × 103 s, which is about 85 minutes. The distractors come from common slips: 2.5 × 103 s is only the one-way trip, which is half a period. 8.1 × 102 s leaves out the factor 2π. 4.1 × 106 s leaves out the square root.
✓ 1
Answer C
Answer: C
Common error: Travelling from one end of the tunnel to the other is only half an oscillation. A full period means returning to the starting point.
44C-1B-11The mass–spring oscillatorPaper 1BEasy9 marks
A student determines the spring constant k of a spring by two methods.
Method 1: a mass of (0.300 ± 0.001) kg is hung from the spring. The position of the bottom of the spring is read on a millimetre scale before and after the mass is added. The extension is (9.6 ± 0.2) cm.
Method 2: the same mass is set oscillating vertically. The time for 20 oscillations is (12.5 ± 0.2) s.
Quantity
Measurement
mass m
(0.300 ± 0.001) kg
extension x (method 1)
(9.6 ± 0.2) cm
time for 20 oscillations (method 2)
(12.5 ± 0.2) s
(a)
Each reading of the scale has an uncertainty of ±0.1 cm. Outline why the uncertainty in the extension is ±0.2 cm.
(1)
(b)
Show that method 1 gives a value of k of about 31 N m−1.
(1)
(c)
Calculate the percentage uncertainty in the value of k from method 1.
(2)
(d)
Determine the value of k from method 2 and its absolute uncertainty.
(3)
(e)
State whether the two methods agree.
(1)
(f)
Suggest one change to method 2 that would reduce the uncertainty in k.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The extension is the difference of two readings, and the absolute uncertainties of the two readings add
✓ 1
OWTTE.
Part (b)
k = mg/x = 0.300 × 9.81/0.096 = 30.7 N m−1
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (c)
Δx/x = 0.2/9.6 = 2.1 % and Δm/m = 0.33 %
✓ 1
Either fraction seen.
2.4 %
✓ 1
Accept 2 % (mass uncertainty ignored). Award [2] for CNA.
Part (d)
T = 12.5/20 = 0.625 s, so k = 4π2m/T2 = 30.3 N m−1
Must double the fractional uncertainty in T. Accept 0.032 (mass ignored).
k = (30 ± 1) N m−1
✓ 1
Accept (30.3 ± 1.1) N m−1. MP3 is for matching the precision of value and uncertainty.
Part (e)
Method 1: 29.9–31.4 N m−1; method 2: 29.2–31.4 N m−1; the ranges overlap, so the methods agree
✓ 1
Allow ECF from (c) and (d). Must refer to the uncertainty ranges.
Part (f)
Time more oscillations (e.g. 50) in each run, OR use a fiducial marker at the equilibrium position to start and stop the timer, OR use a light gate/motion sensor
✓ 1
Do not accept 'repeat the experiment' without a specific reason. Do not accept 'use a more precise stopwatch'.
Answers: (b) k = 30.7 N m−1 · (c) 2.4 % · (d) k = (30 ± 1) N m−1(the remaining parts are explanations — see the table above)
45C-1B-28SHM: period & frequencyPaper 1BEasy9 marks
A tyre hangs from a rope tied to the branch of a tree and is used as a swing. A student pulls the empty tyre a short distance to one side and releases it, then uses the stopwatch on a phone to measure the time t for n complete oscillations. Each timing is repeated three times. The absolute uncertainty in the mean time is taken as half the range of the three readings.
The length from the branch to the centre of the tyre is measured with a tape measure as l = (2.50 ± 0.05) m. The tyre is treated as a simple pendulum.
n
t1 / s
t2 / s
t3 / s
mean t / s
T / s
ΔT / s
1
3.13
3.07
3.37
3.19
3.19
0.15
5
15.84
15.68
15.82
15.78
3.156
0.016
10
31.51
31.57
31.69
20
63.12
63.31
63.32
63.25
3.163
0.005
Diagram not to scale. The tyre is shown released from one side.
(a)
State the resolution of the stopwatch.
(1)
(b)
For n = 10:
(i)
calculate the mean time and its absolute uncertainty
(2)
(ii)
hence calculate the period T and its absolute uncertainty ΔT.
(1)
(c)
Outline why ΔT is smaller when more oscillations are timed.
(1)
(d)
Calculate the period predicted by T = 2π√(l/g), and its absolute uncertainty.
(2)
(e)
Determine whether the period measured for n = 20 agrees with the prediction.
(1)
(f)
The student suggests timing 200 oscillations to reduce the uncertainty further. Suggest why this would not be sensible.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
0.01 s
✓ 1
Do not accept a half-range or a reaction time.
Part (b)(i)
mean = (31.51 + 31.57 + 31.69)/3 = 31.59 s
✓ 1
uncertainty = (31.69 − 31.51)/2 = ±0.09 s
✓ 1
Accept ±0.1 s.
Part (b)(ii)
T = 31.59/10 = 3.159 s AND ΔT = ±0.009 s
✓ 1
Allow ECF from (b)(i). Accept (3.16 ± 0.01) s. Both needed.
Part (c)
The uncertainty in the time «from starting and stopping the stopwatch» is about the same whatever n is, and it is divided by n to find ΔT
✓ 1
OWTTE. Do not accept 'more readings are more accurate'.
Part (d)
T = 2π√(2.50/9.81) = 3.17 s
✓ 1
ΔT/T = ½ × 0.05/2.50 = 1 %, so ΔT = ±0.03 s
✓ 1
Must halve the fractional uncertainty in l. Award [1] max for ±0.06 s.
Part (e)
Measured for n = 20: (3.163 ± 0.005) s, i.e. 3.158–3.168 s; predicted 3.14–3.20 s: the ranges overlap, so they agree
✓ 1
Allow ECF from (d). The comparison must use both ranges (or the predicted range).
Part (f)
The amplitude decreases because of air resistance «damping», so after many swings the tyre almost stops / the turning points are hard to see; it is also easy to miscount so many oscillations
✓ 1
Any one sensible reason. Do not accept 'it takes too long' alone.
Answers: (b)(i) 31.59 s ± 0.09 s · (b)(ii) (3.159 ± 0.009) s · (d) (3.17 ± 0.03) s (the remaining parts are explanations — see the table above)
46C-1B-03The defining equation a = −ω²xPaper 1BMedium10 marks
A glider on a horizontal air track is attached between two stretched springs. A motion sensor measures the displacement x of the glider from its equilibrium position, and a wireless accelerometer fixed to the glider measures its acceleration a. A data-logger records both quantities at the same instants during one oscillation. Seven pairs of values are shown in the table and plotted on the graph.
x / m
a / m s−2
−0.075
+3.18
−0.050
+2.09
−0.025
+1.05
0.000
+0.05
+0.025
−0.91
+0.050
−1.98
+0.075
−3.00
Graph drawn to scale
(a)
Outline how the graph indicates that the glider moves with simple harmonic motion.
(2)
(b)
Draw the line of best fit on the graph.
(1)
(c)
Determine the angular frequency ω of the oscillation.
(3)
(d)
The student also times 20 oscillations with a stopwatch and records 19.6 s ± 0.3 s. Deduce whether this measurement is consistent with your answer to (c).
(2)
(e)
The line of best fit does not pass through the origin. Suggest a cause for this and state its effect on the value of ω.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The graph is a straight line, so the (magnitude of the) acceleration is proportional to the displacement
✓ 1
Accept 'a varies linearly with x'.
The gradient is negative, so the acceleration is always directed opposite to the displacement, towards the equilibrium position
✓ 1
OWTTE.
Part (b)
Single straight line with points evenly distributed on both sides
✓ 1
Must not be forced through the origin.
Part (c)
Uses a triangle covering most of the line to find the gradient
✓ 1
gradient = −41 s−2
✓ 1
Accept −39 to −43 s−2.
ω = √41 = 6.4 rad s−1
✓ 1
Accept 6.2 to 6.6 rad s−1. Allow ECF from the gradient. Award [3] for CNA with a gradient seen.
Part (d)
T = 19.6/20 = 0.980 s ± 0.015 s, i.e. between 0.965 s and 0.995 s
✓ 1
Uncertainty must also be divided by 20.
From (c), T = 2π/ω = 2π/6.4 = 0.98 s, which lies inside this range, so the measurements are consistent
✓ 1
Allow ECF from (c). Conclusion must be supported by comparison with the range.
Part (e)
The line crosses the a axis at about +0.07 m s−2: the accelerometer has a zero error (offset) and gives a non-zero reading when a = 0
✓ 1
Accept an intercept of +0.04 to +0.10 m s−2. Accept any constant systematic offset of the accelerometer or a slightly tilted track.
The offset shifts every value of a by the same amount, so the gradient, and therefore ω, is unaffected
✓ 1
OWTTE.
Answers: (c) ω ≈ 6.4 rad s−1 · (d) T = 0.980 ± 0.015 s; 2π/ω ≈ 0.98 s (the remaining parts are explanations — see the table above)
47C-1B-12The simple pendulumPaper 1BMedium10 marks
A pendulum of length (0.800 ± 0.002) m swings through a light gate placed at the lowest point of the swing. A data-logger connected to the light gate displays the period T directly. The bob is released from rest at an angular amplitude θ0, measured with a protractor, and the mean of the first five periods is recorded. The uncertainty in each mean period is ±0.004 s.
A student's hypothesis is that the period of a pendulum does not depend on its amplitude.
θ0 / °
T / s
5
1.797
10
1.798
15
1.803
20
1.809
30
1.827
40
1.849
50
1.883
60
1.926
Graph drawn to scale
(a)
State the resolution of the period measurements.
(1)
(b)
Draw the curve of best fit on the graph.
(1)
(c)
(i)
Using the graph, estimate the largest amplitude for which the data support the hypothesis.
(1)
(ii)
Explain, with reference to the conditions for simple harmonic motion, why the period increases at large amplitudes.
(2)
(d)
Determine g, and its absolute uncertainty, using the period for θ0 = 5°.
(3)
(e)
Comment on your answer to (d).
(1)
(f)
Suggest why only the first five periods after release were used for each amplitude.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
0.001 s
✓ 1
Do not accept ±0.004 s.
Part (b)
Smooth curve, almost horizontal at small amplitudes and rising increasingly steeply, passing through all error bars
✓ 1
Do not accept a straight line or dot-to-dot.
Part (c)(i)
About 10° to 15°: up to this amplitude the periods agree with the small-amplitude value within the uncertainty
✓ 1
Accept 10° to 15°.
Part (c)(ii)
SHM requires the restoring force (acceleration) to be proportional to the displacement; for a pendulum the restoring force is mg sin θ, which is proportional to θ only for small angles
✓ 1
Accept 'sin θ ≈ θ only for small angles'.
At large amplitudes sin θ < θ, so the restoring force and acceleration are smaller than SHM predicts and the bob takes longer to complete a cycle
✓ 1
OWTTE. Must link smaller acceleration to longer period.
48C-1B-13The defining equation a = −ω²xPaper 1BMedium10 marks
A plastic tube with a flat base is loaded with lead shot so that it floats upright in water. When the tube is pushed down a little and released, it oscillates vertically. A student varies the total mass m of the tube and shot, and uses a video recording to find the time for 10 oscillations, 10T.
The period is predicted to be T = 2π√(m/(ρgA)), where A is the cross-sectional area of the tube and ρ = 1000 kg m−3 is the density of water. The graph shows T2 against m.
m / kg
10T / s
T2 / s2
0.040
4.81
0.231
0.055
5.54
0.307
0.070
6.18
0.085
6.74
0.454
0.100
7.26
0.527
0.115
7.80
0.608
Graph drawn to scale
(a)
When the tube is pushed down by a distance y below its equilibrium position, the upward buoyancy force increases by ρgAy. Outline why the tube performs simple harmonic motion.
(2)
(b)
Show that T2 is about 0.38 s2 for m = 0.070 kg.
(1)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine the gradient of the line of best fit.
(2)
(e)
Determine the external diameter of the tube.
(2)
(f)
The diameter measured with vernier calipers is (32.0 ± 0.1) mm. Comment on your answer to (e).
(1)
(g)
The line of best fit does not pass through the origin. Suggest a physical reason for this.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The resultant force on the tube is ρgAy, which is proportional to the displacement y
✓ 1
Weight and equilibrium buoyancy cancel.
and is directed towards the equilibrium position (opposite to the displacement), so a = −(ρgA/m)y, the defining equation of SHM
✓ 1
OWTTE.
Part (b)
T = 6.18/10 = 0.618 s, T2 = 0.3819 s2
✓ 1
Must see the substitution AND an answer to at least 3 s.f.
Part (c)
Single straight line with points evenly distributed on both sides, not forced through the origin
✓ 1
Part (d)
Uses a triangle covering more than half the line
✓ 1
gradient = 5.0 s2 kg−1
✓ 1
Accept 4.8 to 5.2 s2 kg−1.
Part (e)
Gradient = 4π2/(ρgA), so A = 4π2/(1000 × 9.81 × gradient) = 8.05 × 10−4 m2
✓ 1
Allow ECF from (d).
d = 2√(A/π) = 0.0320 m = 32.0 mm
✓ 1
Accept 31.4 to 32.7 mm.
Part (f)
The value from the graph agrees with the caliper value to within about 1 % (the graph method is much less precise), so the model is supported
✓ 1
Allow ECF. Accept a justified 'not consistent' only if the candidate's (e) is outside 31.4–32.7 mm.
Part (g)
Water around and below the tube moves with it, so the oscillating mass is larger than m by a constant amount
✓ 1
Accept 'the tube drags water with it'. Do not accept 'random error'.
Answers: (b) T2 = 0.382 s2 · (d) 5.0 s2 kg−1 · (e) d ≈ 32.0 mm (the remaining parts are explanations — see the table above)
49C-1B-21The simple pendulumPaper 1BMedium12 marks
A puck of diameter (6.00 ± 0.02) cm floats on a cushion of air on a flat air table, which is tilted at an angle θ to the horizontal. The puck is tied by a light thread to a pin in the table, and it swings to and fro across the sloping surface with a small amplitude. The puck behaves as a simple pendulum in a gravitational field of strength g sin θ, so its period is T = 2π√(l/(g sin θ)), where l is the length of the pendulum.
The student measures l = (0.500 ± 0.002) m from the pin to the point on the edge of the puck where the thread is tied. θ is measured with a protractor and plumb line to ±0.5°, and the time for 10 oscillations, 10T, is read from a video to ±0.2 s. The graph shows T2 against 1/sin θ with error bars.
θ / ° ± 0.5°
10T / s ± 0.2 s
1/sin θ
T2 / s2
5.0
49.4
11.47
24.40
7.5
40.5
7.66
16.40
10.0
35.2
5.76
12.39
15.0
28.8
22.5
23.7
2.61
5.62
35.0
19.3
1.74
3.72
Graph drawn to scale
(a)
Show that the absolute uncertainty in 1/sin θ for θ = 5.0° is about 1.2.
(1)
(b)
Complete the table for θ = 15.0°.
(2)
(c)
Explain why the graph of T2 against 1/sin θ is expected to be a straight line through the origin.
(2)
(d)
Draw the line of best fit on the graph.
(1)
(e)
Determine g, using the gradient of your line and l = 0.500 m.
(2)
(f)
The uncertainty in the gradient is ±2 %. Determine whether the student's value of g is consistent with the accepted value of 9.81 m s−2.
(2)
(g)
Explain how the way in which l was measured accounts for your answer to (f), and show that correcting it gives agreement with 9.81 m s−2.
Must see the calculation AND an answer to at least 2 s.f. Accept the larger half-range 1/sin 4.5° − 1/sin 5.0° = 1.27.
Part (b)
1/sin 15.0° = 3.86
✓ 1
Accept 3.9.
T = 2.88 s, so T2 = 8.29 s2
✓ 1
Accept 2 or 3 s.f.
Part (c)
Squaring: T2 = (4π2l/g) × (1/sin θ)
✓ 1
l and g are constant, so this has the form y = mx: a straight line through the origin with gradient 4π2l/g
✓ 1
OWTTE. Must refer to the constant gradient AND the zero intercept.
Part (d)
Single straight line through all the error bars, passing close to the origin
✓ 1
Do not accept a line joining the first and last points.
Part (e)
gradient = 2.13 s2
✓ 1
Accept 2.08 to 2.17 s2. Must use a large triangle.
g = 4π2 × 0.500/2.13 = 9.27 m s−2
✓ 1
Accept 9.10 to 9.49 m s−2. Allow ECF from the gradient. Award [2] for CNA.
Part (f)
Δg/g = 2 % + 0.002/0.500 (0.4 %) = 2.4 %, so Δg = ±0.22 m s−2
✓ 1
Accept ±2 % alone (length uncertainty ignored) for this mark.
g = (9.27 ± 0.22) m s−2, i.e. 9.05 to 9.49 m s−2: 9.81 m s−2 lies outside this range, so the value is not consistent with the accepted value
✓ 1
Allow ECF from (e). The conclusion must be based on the range.
Part (g)
The length of a pendulum is measured to the centre of mass of the bob, here the centre of the puck, so the true length is longer than 0.500 m by the radius of the puck (3.0 cm); using too small a value of l makes g too small
✓ 1
OWTTE. A systematic error: it affects every reading in the same way.
g = 4π2 × 0.530/2.13 = 9.82 m s−2, which agrees with 9.81 m s−2 within the uncertainty
✓ 1
Allow ECF from (e). Accept 9.6 to 10.0 m s−2.
Answers: (a) 1.16 · (b) 3.86; 8.29 s2 · (e) g ≈ 9.27 m s−2 · (f) (9.27 ± 0.22) m s−2: not consistent · (g) 9.82 m s−2(the remaining parts are explanations — see the table above)
50C-1B-29The mass–spring oscillatorPaper 1BMedium10 marks
A student uses an online automotive database to test whether a car's suspension behaves as a mass–spring system. For six cars the database gives the mass m supported by one front wheel, the wheel rate k (the effective spring constant of the suspension at that wheel) and the frequency f of the free vertical oscillation of the front of the car, measured in a bounce test in which the front is pushed down and released. Each value of f is given to ±0.02 Hz.
The student's model predicts f = (1/2π)√(k/m). The graph shows f against √(k/m) with error bars. The dashed lines are the lines of maximum and minimum gradient that pass through all the error bars.
Car
Type
m / kg
k / kN m−1
√(k/m) / s−1
f / Hz ± 0.02 Hz
A
limousine
520
19.0
6.04
0.88
B
city car
290
17.0
7.66
1.12
C
hatchback
360
26.0
8.50
1.24
D
SUV
560
48.0
1.35
E
estate car
430
45.0
10.23
1.50
F
sports car
330
55.0
12.91
1.90
Graph drawn to scale
(a)
Calculate √(k/m) for car D, giving an appropriate unit.
(2)
(b)
Outline why the student plots f against √(k/m).
(1)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine the gradient of the line of best fit.
(2)
(e)
Using the dashed lines, determine the absolute uncertainty in the gradient.
(1)
(f)
Determine whether the data are consistent with the student's model.
(2)
(g)
Suggest one physical reason, other than measurement error, for your conclusion in (f).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
√(48.0 × 103/560) = 9.26
✓ 1
Must convert kN m−1 to N m−1. Award [0] for 0.293.
s−1
✓ 1
Accept rad s−1. Do not accept Hz «√(k/m) is an angular frequency, not f». Unit mark independent of the value.
Part (b)
The model gives f ∝ √(k/m), so this graph should be a straight line through the origin with gradient 1/2π
✓ 1
OWTTE. Must mention straight line AND gradient or origin.
Part (c)
Single straight line lying between the two dashed lines and passing through all the error bars
✓ 1
Part (d)
Uses two well-separated points on the line «triangle over more than half of the line»
✓ 1
gradient = 0.149
✓ 1
Accept 0.145 to 0.153. Unit not required; if given, accept none or Hz s.
Part (e)
gradients of dashed lines 0.154 and 0.143, so uncertainty = ±0.006
✓ 1
Accept ±0.005 to ±0.007. Accept the larger difference from the best value.
Part (f)
Model gradient 1/2π = 0.159
✓ 1
this lies outside the range 0.143–0.154, so the data are not consistent with the model «the measured frequencies are lower than predicted»
✓ 1
Allow ECF from (d) and (e). The conclusion must follow from a comparison with the range.
Part (g)
The shock absorbers damp the oscillation, and damping lowers the frequency of free oscillation; OR the tyre also acts as a spring in series with the suspension, so the effective spring constant is smaller than k
✓ 1
Either reason. Do not accept 'friction' without reference to the frequency or the spring constant.
Answers: (a) 9.26 s−1 · (d) 0.149 · (e) ±0.006 (the remaining parts are explanations — see the table above)
A student investigates how the period T of a pendulum depends on its length l. The bob is a steel nut tied to a light thread, and l is measured from the point of suspension to the centre of the nut with a metre rule. For each length the time for 10 oscillations, 10T, is measured with a hand-held stopwatch. The uncertainty in each value of 10T is ±0.2 s.
The student's hypothesis is that T = Kln, where K and n are constants and n = 0.5. The table shows the data and the graph shows log T against log l for all six lengths.
l / m ± 0.002 m
10T / s ± 0.2 s
log( l / m )
log( T / s )
0.150
7.78
−0.824
−0.109
0.250
9.97
−0.602
−0.001
0.400
12.64
−0.398
0.102
0.600
15.52
0.850
18.57
−0.071
0.269
1.200
22.02
0.079
0.343
Graph drawn to scale
(a)
State the absolute uncertainty in the period T.
(1)
(b)
Complete the table for l = 0.600 m.
(2)
(c)
Explain why a graph of log T against log l is used to test the hypothesis.
(2)
(d)
Draw the line of best fit on the graph.
(1)
(e)
(i)
Determine the gradient of the line of best fit.
(2)
(ii)
State whether the data support the hypothesis.
(1)
(f)
The theory of the simple pendulum gives K = 2π/√g. Determine g using the line of best fit.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
±0.02 s
✓ 1
Accept ±0.02 s written with T, e.g. (1.55 ± 0.02) s.
Part (b)
log( l / m ) = log 0.600 = −0.222
✓ 1
Accept 2 or 4 decimal places.
T = 15.52/10 = 1.552 s AND log( T / s ) = 0.191
✓ 1
Accept 0.190 to 0.191. Both values needed for the second mark.
Part (c)
log T = n log l + log K
✓ 1
Taking logs of both sides of the hypothesis.
This has the form y = mx + c, so a straight line is expected, with gradient equal to n (and intercept log K)
✓ 1
OWTTE. Must link the gradient to n.
Part (d)
A single straight line with points distributed evenly on both sides, extended at least to log( l / m ) = 0
✓ 1
Do not accept a line joining the first and last points, or a curve.
Part (e)(i)
Uses two well-separated points on the drawn line (triangle covering more than half the line)
✓ 1
gradient = 0.50
✓ 1
Accept 0.48 to 0.52. No unit.
Part (e)(ii)
Yes: the gradient is equal to 0.5 to within the precision of the graph (and the line is straight)
✓ 1
Allow ECF from (e)(i): a gradient outside 0.47–0.53 leads to 'not supported'.
Part (f)
Intercept at log( l / m ) = 0 read as 0.303, so K = 100.303 = 2.01 s m−0.5
✓ 1
Accept an intercept of 0.298 to 0.308.
g = 4π2/K2 = 9.79 m s−2
✓ 1
Accept 9.6 to 10.0 m s−2. Allow ECF from the intercept. Award [2] for CNA.
Answers: (a) ±0.02 s · (b) −0.222; 0.191 · (e)(i) 0.50 · (f) g ≈ 9.79 m s−2(the remaining parts are explanations — see the table above)
52C-1B-02The mass–spring oscillatorPaper 1BHard11 marks
A spring hangs from a clamp and a mass hanger carrying slotted masses is attached to its lower end. The total mass m on the spring is measured with a top-pan balance. The mass is pulled down a short distance and released, and the time for 10 oscillations, 10T, is measured with a stopwatch. The uncertainty in each value of 10T is ±0.1 s.
The graph shows T2 against m with error bars for T2. The dashed lines are the lines of maximum and minimum gradient that pass through all the error bars.
m / kg
10T / s ± 0.1 s
0.100
4.27
0.150
5.14
0.200
5.80
0.250
6.48
0.300
7.06
0.350
7.64
Graph drawn to scale
(a)
For m = 0.200 kg, calculate T2 and its absolute uncertainty.
(2)
(b)
Draw the line of best fit on the graph.
(1)
(c)
(i)
Show that the gradient of your line of best fit is about 1.6 s2 kg−1.
(1)
(ii)
Determine the spring constant k of the spring.
(2)
(d)
Using the dashed lines, determine the absolute uncertainty in k.
(2)
(e)
The manufacturer states that k = 25.0 N m−1 ± 2 %. Comment on the student's result.
(1)
(f)
A computer fit shows that the line of best fit meets the m axis at m = −0.014 kg rather than at the origin. Suggest a reason for this, and explain why it does not affect the value of k found in (c)(ii).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
T = 0.580 s, so T2 = 0.336 s2
✓ 1
Δ(T2) = 2 × (0.01/0.580) × 0.336 = ±0.012 s2
✓ 1
Accept ±0.01 s2. Uses the fractional uncertainty in T doubled.
Part (b)
Single straight line lying between the two dashed lines and passing through all error bars
✓ 1
Part (c)(i)
Gradient from two well-separated points on the line, e.g. (0.580 − 0.181)/(0.350 − 0.100) = 1.60
✓ 1
Must see the working AND an answer to at least 3 s.f. (between 1.55 and 1.67).
Part (c)(ii)
T2 = (4π2/k) m, so gradient = 4π2/k
✓ 1
k = 4π2/1.60 = 24.7 N m−1
✓ 1
Accept 23.9 to 25.5 N m−1. Allow ECF from (c)(i). Award [2] for CNA.
Part (d)
Gradients of the dashed lines read as 1.69 and 1.51 s2 kg−1, giving k = 23.3 and 26.1 N m−1
✓ 1
Accept ±0.03 on each gradient.
Δk = (26.1 − 23.3)/2 ≈ ±1.4 N m−1, so k = (24.7 ± 1.4) N m−1 OR (25 ± 1) N m−1
✓ 1
Accept ±1.1 to ±1.7 N m−1. Accept the largest difference from the best value.
Part (e)
The range 23.3–26.1 N m−1 overlaps the manufacturer's range 24.5–25.5 N m−1, so the result is consistent with the stated value
✓ 1
Allow ECF from (c)(ii) and (d). Must refer to both ranges or to the uncertainty.
Part (f)
The spring itself has mass, and part of it oscillates with the load, so the effective oscillating mass is greater than m
✓ 1
Accept 'the mass of the spring is not included in m'.
This adds the same mass to every reading, which shifts the line sideways but does not change its gradient
✓ 1
OWTTE. A systematic offset changes the intercept only.
Answers: (a) T2 = (0.336 ± 0.012) s2 · (c)(ii) k ≈ 24.7 N m−1 · (d) Δk ≈ ±1.4 N m−1(the remaining parts are explanations — see the table above)
53C-1B-27The mass–spring oscillatorPaper 1BHard11 marks
A wooden metre rule is clamped horizontally to the edge of a bench. A small wireless accelerometer of mass 60 g is taped to the free end, which is pulled down slightly and released. The accelerometer app displays the frequency f of the vertical oscillations to the nearest 0.01 Hz. The free length L, from the edge of the clamp to the centre of the accelerometer, is varied and measured to ±0.002 m. Three readings of f are taken for each length; their spread gives an uncertainty of ±0.05 Hz in f.
The ruler acts as a spring whose spring constant is proportional to 1/L3. If the mass of the ruler is negligible, the period is given by T = 2π√(m/k) with m the mass of the accelerometer, and the student's hypothesis is that f = KLn, where K is a constant and n = −1.5. The graph shows log f against log L. The error bars are drawn but are too small to be seen clearly at this scale.
L / m ± 0.002 m
f / Hz ± 0.05 Hz
log(L / m)
log(f / Hz)
0.300
13.81
−0.523
1.140
0.400
8.85
−0.398
0.947
0.500
6.23
−0.301
0.794
0.600
4.65
0.700
3.64
−0.155
0.561
0.800
2.95
−0.097
0.470
Graph drawn to scale
(a)
Suggest why the uncertainty in f is larger than the resolution of the app.
(1)
(b)
Complete the table for L = 0.600 m.
(2)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine the gradient of the line of best fit.
(2)
(e)
The lines of maximum and minimum gradient that pass through all the error bars have gradients −1.54 and −1.61. Determine whether the data support the hypothesis.
(2)
(f)
The mass of the ruler is 90 g. Suggest why the gradient differs from −1.5 in the way that it does.
(2)
(g)
Suggest one change to the apparatus that would make the hypothesis a better model of the oscillations.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Repeated readings of f vary «by more than 0.01 Hz» because of random effects, e.g. the way the ruler is released or the accelerometer slightly changing position
✓ 1
OWTTE. Do not accept 'human error' alone.
Part (b)
log(0.600) = −0.222
✓ 1
Accept −0.22.
log(4.65) = 0.667
✓ 1
Accept 0.67. Do not accept ln values.
Part (c)
Single straight line through all (or all but one of) the error bars, with points on both sides
✓ 1
Do not accept a line joining the first and last points.
Part (d)
Uses a triangle covering more than half of the line «Δlog f/Δlog L»
✓ 1
gradient = −1.58
✓ 1
Accept −1.60 to −1.56. Negative sign required. No unit.
Part (e)
The range of gradients is −1.61 to −1.54 «≈ −1.58 ± 0.04»
✓ 1
−1.5 lies outside this range, so the data do not support the hypothesis «n is more negative than −1.5»
✓ 1
Allow ECF from (d) only if the conclusion is based on the given range. A bare 'no' scores 0.
Part (f)
Part of the ruler moves with the accelerometer, so the oscillating mass is greater than 60 g «the ruler's mass is not negligible»
✓ 1
Accept 'the effective mass includes part of the ruler'.
The longer the free length, the more of the ruler oscillates, so f falls more quickly with L than L−1.5, making the gradient more negative
✓ 1
OWTTE. Must link the extra mass to increasing L.
Part (g)
Use a heavier mass at the free end «e.g. add 200 g of slotted masses» so that the ruler's mass is a much smaller fraction of the oscillating mass
✓ 1
Accept 'use a lighter but equally stiff strip'. Do not accept 'repeat more times' or 'use a longer range'.
Answers: (b) −0.222, 0.667 · (d) gradient = −1.58 (the remaining parts are explanations — see the table above)
A long pendulum in a science museum consists of a bob of mass 28 kg on a steel wire. The distance from the point of suspension to the centre of the bob is 11.0 m.
The bob is pulled aside so that its centre rises 1.20 cm above its lowest position, and is released from rest at time t = 0. Treat the pendulum as a simple pendulum. Ignore air resistance except in part (d). Take g = 9.81 m s−2.
Figure 1 — for part (c)
(a)
(i)
Show that the period of the pendulum is about 6.7 s.
(1)
(ii)
State the time t at which the speed of the bob is first a maximum.
(1)
(b)
Calculate the maximum speed of the bob.
(2)
(c)
Sketch, on the axes in Figure 1, graphs to show how the kinetic energy Ek and the gravitational potential energy Ep of the bob vary with time from t = 0 to t = T, where T is the period. Ep is measured from the lowest position of the bob. Label your graphs.
(3)
(d)
In practice the amplitude of the swing slowly decreases.
(i)
Outline why the amplitude decreases.
(1)
(ii)
An electromagnet under the floor gives the bob a small push once in each cycle so that the amplitude stays constant. Suggest where in the swing, and in which direction, the push should be given.
(2)
(e)
Suggest why the museum uses a heavy bob rather than a light bob of the same size.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
T = 2π√(11.0/9.81) «= 6.65 s»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (a)(ii)
T/4 = 1.7 s
✓ 1
Accept 1.6–1.7 s. Allow ECF from (a)(i)
Part (b)
½mv² = mgh so v = √(2 × 9.81 × 0.0120)
✓ 1
Award [1] max if h = 1.20 is used «4.85 m s−1»
v = 0.49 m s−1
✓ 1
Accept 0.48–0.49 m s−1. Award [2] for CNA
Part (c)
Ep = E at t = 0, T/2 and T, and Ep = 0 at T/4 and 3T/4
✓ 1
Labels required for MP1 and MP2
Ek = 0 when Ep = E, and Ek = E when Ep = 0
✓ 1
smooth curves «not straight lines» that are never negative, with Ek + Ep = E at every instant «the curves cross at E/2»
✓ 1
Do not award MP3 for curves with sharp corners at the maxima
Part (d)(i)
work is done against air resistance «and friction at the support», so energy of the oscillation is transferred to the surroundings as thermal energy
✓ 1
OWTTE
Part (d)(ii)
at «or near» the lowest point of the swing, in the direction in which the bob is moving
✓ 1
Accept 'in the direction of the velocity of the bob'
so the force does work on the bob / supplies energy in step with the motion, replacing the energy lost in each cycle
✓ 1
OWTTE. Do not accept a push against the motion
Part (e)
the resistive forces on bobs of the same size are about the same, but the heavier bob has more energy «and inertia», so a smaller fraction of its energy is lost in each swing «the amplitude decreases more slowly»
✓ 1
OWTTE
Answers: (a)(i) T = 6.65 s · (a)(ii) t = 1.7 s · (b) v = 0.49 m s−1(the remaining parts are explanations — see the table above)
55IBP2S-018-00-11The mass–spring oscillatorPaper 2Easy5 marks
In a simple model of a racing car's suspension, one corner of the car supports a mass of 300 kg on a spring of spring constant 2.4 × 104 N m−1. When the car passes over a kerb, this corner oscillates vertically.
(a)
State the two conditions on the acceleration of an object for its motion to be simple harmonic.
(2)
(b)
Calculate the period of the vertical oscillation.
(2)
(c)
Calculate the frequency of the oscillation.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The acceleration is proportional to the displacement from the equilibrium position . The acceleration is directed towards the equilibrium position, opposite to the displacement .
✓ 2
Part (b)
Substitution into \(T = 2\pi\sqrt{m/k} = 2\pi\sqrt{300/(2.4\times10^4)}\) . T = 0.70 s .
✓ 2
Part (c)
f = 1/T = 1/0.70 ≈ 1.4 Hz (accept 1.42 Hz; allow error carried forward from the previous part) .
✓ 1
Exam tip: When asked to state the conditions for simple harmonic motion, give both proportionality and direction. 'Acceleration proportional to displacement' on its own earns only one mark.
56IBP2S-018-00-13SHM: period & frequencyPaper 2Easy5 marks
A simple vertical seismometer contains a 0.50 kg mass suspended from a spring. The natural frequency of its oscillation is 1.0 Hz.
(a)
Calculate the angular frequency of the oscillation.
(1)
(b)
Calculate the spring constant of the spring.
(2)
(c)
During a small tremor the mass oscillates with this frequency. Calculate the magnitude of its acceleration at the moment its displacement from equilibrium is 4.0 mm.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
ω = 2πf = 2π × 1.0 = 6.3 rad s−1 .
✓ 1
Part (b)
Uses k = mω2, or rearranges \(T = 2\pi\sqrt{m/k}\) . k = 0.50 × (6.28)2 = 20 N m−1 . ECF from (a).
✓ 2
Part (c)
Uses a = ω2x with x = 4.0 × 10−3 m . a = 39.5 × 4.0 × 10−3 = 0.16 m s−2 . ECF from (a).
✓ 2
Exam tip: Convert millimetres to metres before substituting into a = ω²x.
57IBP2S-018-02-11The simple pendulumPaper 2Easy7 marks
A demolition crane swings a heavy steel ball on the end of a cable. For small oscillations the ball can be modelled as a simple pendulum. The distance from the pivot to the centre of the ball is 12 m.
(a)
State the condition for an object to perform simple harmonic motion.
(1)
(b)
Calculate the period of oscillation of the ball.
(2)
(c)
The ball is replaced by one of twice the mass, with its centre at the same distance from the pivot. Outline why the period is unchanged.
(2)
(d)
The operator wants a period of 5.0 s. Determine the distance from the pivot to the centre of the ball that is needed.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The acceleration (or net force) is proportional to the displacement from equilibrium and directed towards equilibrium, i.e. opposite to the displacement
✓ 1
Part (b)
Uses T = 2π√(L/g) = 2π√(12/9.8)
✓ 1
T = 6.95 s (accept 7.0 s; accept g = 9.81 m s−2)
✓ 1
Part (c)
The restoring force (mg sin θ ≈ mgθ for small angles) and the inertia (m) are both proportional to the mass
✓ 1
so the acceleration a = −(g/L)x, and therefore ω and T, do not depend on the mass
✓ 1
Part (d)
Rearranges to L = gT²/(4π²)
✓ 1
L = 9.8 × 5.0²/(4π²) = 6.2 m
✓ 1
Exam tip: Rearrange T = 2π√(L/g) symbolically before substituting. Squaring first gives L = gT²/4π², which avoids slips with the square root.
58IBP2S-018-02-12The defining equation a = −ω²xPaper 2Easy6 marks
A vibration-isolation platform in a laboratory is modelled as a mass of 2.5 kg on a spring of spring constant 400 N m−1. It oscillates horizontally with simple harmonic motion of amplitude 3.0 cm. Friction can be ignored.
(a)
Calculate the total energy of the oscillation.
(2)
(b)
Calculate the maximum speed of the platform.
(2)
(c)
Determine the speed of the platform when its displacement is 1.5 cm.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
E = ½kx0² = ½ × 400 × 0.030²
✓ 1
E = 0.18 J
✓ 1
Part (b)
½mvmax² = 0.18 J, so vmax = √(2 × 0.18/2.5) or ω = √(k/m) = √(400/2.5) = 12.6 rad s−1 and vmax = ωx0
✓ 1
vmax = 0.38 m s−1 (allow error carried forward from the total energy calculated earlier)
✓ 1
Part (c)
EP = ½ × 400 × 0.015² = 0.045 J, so EK = 0.18 − 0.045 = 0.135 J or v = ω√(x0² − x²) = 12.6 × √(0.030² − 0.015²)
✓ 1
v = 0.33 m s−1 (allow error carried forward from earlier parts)
✓ 1
Common error: Convert centimetres to metres before squaring. Using 3.0 instead of 0.030 makes the energy 10⁴ times too large.
59IBP2S-018-01-11The defining equation a = −ω²xPaper 2Easy6 marks
A car body of mass 1200 kg is supported by its suspension, which behaves as a single spring of spring constant 4.8 × 104 N m−1. After the car passes over a bump, the body oscillates vertically. Assume the dampers are faulty, so damping can be ignored.
(a)
State the conditions required for the car body to perform simple harmonic motion.
(2)
(b)
Show that the angular frequency of the oscillation is about 6.3 rad s−1.
(2)
(c)
The amplitude of the oscillation is 3.0 cm. Calculate the maximum acceleration of the car body.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Acceleration (or resultant force) is proportional to the displacement from the equilibrium position . Acceleration is directed towards the equilibrium position, i.e. opposite to the displacement . Accept a = −ω²x with a and x identified as acceleration and displacement from equilibrium for both marks.
✓ 2
Part (b)
ω = √(k/m) seen or used, or T = 2π√(m/k) with ω = 2π/T . ω = √(4.8 × 104/1200) = √40 = 6.32 rad s−1 .
✓ 2
Part (c)
amax = ω²A, or amax = kA/m . amax = 40 × 0.030 = 1.2 m s−2 . Using ω = 6.3 gives 1.19 m s−2, also accepted. ECF from (b).
✓ 2
Exam tip: In a 'show that' question, give the calculated value to at least one more significant figure than the value quoted (6.32, not 6.3).
60IBP2S-018-01-12The defining equation a = −ω²xPaper 2Easy5 marks
A vibration-isolation platform for a laboratory laser is disturbed and oscillates horizontally. A sensor records the platform's displacement x from equilibrium and its acceleration a.
x / mm
−4.0
−2.0
0.0
+2.0
+4.0
a / m s−2
+3.2
+1.6
0.0
−1.6
−3.2
(a)
Outline how the data show that the platform performs simple harmonic motion.
(2)
(b)
Determine the angular frequency of the oscillation.
(2)
(c)
Calculate the frequency of the oscillation.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The ratio a/x is constant (−800 s−2), or doubling x doubles a, so a ∝ x . a always has the opposite sign to x, so the acceleration is directed towards equilibrium .
✓ 2
Part (b)
ω² = a/x = 3.2/(4.0 × 10−3) = 800 s−2, with the mm converted to m . ω = 28 rad s−1 .
✓ 2
Part (c)
f = ω/2π = 28.3/2π = 4.5 Hz . ECF from (b).
✓ 1
Common error: Leaving x in millimetres gives ω² = 0.8 s⁻². Convert to metres before dividing.
A small puck moves without friction on a track made from two straight ramps. Each ramp is inclined at 6.0° to the horizontal, and the ramps meet at a smoothly rounded bottom of negligible length, as shown in Figure 1. The displacement x of the puck is measured along the track from the bottom, and is positive to the right.
The puck is released from rest at x = +0.40 m. Take g = 9.81 m s−2.
Figure 1 — Diagram NOT to scale (the angle is exaggerated)Figure 2 — for part (b)(ii)
(a)
State the two conditions on the acceleration of a body that are required for its motion to be simple harmonic.
(2)
(b)
(i)
Show that the magnitude of the acceleration of the puck on either ramp is about 1.0 m s−2.
(1)
(ii)
Sketch, on the axes in Figure 2, a graph to show how the acceleration a of the puck varies with x for −0.40 m ≤ x ≤ +0.40 m.
(2)
(iii)
Hence explain why the motion of the puck is not simple harmonic.
(2)
(c)
Determine the period of the oscillation of the puck.
(3)
(d)
The ramps are replaced by a horizontal track on which the same puck, of mass 0.050 kg, is held between two stretched springs. The combined spring constant of the springs is 0.16 N m−1.
(i)
Calculate the period of oscillation of the puck on the springs.
(1)
(ii)
Both oscillators are now started with an amplitude of 0.10 m instead of 0.40 m. Compare the periods of the two oscillators at this amplitude with their periods at the larger amplitude. Support your answer with a calculation.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the acceleration is «directly» proportional to the displacement from the equilibrium position
✓ 1
OWTTE. Accept a ∝ x
the acceleration is «always» directed towards the equilibrium position / opposite to the displacement
✓ 1
Award [2] for a = −ω²x with both the proportionality and the meaning of the minus sign explained
Part (b)(i)
a = g sin 6.0° = 9.81 × sin 6.0° «= 1.03 m s−2»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)(ii)
horizontal line at a ≈ −1.0 m s−2 for positive x AND horizontal line at a ≈ +1.0 m s−2 for negative x, each extending to ±0.40 m
✓ 1
Allow ECF from (b)(i) for the size of a
a changes sign abruptly at x = 0
✓ 1
A vertical line at x = 0 is not required. Do not award MP2 for a sloping straight line through the origin
Part (b)(iii)
the magnitude of the acceleration is constant «≈ 1.0 m s−2» / does not depend on the displacement
✓ 1
so a is not proportional to x «the graph is not a straight line through the origin», which SHM requires, even though a is always directed towards the equilibrium position
✓ 1
OWTTE. Award [1] max for 'the period depends on the amplitude' with no reference to the acceleration
Part (c)
time to reach the bottom from s = ½at²: 0.40 = ½ × 1.03 × t²
✓ 1
Allow use of a = 1.0 m s−2 throughout, giving T = 3.58 s
t = 0.883 s
✓ 1
one oscillation consists of four such journeys, so T = 4t = 3.53 s
✓ 1
Award [3] for a bald correct answer. Accept 3.5–3.6 s. Award [2] max for T = 2t
Part (d)(i)
T = 2π√(0.050/0.16) = 3.51 s
✓ 1
Accept 3.5 s
Part (d)(ii)
spring oscillator: the period is unchanged «at 3.51 s» because T = 2π√(m/k) does not depend on the amplitude
✓ 1
ramps: t ∝ √s, so when the distance is quartered the time «and the period» is halved
✓ 1
Accept a full recalculation: t = √(2 × 0.10/1.03) = 0.442 s
T = 1.77 s for the ramps «so the two oscillators no longer have the same period»
✓ 1
Allow ECF from (c). Accept 1.76–1.79 s
Answers: (b)(i) a = 1.03 m s−2 · (c) T = 3.53 s · (d)(i) T = 3.51 s · (d)(ii) springs 3.51 s (unchanged); ramps 1.77 s (the remaining parts are explanations — see the table above)
A tall building sways from side to side in strong winds. The natural period of this sway is 6.0 s.
To reduce the sway, engineers hang a large steel ball from cables inside the top of the building. The ball and cables behave as a simple pendulum whose natural period is equal to that of the building. Hydraulic dampers connect the ball to the building. Take g = 9.81 m s−2.
(a)
Show that the length of the pendulum must be about 9 m.
(1)
(b)
Explain, in terms of resonance, why gusts of wind that push on the building every 6.0 s are dangerous.
(3)
(c)
During a storm the ball swings with a horizontal amplitude of 1.5 m. The equation T = 2π√(l/g) is accurate to within 1 % for angular amplitudes up to about 23°. Determine whether the equation may be used for the pendulum during this storm.
(3)
(d)
The mass of the ball is 3.3 × 105 kg. State and explain whether a ball of smaller mass, hung from cables of the same length, would have a different natural period.
(2)
(e)
The hydraulic dampers remove energy from the swinging ball. State the form of energy into which this energy is transferred.
(1)
(f)
Suggest one reason why the period of the real ball-and-cable system may differ from the value given by T = 2π√(l/g).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
l = gT²/4π² = 9.81 × 6.0²/4π² «= 8.95 m»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)
the driving frequency «1/6.0 s» equals the natural frequency of the building, so resonance occurs
✓ 1
energy is transferred to the building in every cycle «because each push is in step with the motion»
✓ 1
OWTTE
so the amplitude of the sway builds up to a large value that could damage the structure
✓ 1
Part (c)
sin θ = 1.5/8.95
✓ 1
Allow ECF from (a). Accept tan θ or the small-angle value 1.5/8.95 rad
θ = 9.7°
✓ 1
Accept 9.5°–9.8° «9.6° if l = 9.0 m is used»
this is less than 23°, so the equation may be used
✓ 1
MP3 only scores if a comparison is made
Part (d)
no, the period would be the same
✓ 1
T = 2π√(l/g) does not contain the mass, because the restoring force and the inertia are both proportional to the mass
✓ 1
OWTTE. Award [1] max for quoting the equation without explanation
Part (e)
thermal energy «internal energy» of the oil in the dampers and the surroundings
✓ 1
Accept 'heat'
Part (f)
the ball is large, not a point mass / the cables have mass / the dampers exert forces on the ball / the amplitude may not be small
✓ 1
Any one reasonable suggestion
Answers: (a) l = 8.95 m · (c) θ = 9.7° < 23°: yes (the remaining parts are explanations — see the table above)
63C-2-18The mass–spring oscillatorPaper 2Medium12 marks
An inertial balance consists of a tray clamped between two flexible horizontal steel strips. When the tray is pulled sideways and released, it oscillates horizontally with simple harmonic motion. An object of mass m clamped to the tray changes the period T according to
T = 2π√((m + mt)/k)
where mt is the mass of the tray and k is the effective spring constant of the strips. The table shows two calibration measurements.
Added mass m / kg
Time for 10 oscillations / s
0
4.20
0.200
6.30
Figure 1 — for part (c)
(a)
An inertial balance can be used to find the mass of an object on board an orbiting space station, where a spring balance (newton meter) cannot be used. Explain why.
(3)
(b)
(i)
Show that k is about 36 N m−1.
(2)
(ii)
Determine mt.
(2)
(c)
Draw, on the grid in Figure 1, the line that shows how T² varies with m for 0 ≤ m ≤ 0.30 kg.
(2)
(d)
The tray, carrying an unknown object, oscillates with a period of 0.560 s. Determine the mass of the object.
(2)
(e)
Suggest why the object must be clamped firmly to the tray.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
in orbit the station and everything in it are in free fall with the same acceleration
✓ 1
OWTTE
so no supporting force is needed and a spring balance reads zero «for any mass»
✓ 1
the inertial balance depends on the force needed to accelerate the mass «F = ma» / on the inertia of the mass, which does not depend on g
✓ 1
Do not accept 'there is no gravity in orbit'
Part (b)(i)
T0 = 0.420 s and T1 = 0.630 s, and subtracting: T1² − T0² = 4π² × 0.200/k
✓ 1
Must see the two periods used
k = 4π² × 0.200/(0.630² − 0.420²) «= 35.8 N m−1»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)(ii)
mt = kT0²/4π² = 35.8 × 0.420²/4π²
✓ 1
ALTERNATIVE: m_t/(m_t + 0.200) = 0.420²/0.630²
mt = 0.160 kg
✓ 1
Accept 0.16 kg. Using k = 36 N m−1 gives 0.161 kg
Part (c)
straight line with an intercept of T² = 0.176 s² at m = 0
✓ 1
Accept 0.17–0.18 s²
line passing through (0.200 kg, 0.397 s²) / gradient 1.10 s² kg−1, drawn to m = 0.30 kg «T² = 0.507 s²»
✓ 1
Allow ECF from (b)
Part (d)
T² = 0.560² = 0.314 s²
✓ 1
Allow reading from the line in (c)
m = kT²/4π² − mt = 0.124 kg
✓ 1
Accept 0.12–0.13 kg. Allow ECF from (b)
Part (e)
if it slides on the tray, not all of its mass is accelerated with the tray «and friction removes energy», so the period no longer corresponds to the full mass
✓ 1
OWTTE
Answers: (b)(i) k = 35.8 N m−1 · (b)(ii) mt = 0.160 kg · (d) m = 0.124 kg (the remaining parts are explanations — see the table above)
64C-2-19SHM: period & frequencyPaper 2Medium12 marks
A smartphone contains an accelerometer. Inside it, a tiny silicon block of mass 4.0 × 10−9 kg is held between thin silicon beams that act as a spring, so the block can move along one horizontal line. The block and beams behave as a mass on a spring. The displacement x of the block from its equilibrium position, relative to the case of the phone, is measured electrically.
In a test in which damping is negligible, the phone is tapped and the block oscillates with simple harmonic motion. The graph shows how x varies with time t after the tap.
Graph drawn to scale
(a)
Determine the frequency of the oscillation.
(2)
(b)
Show that the spring constant of the beams is about 2.5 N m−1.
(1)
(c)
The phone is now given a constant acceleration of 3.0 m s−2 along the line in which the block can move.
(i)
Explain why the block is displaced from its equilibrium position relative to the case of the phone.
(2)
(ii)
Calculate this displacement.
(2)
(d)
The electronics can detect a displacement of the block of 0.20 nm or more. Determine whether the accelerometer can detect a constant acceleration of 0.10 m s−2.
(3)
(e)
The designer wants a larger displacement of the block for the same acceleration. Suggest, with reference to your answer to (c)(ii), how the natural frequency of the block and beams must change.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Four cycles take 1.00 ms, so T = 0.25 ms «2.5 × 10−4 s»
✓ 1
Accept T read from a single cycle.
f = 1/T = 4.0 × 103 Hz
✓ 1
Accept 3.9–4.1 kHz. Allow ECF from an incorrect T. Award [2] for CNA.
Must see full substitution OR answer to at least 3 s.f. Accept a rearrangement of T = 2π√(m/k).
Part (c)(i)
The block must accelerate with the phone, so a resultant force F = ma must act on it
✓ 1
This force can only be provided by the beams, which exert a force only when they are deformed «F = kx», so the block is displaced «towards the rear, opposite to the acceleration»
✓ 1
OWTTE. Do not accept 'the block is left behind' without reference to the force from the beams.
Part (c)(ii)
F = ma = 4.0 × 10−9 × 3.0 = 1.2 × 10−8 N
✓ 1
ALTERNATIVE: x = a/ω² with ω = 2π × 4.0 × 103 rad s−1.
x = F/k = 1.2 × 10−8/2.53 = 4.7 × 10−9 m
✓ 1
Accept 4.7–4.8 × 10−9 m «4.8 nm from k = 2.5 N m−1». Award [2] for CNA.
0.16 nm is less than 0.20 nm, so this acceleration cannot be detected
✓ 1
MP3 only scores if a displacement «or a smallest acceleration» has been calculated. Allow ECF for the conclusion.
Part (e)
For a given acceleration x = ma/k = a/ω² «= a/(2πf)²»
✓ 1
Allow ECF from (c)(ii).
so the natural frequency must be lower «e.g. a larger mass or less stiff beams»
✓ 1
MP2 needs a reason. OWTTE.
Answers: (a) f = 4.0 × 103 Hz · (b) k = 2.53 N m−1 · (c)(ii) 4.7 × 10−9 m · (d) 1.6 × 10−10 m — cannot be detected (the remaining parts are explanations — see the table above)
The air suspension of a bus uses air springs instead of steel springs. One air spring is modelled as a vertical cylinder fixed to an axle. Air is trapped below a piston of area A = 0.0500 m2, which supports a load of mass M = 2.00 × 103 kg, as shown in Figure 1. The top of the piston is open to the atmosphere. At equilibrium the height of the trapped air is h = 0.250 m.
Atmospheric pressure = 1.01 × 105 Pa. Assume that the temperature of the trapped air stays constant, and ignore friction and the mass of the piston.
Figure 1 — Diagram NOT to scale
(a)
Show that the pressure p0 of the trapped air at equilibrium is about 4.9 × 105 Pa.
(2)
(b)
The load is pushed down by a small distance y from its equilibrium position.
(i)
Show that the pressure of the trapped air becomes p0h/(h − y).
(1)
(ii)
Hence show that, when y is much smaller than h, the resultant force on the load is upwards and has magnitude (p0A/h)y.
(2)
(iii)
Explain why the vertical oscillations of the load are simple harmonic when their amplitude is small.
(2)
(c)
Calculate the frequency of the vertical oscillations of the load.
(2)
(d)
When the bus is full, the load on this air spring is 3.00 × 103 kg. A pump adds air so that the height of the trapped air is again 0.250 m. Determine the new frequency, and compare the change with the change that would occur for a steel spring of fixed spring constant carrying the same loads.
(3)
(e)
The bus hits a pothole and the load oscillates with an amplitude of 5.0 cm. Comment on whether the motion is still well described as simple harmonic, using a calculation at the lowest point.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
At equilibrium p0A = Mg + «atmospheric pressure» × A
Steel spring: f ∝ 1/√M, so f = 1.12 × √(2.00/3.00) = 0.91 Hz; the air spring's frequency falls by only about 3.5 % compared with about 18 %
✓ 1
OWTTE. The comparison must be supported by both values.
Part (e)
At y = 0.050 m the actual force is p0Ay/(h − y) = p0Ay/0.200, which is 1.25 times «25 % greater than» (p0A/h)y
✓ 1
Accept a calculation of the two forces: 4.93 × 103 N and 6.17 × 103 N.
So the restoring force is not proportional to the displacement «y is not much smaller than h»: the motion is not well described as simple harmonic
✓ 1
Allow ECF from (b)(ii). The comment must follow from the calculation.
Answers: (a) p0 = 4.93 × 105 Pa · (c) f = 1.12 Hz · (d) 1.08 Hz (steel spring: 0.91 Hz) · (e) force 25 % larger than the SHM value (the remaining parts are explanations — see the table above)
66C-2-39The mass–spring oscillatorPaper 2Medium15 marks
In a laboratory that makes microchips, a granite tabletop of mass 450 kg rests on four identical vertical springs to isolate it from vibrations of the floor. When the tabletop is lowered onto the springs, each spring is compressed by 12.0 mm. Treat the tabletop and springs as a single mass–spring system that moves only vertically. Take g = 9.81 m s−2.
The tabletop is struck lightly and oscillates. Figure 1 shows how its displacement x from the equilibrium position varies with time t. Upward displacements are positive. Damping can be neglected over the time shown.
Figure 1 — drawn to scaleFigure 2 — for part (c)(ii)
(a)
Show that the spring constant of each spring is about 9.2 × 104 N m−1.
(1)
(b)
Calculate the natural frequency of the vertical oscillations of the tabletop.
(2)
(c)
(i)
Determine the maximum acceleration of the tabletop.
(2)
(ii)
Sketch, on the axes in Figure 2, a graph to show how the acceleration a of the tabletop varies with t from t = 0 to t = 0.50 s.
(2)
(d)
Outline the energy changes as the tabletop moves from its lowest position to its equilibrium position.
(2)
(e)
The floor vibrates because of footsteps in the corridor (about 2 Hz), a pump in the next room (4.5 Hz) and a ventilation fan (50 Hz). Explain which source is most likely to cause large oscillations of the tabletop.
(3)
(f)
An engineer proposes replacing the springs with softer springs that are compressed by 48.0 mm when the tabletop rests on them. Determine the new natural frequency and comment on the proposal.
Must see full substitution OR answer to at least 3 s.f.
Part (b)
Total spring constant 3.68 × 105 N m−1; T = 2π√(450/3.68 × 105) = 0.220 s
✓ 1
ALTERNATIVE: T = 2π√(Δx/g) because k = mg/Δx.
f = 1/T = 4.55 Hz
✓ 1
Award [2] for CNA. Allow ECF from (a). Award [1 max] for 2.28 Hz «one spring only».
Part (c)(i)
amax = ω2x0 with x0 = 0.40 mm read from the graph
✓ 1
amax = (2π × 4.55)2 × 0.40 × 10−3 = 0.33 m s−2
✓ 1
Accept 0.32–0.33 m s−2. Allow ECF from (b), or use of T = 0.22 s from the graph.
Part (c)(ii)
Sinusoidal curve of the same period as Figure 1, with a = 0 where x = 0 and a positive where x is negative «antiphase / starting at zero and rising first»
✓ 1
Do not award MP1 for a curve in phase with x.
Peaks at ±0.33 m s−2
✓ 1
Allow ECF from (c)(i). Judge peaks within half a small square.
Part (d)
The elastic potential energy of the springs decreases
✓ 1
The kinetic energy «and the gravitational potential energy» of the tabletop increases, the kinetic energy reaching a maximum at the equilibrium position
✓ 1
Total energy stays constant.
Part (e)
The pump: its frequency «4.5 Hz» is very close to the natural frequency «4.55 Hz»
✓ 1
Allow ECF from (b). MP1 can be awarded only with a reason.
so resonance occurs: energy is transferred to the tabletop efficiently on every cycle and the amplitude becomes large
✓ 1
OWTTE.
The footsteps «2 Hz» and the fan «50 Hz» drive at frequencies far from the natural frequency, so the amplitudes they produce stay small
✓ 1
Part (f)
f ∝ 1/√Δx «f = (1/2π)√(g/Δx)», so four times the compression halves the frequency
✓ 1
f = 2.28 Hz
✓ 1
Allow ECF from (b). Accept 2.27–2.28 Hz.
Comment: far from the pump frequency, so the pump no longer causes resonance, BUT now close to the ≈ 2 Hz of footsteps, which could now cause resonance «so the proposal needs damping as well»
✓ 1
Both halves of the comment are needed for MP3.
Answers: (a) k = 9.20 × 104 N m−1 · (b) f = 4.55 Hz · (c)(i) 0.33 m s−2 · (f) f = 2.28 Hz (the remaining parts are explanations — see the table above)
67C-2-49The defining equation a = −ω²xPaper 2Medium14 marks
A subwoofer loudspeaker is driven by an electrical signal of constant frequency. The centre of its cone moves along the axis of the loudspeaker with simple harmonic motion. A laser vibrometer measures the velocity v of the centre of the cone; v is positive when the cone moves outwards. At t = 0 the cone passes through its equilibrium position.
Figure 1 shows how v varies with time t.
Figure 1 — graph drawn to scale
(a)
Determine the frequency of the oscillation of the cone.
(2)
(b)
(i)
State a time at which the cone is at its maximum outward displacement.
(1)
(ii)
Explain your answer to (b)(i).
(1)
(c)
By drawing a tangent on Figure 1, determine the magnitude of the maximum acceleration of the cone.
(3)
(d)
Estimate, using the area under the graph, the amplitude of the oscillation of the cone.
(2)
(e)
Show that your answers to (a), (c) and (d) are consistent with the defining equation of simple harmonic motion.
(2)
(f)
The cone is held by a flexible suspension that behaves as a spring. Outline when, between t = 0 and t = 25 ms, the kinetic energy of the cone is greatest and when the elastic potential energy of the suspension is greatest.
(2)
(g)
The speed of sound in air is 340 m s−1. Calculate the wavelength of the sound emitted by the loudspeaker.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
T = 25 ms «read over two cycles: 50 ms/2»
✓ 1
Accept 24–26 ms.
f = 1/T = 40 Hz
✓ 1
Allow ECF from T. Award [2] for CNA.
Part (b)(i)
t = 6.25 ms «or 31.25 ms»
✓ 1
Accept 6.0–6.5 ms or 31.0–31.5 ms.
Part (b)(ii)
The velocity is zero there «so the cone is at an extreme», and just before this it was moving outwards «v positive»
✓ 1
OWTTE. Do not accept 'v = 0' alone.
Part (c)
Tangent drawn at a point where v = 0 «e.g. t = 6.25 ms»
✓ 1
The tangent must touch the curve at a zero crossing; MP1 is for the position.
Gradient calculated from a large triangle on the tangent, e.g. 0.50 m s−1/4.0 ms
✓ 1
|a|max = 126 m s−2
✓ 1
Accept 110–140 m s−2. Accept 1.3 × 102 m s−2.
Part (d)
Area between t = 0 and the next zero of v «a quarter of a cycle» is equal to the displacement from equilibrium to the extreme, about 50 small squares
✓ 1
Accept 45–55 squares.
each square = 1 ms × 0.04 m s−1 = 4 × 10−5 m, so x0 ≈ 2.0 × 10−3 m
✓ 1
Accept 1.8–2.2 mm. Award [1] max for 1.6 mm «area as a triangle» with no correction.
Part (e)
ω = 2πf = 2π × 40 = 251 rad s−1
✓ 1
Allow ECF from (a).
ω2x0 = 2512 × 2.0 × 10−3 = 126 m s−2, which agrees with (c) «within the precision of the graph», so |a|max = ω2x0
✓ 1
Must see the comparison. Allow ECF from (c) and (d).
Part (f)
Kinetic energy greatest at t = 0, 12.5 ms and 25 ms, when the speed is greatest «at the equilibrium position»
✓ 1
Accept any one of these times with the reason.
Elastic potential energy greatest at t = 6.25 ms and 18.75 ms, when v = 0 «at maximum displacement» and the kinetic energy is zero
✓ 1
Accept 6–6.5 ms and 18.5–19 ms.
Part (g)
λ = v/f = 340/40 = 8.5 m
✓ 1
Allow ECF from (a).
Answers: (a) 40 Hz · (b)(i) 6.25 ms · (c) ≈ 126 m s−2 · (d) ≈ 2.0 mm · (g) 8.5 m (the remaining parts are explanations — see the table above)
68C-2-51The mass–spring oscillatorPaper 2Medium15 marks
A cargo spacecraft of mass 1.20 × 104 kg approaches a space station along a straight line at a speed of 0.080 m s−1 relative to the station. When it makes contact, it pushes against a capture ring mounted on springs. The springs behave as a single spring of spring constant 3.0 × 104 N m−1 that is at its natural length at the moment of first contact. The springs push on the spacecraft but are not attached to it.
The station is so massive that its motion can be ignored. Treat the spacecraft as a particle and ignore all forces other than the force from the springs.
Figure 1 — for part (e)(ii)
(a)
Explain why the spacecraft moves with simple harmonic motion while it is in contact with the springs.
(2)
(b)
(i)
Show that the period of this simple harmonic motion is about 4.0 s.
(1)
(ii)
Explain why the spacecraft stays in contact with the springs for half of this period, and state the time of contact.
(2)
(c)
Latches lock the spacecraft to the station 1.5 s after first contact, provided the spacecraft is still in contact. Determine whether the latches lock.
(1)
(d)
A second, identical spacecraft approaches at 0.16 m s−1. State and explain whether the latches lock this spacecraft.
(2)
(e)
Suppose that the latches fail, so that the first spacecraft leaves the springs moving away from the station at 0.080 m s−1.
(i)
Calculate the average force exerted on the spacecraft by the springs during the contact.
(2)
(ii)
Sketch, on Figure 1, a graph to show how the force exerted by the springs on the spacecraft varies with time during the contact.
(2)
(f)
Outline the energy changes from the moment of first contact to the moment of maximum compression of the springs.
(2)
(g)
Suggest why real docking systems also contain dampers.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The only force on the spacecraft is the spring force, which is proportional to the compression «F = −kx, Hooke's law»
✓ 1
and directed opposite to the displacement, towards the position where the springs have their natural length, so a = −(k/m)x
✓ 1
OWTTE. Accept ω2 = k/m with the minus sign explained.
Part (b)(i)
T = 2π√(1.20 × 104/3.0 × 104) «= 3.97 s»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)(ii)
Contact starts and ends with the springs at their natural length «the equilibrium position»: the spacecraft moves to maximum compression and back, which is half an oscillation; the springs cannot pull it back, so it then leaves
✓ 1
OWTTE.
time of contact = 3.97/2 = 2.0 s
✓ 1
Allow ECF from (b)(i). Accept 2 s.
Part (c)
1.5 s is less than the contact time of 2.0 s, so the spacecraft is still in contact and the latches lock
✓ 1
Allow ECF from (b)(ii). A comparison must be seen.
Part (d)
Yes, the latches lock
✓ 1
MP1 may be inferred from MP2.
The period «and so the contact time» depends only on m and k, not on the amplitude «or the approach speed», so the contact time is still 2.0 s
✓ 1
Do not accept 'it compresses the springs further' alone.
Part (e)(i)
Δp = 2mv = 2 × 1.20 × 104 × 0.080 = 1920 N s
✓ 1
Award [1] max for mv = 960 N s «gives 483 N».
F = Δp/Δt = 1920/1.987 = 966 N
✓ 1
Allow ECF from (b)(ii). Accept 960–970 N. Award [2] for CNA.
Part (e)(ii)
Single smooth hump «half a sine curve» that starts at F = 0 at t = 0 and returns to zero at t ≈ 2.0 s, with its maximum at t ≈ 1.0 s
✓ 1
Allow ECF from (b)(ii) for the end time.
Maximum clearly above the average force from (e)(i) «about 1.5 × 103 N», so that the area under the curve equals the impulse
✓ 1
A value for the peak is not required. Do not award MP2 for a rectangle or a triangle.
Part (f)
The kinetic energy of the spacecraft decreases, to zero at maximum compression
✓ 1
The elastic potential energy of the springs increases by the same amount «total energy constant», to a maximum at maximum compression
✓ 1
OWTTE.
Part (g)
Dampers transfer the energy to internal energy «thermal energy» so that the spacecraft does not rebound or keep oscillating
✓ 1
Accept 'to stop the oscillations quickly'.
Answers: (b)(ii) 2.0 s · (e)(i) 966 N (the remaining parts are explanations — see the table above)
In a science-centre exhibit, a small heavy bob hangs on a light string of length 1.60 m. A thin horizontal peg Q is fixed 1.20 m vertically below the pivot. When the bob is to the right of the vertical through the pivot, the whole string swings. When the bob is to the left, the string presses against the peg and only the lowest 0.40 m swings, as shown in Figure 1. Air resistance and friction are negligible.
The bob is released from rest with a horizontal displacement x = +0.080 m to the right of the vertical, at time t = 0. For small displacements the height h of the bob above its lowest point is related to x by h ≈ x2/(2l), where l is the length of string that is swinging.
Figure 1 — Diagram NOT to scale (angles exaggerated)Figure 2 — for part (f)
(a)
Calculate the period of the pendulum if the peg is removed.
(1)
(b)
Show that the period of the pendulum with the peg is about 1.9 s.
(2)
(c)
Calculate the frequency of the pendulum with the peg.
(1)
(d)
Explain why the bob rises to the same height on the left as the height from which it was released.
(2)
(e)
Show that the greatest horizontal displacement of the bob to the left of the vertical is about 0.040 m.
(2)
(f)
Figure 2 shows how x varies with t from t = 0 until the bob first reaches the vertical. Sketch, on Figure 2, the variation of x with t for the rest of one complete oscillation.
(3)
(g)
State and explain whether the motion of the bob over one complete oscillation is simple harmonic.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
T = 2π√(1.60/9.81) = 2.54 s
✓ 1
Accept 2.5 s.
Part (b)
The bob spends half a period of the 1.60 m pendulum on the right and half a period of the 0.40 m pendulum on the left
Gravitational potential energy is transferred to kinetic energy and back; the tension «and the peg» does no work because it is perpendicular to the motion, and there are no resistive forces
✓ 1
OWTTE.
At both highest points the bob is momentarily at rest, so it has the same gravitational potential energy and therefore the same height
✓ 1
Do not accept 'energy is conserved' alone for MP2.
Part (e)
h = 0.0802/(2 × 1.60) = 2.0 × 10−3 m
✓ 1
x = √(2 × 0.40 × 2.0 × 10−3) «= 0.040 m»
✓ 1
Must see full substitution OR answer to 3 s.f. ALTERNATIVE: x ∝ √l for the same h, so x = 0.080 × √(0.40/1.60).
Part (f)
Negative half-loop from t = 0.63 s to 1.27 s with minimum x = −0.040 m at t ≈ 0.95 s
✓ 1
Allow ECF from (e). Judge to within one small square.
Smooth, symmetric «sinusoidal» shape on the left, crossing x = 0 at both ends with the steepest gradient there
✓ 1
Do not award for straight-line segments or sharp corners at x = 0.
From 1.27 s the curve rises as a quarter cycle of the longer pendulum to x = +0.080 m at t ≈ 1.90 s, where its gradient is zero
✓ 1
Allow ECF from (b).
Part (g)
On the right a = −(g/1.60)x, but on the left a = −(g/0.40)x «the restoring acceleration per unit displacement is four times larger on the left»
✓ 1
OWTTE. Accept a comparison of the two values of ω2.
so a is not proportional to x with a single constant: the motion is periodic but not simple harmonic
✓ 1
MP2 only scores if MP1 scores. Do not accept 'not SHM because the amplitudes differ' alone.
Answers: (a) 2.54 s · (b) 1.90 s · (c) 0.53 Hz · (e) 0.040 m (the remaining parts are explanations — see the table above)
70IBP2S-018-00-12The mass–spring oscillatorPaper 2Medium7 marks
A student investigates how the period of a mass–spring oscillator depends on the hanging mass m. For each mass, she measures the time t10 for 10 complete oscillations.
m / kg — t10 / s 0.100 — 5.0 0.200 — 7.0 0.300 — 8.6 0.400 — 9.9
(a)
Explain why the student times 10 oscillations rather than one.
(2)
(b)
Outline how a straight-line graph could be drawn from these data, and how k could be found from it.
(2)
(c)
Determine the spring constant k using the data for m = 0.400 kg.
(2)
(d)
The student's T2 against m graph has a small positive intercept on the T2 axis. Suggest a reason for this.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The absolute uncertainty in timing, mainly from reaction time, is about the same whatever the length of the measured interval . Dividing the total time by 10 divides this uncertainty by 10, so the percentage uncertainty in T is smaller .
✓ 2
Part (b)
Calculate T = t10/10 and plot T2 on the y-axis against m on the x-axis . Since T2 = (4π2/k)m, the gradient is 4π2/k, so k = 4π2/gradient. Ideally the line passes through the origin .
✓ 2
Part (c)
T = 9.9/10 = 0.99 s . k = 4π2m/T2 = 4π2 × 0.400/0.992 = 16 N m−1 .
✓ 2
Part (d)
Either reason earns the mark: the spring itself has mass and also oscillates, or the mass hanger is not included in m. In both cases the effective oscillating mass is greater than m .
✓ 1
Common error: Using t₁₀ directly as the period gives a k that is 100 times too small. Divide by the number of oscillations first.
71IBP2S-018-02-13The simple pendulumPaper 2Medium9 marks
Students investigate how the period of a simple pendulum depends on its length L. For each length they measure the time t10 for 10 oscillations, starting from a small amplitude.
L / m : 0.20 | 0.40 | 0.60 | 0.80 | 1.00 t10 / s : 8.9 | 12.7 | 15.6 | 17.9 | 20.1
(a)
Outline why the students timed 10 oscillations rather than one.
(2)
(b)
Show that a graph of T² against L should be a straight line through the origin, and state how g is found from its gradient.
(2)
(c)
Determine g using the first and last rows of the data.
(3)
(d)
Another group releases each pendulum from an angle of 35°. Explain the effect on their value of g.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The absolute (reaction-time) uncertainty in a stopwatch reading is roughly fixed
✓ 1
so timing a longer interval and dividing by 10 reduces the percentage uncertainty in T
✓ 1
Part (b)
Squaring T = 2π√(L/g) gives T² = (4π²/g)L, which has the form y = mx, so the graph is a straight line through the origin
✓ 1
gradient = 4π²/g, so g = 4π²/gradient
✓ 1
Part (c)
T = 0.89 s and 2.01 s, so T² = 0.792 s² and 4.04 s²
g = 4π²/4.06 = 9.7 m s−2 (allow ECF: correct use of g = 4π²/gradient with an incorrect gradient)
✓ 1
Part (d)
At large amplitude the small-angle approximation fails and the period is longer than 2π√(L/g) predicts
✓ 1
so the measured T² values are too large, the gradient is too large and g is underestimated
✓ 1
Exam tip: When you determine a gradient, use widely separated points and show the T² values you calculate. Each intermediate value can earn credit even if the final answer slips.
72IBP2S-018-01-13The defining equation a = −ω²xPaper 2Medium7 marks
A liquid-filled U-tube manometer is used in a hospital to monitor gas pressure. Both arms of the tube are vertical. The liquid column has total length L, measured along the tube, density ρ and uniform cross-sectional area S. The liquid surface in one arm is pushed down by a distance y, so the surface in the other arm rises by y. The liquid is then released.
Diagram not to scale
(a)
Show that the acceleration of the liquid column is given by a = −(2g/L)y.
(3)
(b)
Explain why the motion is simple harmonic.
(1)
(c)
The liquid column has length 0.48 m. Calculate the period of oscillation.
(2)
(d)
Suggest why the real liquid does not perform perfect simple harmonic motion.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The difference in levels between the arms is 2y, so the unbalanced column has weight 2ySρg . The mass of the whole moving column is LSρ . a = F/m = −2ySρg/(LSρ) = −(2g/L)y. The minus sign shows that the force acts towards the equilibrium position .
✓ 3
Part (b)
2g/L is constant, so the acceleration is proportional to the displacement y and opposite in direction to it, matching a = −ω²x .
✓ 1
Part (c)
ω² = 2g/L, so T = 2π√(L/2g) . T = 2π√(0.48/19.62) = 0.98 s .
✓ 2
Part (d)
Viscous friction between the liquid and the tube walls, and within the liquid itself, dissipates energy. The oscillations are therefore damped and the amplitude decreases . Also accept: a resistive force that is not of the form −kx, or surface tension or turbulence effects, provided the answer explains the departure from ideal SHM.
✓ 1
Exam tip: When deriving a = −ω²x, state clearly which mass is accelerated. Here it is the whole column of liquid, not just the raised part.
73IBP2S-018-01-14The mass–spring oscillatorPaper 2Medium7 marks
On an orbiting space station, an astronaut's mass is measured by strapping them into a chair of mass 15 kg. The chair is attached to springs of combined spring constant 2.4 × 103 N m−1 and oscillates horizontally in simple harmonic motion. With the astronaut in the chair, the period is 1.30 s.
(a)
Explain why ordinary bathroom scales cannot measure the astronaut's mass on the station.
(2)
(b)
Determine the mass of the astronaut.
(3)
(c)
The amplitude of the oscillation is 0.15 m. Calculate the maximum acceleration experienced by the astronaut.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The astronaut and the scales are in free fall together, so they have the same acceleration . The contact force is zero, so the scales read zero / weight cannot be used to infer mass .
✓ 2
Part (b)
T = 2π√(m/k), so m = kT²/4π² . Total mass = 2400 × 1.30²/4π² = 103 kg . Astronaut's mass = 103 − 15 = 88 kg .
✓ 3
Part (c)
ω = 2π/1.30 = 4.83 rad s−1 . amax = ω²A = 4.83² × 0.15 = 3.5 m s−2 .
✓ 2
Common error: The period gives the total oscillating mass. Subtract the chair's mass to find the astronaut's mass.
A robotic lander on the surface of Mars carries a simple pendulum of length 0.500 m. A camera on the lander records that the pendulum makes 20 complete oscillations of small amplitude in 46.1 s.
(a)
(i)
Determine the gravitational field strength gM at the surface of Mars.
(3)
(ii)
Outline why the time for 20 oscillations is measured rather than the time for one oscillation.
(1)
(b)
The radius of Mars is 3.39 × 106 m.
(i)
Calculate the mass of Mars.
(2)
(ii)
The same pendulum is taken to the Earth. Calculate its period on the Earth.
(1)
(c)
Phobos, a moon of Mars, orbits at a distance of 9.38 × 106 m from the centre of Mars.
(i)
Determine the gravitational field strength due to Mars at the orbit of Phobos.
(2)
(ii)
Determine the orbital period of Phobos, in hours.
(2)
(d)
The length of the pendulum is known to ±0.005 m and the time for 20 oscillations to ±0.2 s. Determine the absolute uncertainty in gM, and state gM with its uncertainty.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
T = 46.1/20 = 2.305 s
✓ 1
gM = 4π²l/T² = 4π² × 0.500/2.305²
✓ 1
gM = 3.72 m s−2
✓ 1
Award [3] for a bald correct answer. Accept 3.71–3.72 m s−2. Award [2] max for using 46.1 s as the period
Part (a)(ii)
the «absolute» uncertainty in starting and stopping the timer is shared over 20 periods, so the percentage uncertainty in T is smaller
✓ 1
OWTTE. Do not accept 'more accurate' alone
Part (b)(i)
M = gMr²/G = 3.72 × (3.39 × 106)²/6.67 × 10−11
✓ 1
Allow ECF from (a)(i)
M = 6.40 × 1023 kg
✓ 1
Accept 6.3–6.5 × 1023 kg
Part (b)(ii)
T = 2π√(0.500/9.81) = 1.42 s
✓ 1
Part (c)(i)
g ∝ 1/r², so g = 3.72 × (3.39/9.38)²
✓ 1
ALTERNATIVE: g = GM/r² with M from (b)(i). Allow ECF
g = 0.485 N kg−1
✓ 1
Accept 0.48–0.49 N kg−1 or m s−2
Part (c)(ii)
g = v²/r so v = √(0.485 × 9.38 × 106) = 2.13 × 103 m s−1
✓ 1
ALTERNATIVE: T = 2π√(r/g). Allow ECF from (c)(i)
T = 2πr/v = 2.76 × 104 s = 7.7 h
✓ 1
Accept 7.6–7.7 h. Award [2] for CNA
Part (d)
fractional uncertainty in l = 0.005/0.500 = 0.010
✓ 1
fractional uncertainty in T² = 2 × 0.2/46.1 = 0.0087
✓ 1
Award this mark for 2 × the fractional uncertainty in T
ΔgM = (0.010 + 0.0087) × 3.72 = 0.069 m s−2
✓ 1
Allow ECF from (a)(i)
gM = (3.72 ± 0.07) m s−2
✓ 1
MP4 is for matching the precision of the value and the uncertainty
Answers: (a)(i) gM = 3.72 m s−2 · (b)(i) 6.40 × 1023 kg · (b)(ii) 1.42 s · (c)(i) 0.485 N kg−1 · (c)(ii) 7.7 h · (d) (3.72 ± 0.07) m s−2(the remaining parts are explanations — see the table above)
75C-2-16The defining equation a = −ω²xPaper 2Hard13 marks
A laboratory shaker has a horizontal platform that oscillates vertically with simple harmonic motion. A tray of mass 0.150 kg rests on the platform without being fixed to it. The platform oscillates with a frequency of 8.0 Hz and an amplitude of 2.0 mm. Take g = 9.81 m s−2.
Figure 1 — Diagram NOT to scale
(a)
(i)
Show that the maximum acceleration of the platform is about 5 m s−2.
(1)
(ii)
State the direction of the acceleration of the tray when the platform is at its highest point.
(1)
(b)
(i)
On Figure 1, draw and label arrows to represent the forces acting on the tray when the platform is at its highest point. The lengths of your arrows should show the relative sizes of the forces.
(2)
(ii)
Calculate the magnitude of the normal force on the tray at the highest point.
(2)
(c)
(i)
Determine the frequency at which the tray just loses contact with the platform when the amplitude is 2.0 mm.
(3)
(ii)
Explain why this frequency does not depend on the mass of the tray.
(1)
(d)
The frequency is returned to 8.0 Hz and the amplitude is increased to 4.0 mm. Determine whether the tray stays in contact with the platform throughout the oscillation.
two arrows only, starting on the tray: weight «W, mg, Fg» vertically downwards and normal force «N, R, FN» vertically upwards, both labelled
✓ 1
Do not credit an extra 'restoring force' or 'acceleration' arrow
weight arrow clearly longer than the normal force arrow
✓ 1
MP2 only scores if MP1 scores. Accept lengths in the ratio of about 2 : 1
Part (b)(ii)
mg − N = ma, so N = 0.150 × (9.81 − 5.05)
✓ 1
Allow ECF from (a)(i); using a = 5 gives 0.72 N
N = 0.71 N
✓ 1
Accept 0.71–0.72 N
Part (c)(i)
contact is lost when N = 0 at the highest point, i.e. when the required downward acceleration ω²A equals g
✓ 1
ω = √(9.81/2.0 × 10−3) = 70.0 rad s−1
✓ 1
f = ω/2π = 11.1 Hz
✓ 1
Award [3] for a bald correct answer. Accept 11 Hz
Part (c)(ii)
the weight and the resultant force needed for the acceleration are both proportional to the mass, so the condition ω²A = g does not involve m
✓ 1
OWTTE
Part (d)
ω²A = (2π × 8.0)² × 4.0 × 10−3 = 10.1 m s−2
✓ 1
ALTERNATIVE: largest amplitude for contact at 8.0 Hz = g/ω² = 3.88 mm
this is greater than g «= 9.81 m s−2»
✓ 1
ALTERNATIVE: 4.0 mm > 3.88 mm
so the tray does not stay in contact; it leaves the platform near the top of the motion
✓ 1
MP3 only scores if a comparison with g (or the limiting amplitude) is made
Answers: (a)(i) 5.05 m s−2 · (b)(ii) N = 0.71 N · (c)(i) 11.1 Hz · (d) ω²A = 10.1 m s−2 > g: contact is lost (the remaining parts are explanations — see the table above)
76C-2-50The defining equation a = −ω²xPaper 2Hard14 marks
Figure 1 shows the mechanism of an electric jigsaw. A pin P is fixed to a gear wheel at a distance r = 12.5 mm from the axle O. The pin slides in a horizontal slot in a yoke. The yoke is fixed to a vertical bar that carries the blade and can move only vertically, in guides. The gear wheel rotates at a constant 3000 revolutions per minute. The vertical displacement y of the blade from its central position is equal to the height of P above O.
In this question the blade is not cutting, and friction is negligible.
Figure 1 — Diagram NOT to scale
(a)
(i)
State the amplitude of the oscillation of the blade.
(1)
(ii)
Show that the period of the oscillation is 0.020 s.
(1)
(b)
(i)
State the magnitude and the direction of the acceleration of the pin P.
(2)
(ii)
Hence explain why the blade moves with simple harmonic motion.
(3)
(c)
Show that the maximum acceleration of the blade is about 1.2 × 103 m s−2.
(1)
(d)
The yoke, bar and blade have a total mass of 0.060 kg. Calculate the maximum resultant force on them.
(1)
(e)
The jigsaw is switched to a lower setting, at which the gear wheel rotates at 1800 revolutions per minute. Calculate the new maximum resultant force on the yoke, bar and blade.
(1)
(f)
Outline the energy transfers between the gear wheel and the blade as the blade moves from the top of its stroke to the bottom.
(2)
(g)
Explain why the body of the jigsaw vibrates in the user's hand, even when the blade is not cutting.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
12.5 mm «= r»
✓ 1
Do not accept 25 mm.
Part (a)(ii)
f = 3000/60 = 50 Hz «one oscillation per revolution» so T = 1/50 = 0.020 s
✓ 1
Must see full substitution OR the frequency 50 Hz.
Part (b)(i)
ω2r «= v2/r»
✓ 1
Accept a numerical value of 1.2 × 103 m s−2.
towards the centre O «of the circle»
✓ 1
Accept 'centripetal'.
Part (b)(ii)
The slot allows P to move horizontally relative to the yoke, so the blade has the same vertical displacement and vertical acceleration as P
✓ 1
OWTTE.
The vertical component of the acceleration of P is ω2r sin θ = ω2y
✓ 1
Accept a correct component shown on a diagram.
It is directed towards the level of O, i.e. opposite to y, and ω is constant, so a = −ω2y «the condition for SHM»
✓ 1
MP3 needs both the direction and the proportionality to y.
Part (c)
a = ω2r = (2π × 50)2 × 0.0125 «= 1234 m s−2»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (d)
F = 0.060 × 1234 = 74 N
✓ 1
Allow ECF from (c). Accept 72 N «from 1.2 × 103».
Part (e)
F ∝ ω2, so F = 74 × (1800/3000)2 = 27 N
✓ 1
Allow ECF from (d). Accept 26–27 N.
Part (f)
From the top to the central position the blade speeds up: the pin does work on the yoke and the kinetic energy of the blade increases «from zero to a maximum; the small decrease in the gravitational potential energy of the blade also contributes»
✓ 1
OWTTE.
From the central position to the bottom the blade slows down: the yoke does work on the pin, so the kinetic energy of the blade is transferred back to the gear wheel «and is zero at the bottom»
✓ 1
Do not accept 'energy is lost' for MP2.
Part (g)
The pin exerts a force on the yoke that changes direction twice in every cycle «up to 74 N»
✓ 1
By Newton's third law the yoke exerts an equal and opposite force on the pin and so on the gear wheel and the body of the jigsaw, which therefore vibrates «at 50 Hz»
✓ 1
Must refer to Newton's third law or to an equal and opposite force.
Answers: (a)(i) 12.5 mm · (c) 1234 m s−2 · (d) 74 N · (e) 27 N (the remaining parts are explanations — see the table above)
In a simple model of an ion trap, two fixed point charges, each +4.0 × 10−13 C, are 2.00 mm apart in a vacuum. A calcium ion of charge +e and mass 6.64 × 10−26 kg is on the straight line joining them, as shown in the diagram. Other fields, not considered here, keep the ion on this line. M is the midpoint between the charges and x is the displacement of the ion from M.
The table gives the magnitude F of the resultant electric force on the ion for several values of x.
x / mm
F / 10−16 N
0.050
1.16
0.100
0.200
4.99
0.400
13.0
0.600
33.7
Diagram NOT to scale
(a)
Explain why M is an equilibrium position for the ion.
(1)
(b)
Show that F is about 2.3 × 10−16 N when x = 0.100 mm.
(2)
(c)
State the direction of the resultant force on the ion when it is displaced to the right of M.
(1)
(d)
(i)
Deduce, using the table, how F depends on x when x is small.
(2)
(ii)
Hence explain why the oscillations of the ion about M are simple harmonic only when their amplitude is small.
(2)
(e)
Determine the frequency of small-amplitude oscillations of the ion.
(2)
(f)
Outline why the gravitational force on the ion can be ignored.
(1)
(g)
Both fixed charges are doubled. Deduce the new frequency of small-amplitude oscillations.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The two forces on the ion at M are equal in magnitude «equal charges at equal distances» and opposite in direction, so the resultant force is zero
✓ 1
OWTTE.
Part (b)
F = kQe/(0.90 × 10−3)2 − kQe/(1.10 × 10−3)2 = 7.103 × 10−16 − 4.755 × 10−16 N
✓ 1
«= 2.35 × 10−16 N»
✓ 1
Must see full substitution of both forces OR answer to 3 s.f.
Part (c)
To the left, towards M «away from the nearer charge»
✓ 1
Part (d)(i)
F/x calculated for at least two values with x ≤ 0.200 mm: 2.32, 2.35, 2.50 × 10−12 N m−1
✓ 1
Allow ECF from (b).
F/x is approximately constant «within about 8 %», so for small x, F is «approximately» proportional to x
✓ 1
Do not accept 'F increases with x' alone.
Part (d)(ii)
For small x the force is proportional to x and always directed towards M, so a = −(F/x)(1/m)x, the defining equation of SHM
✓ 1
OWTTE.
For larger x, F/x increases «to 5.6 × 10−12 N m−1 at 0.600 mm», so F is not proportional to x and the motion is not simple harmonic
78IBP2S-018-00-16The mass–spring oscillatorPaper 2Hard7 marks
When a loaded trailer is lowered onto its suspension springs, the springs compress by 3.5 cm from their natural length. Assume the springs obey Hooke's law and the trailer oscillates vertically as a single mass.
(a)
Show that the period of vertical oscillation is given by \(T = 2\pi\sqrt{\frac{\Delta x}{g}}\), where Δx is the compression of the springs.
(2)
(b)
Calculate the period of oscillation.
(1)
(c)
The trailer is towed along a road with transverse ridges spaced 4.5 m apart. Determine the towing speed at which the trailer meets one ridge every period of its natural oscillation.
(2)
(d)
A heavier load is placed in the trailer. Explain how the speed found in (c) changes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
At equilibrium kΔx = mg, so m/k = Δx/g . Substituting into \(T = 2\pi\sqrt{m/k}\) gives \(T = 2\pi\sqrt{\Delta x/g}\) .
✓ 2
Part (b)
\(T = 2\pi\sqrt{0.035/9.8}\) = 0.38 s .
✓ 1
Part (c)
One ridge is met per period, so v = d/T . v = 4.5/0.38 = 12 m s−1 . ECF from (b).
✓ 2
Part (d)
The springs compress more, so Δx is larger and the period is longer . The ridges must be met less often, so the speed is lower .
✓ 2
Exam tip: T = 2π√(Δx/g) lets you find the period from a single static measurement. You do not need to know the mass or the spring constant.
79IBP2S-018-01-16The defining equation a = −ω²xPaper 2Hard9 marks
A vibration-test rig consists of a light platform hanging from a vertical spring. A component of mass 0.80 kg is fixed to the platform. In equilibrium, the spring is extended by 0.0392 m. The platform is pulled down and released, and oscillates vertically.
Diagram not to scale. Equilibrium position shown.
(a)
Show that the spring constant is 200 N m−1.
(1)
(b)
The platform is displaced a distance x below the equilibrium position. Taking downward as positive, show that the resultant force on the oscillating mass is −kx, and hence explain why the motion is simple harmonic.
(3)
(c)
Calculate the period of oscillation.
(2)
(d)
The component is replaced by a loose sensor of the same mass that simply rests on top of the platform. Determine the maximum amplitude for which the sensor stays in contact with the platform throughout the motion.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
k = mg/e = 0.80 × 9.8/0.0392 = 200 N m−1 .
✓ 1
Part (b)
Weight mg acts downward (+) and spring force k(e + x) acts upward (−) . Since ke = mg, resultant force = mg − k(e + x) = −kx . a = −(k/m)x with k/m constant, so acceleration is proportional to displacement from equilibrium and directed towards it, so the motion is SHM .
✓ 3
Part (c)
ω = √(k/m) = √(200/0.80) = 15.8 rad s−1, or T = 2π√(m/k) . T = 0.40 s .
✓ 2
Part (d)
Contact is lost first at the top of the motion, where the platform's downward acceleration is greatest, ω²A . The sensor can accelerate downwards at most at g, so the condition is ω²A ≤ g . Amax = g/ω² = 9.8/250 = 0.039 m (0.0392 m), equal to the static extension . Allow ECF from (c).
✓ 3
Common error: Measuring x from the spring's natural length instead of from the equilibrium position leaves an extra mg term and hides the −kx form.
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