ExaminerPrep › Question bank › IAL Physics › 5.5 Simple harmonic motion
Edexcel International A Level Physics · Unit 5 · Thermodynamics, Radiation, Oscillations and Cosmology
40 free, high-quality past-paper-style questions on Simple harmonic motion for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
Which statement describes the acceleration of an object moving with simple harmonic motion?
| Answer | Mark | Notes |
|---|---|---|
| SHM requires a = −ω2x: the acceleration is proportional to the displacement and directed towards the equilibrium position (the minus sign). | ✓ 1 | Answer B |
Answer: B
Exam tip: A definition of SHM needs both parts: proportional to displacement, and directed towards equilibrium.
An oscillator vibrates with a frequency of 0.80 Hz.
What is its angular frequency?
| Answer | Mark | Notes |
|---|---|---|
| ω = 2πf = 2π × 0.80 = 5.0 rad s−1 (5.03). | ✓ 1 | Answer C |
Answer: C
Common error: Dividing by 2π instead of multiplying. Angular frequency is always the larger number: ω = 2πf.
An object moves with simple harmonic motion. At one instant its displacement from the equilibrium position is zero.
Which row describes the object at this instant?
| Answer | Mark | Notes |
|---|---|---|
| At equilibrium x = 0, so a = −ω2x = 0. The speed is greatest there, Aω: speed maximum, acceleration zero. | ✓ 1 | Answer A |
Answer: A
Common error: Linking zero acceleration with zero speed. In SHM they happen at different places: zero speed at the extremes, zero acceleration at equilibrium.
A puck on a level air table is attached between two stretched springs. When the puck is displaced 0.025 m from its equilibrium position, the resultant force on it is 0.60 N towards the equilibrium position.
State the conditions for an object to move with simple harmonic motion.
The puck moves with simple harmonic motion. Calculate the constant k in F = −kx for this system.
Calculate the size of the resultant force on the puck when its displacement is 0.040 m.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| acceleration (or resultant force) proportional to the displacement from the equilibrium position | ✓ 1 | |
| and always directed towards the equilibrium position / opposite to the displacement | ✓ 1 | |
| Part (b) | ||
| k = 0.60 ÷ 0.025 = 24 N m−1 | ✓ 1 | |
| Part (c) | ||
| 24 × 0.040 = 0.96 N (towards equilibrium) | ✓ 1 | |
Exam tip: In SHM the restoring force scales with the displacement, so a ratio is often quicker: 0.040/0.025 = 1.6, and 1.6 × 0.60 N = 0.96 N.
The prong of a tuning fork vibrates with simple harmonic motion at 440 Hz. The amplitude of the tip of the prong is 0.80 mm.
Calculate the angular frequency of the vibration.
Calculate the maximum speed of the tip of the prong.
Calculate the maximum acceleration of the tip of the prong.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ω = 2π × 440 = 2.76 × 103 rad s−1 | ✓ 1 | |
| Part (b) | ||
| vmax = Aω = 0.80 × 10−3 × 2765 | ✓ 1 | |
| = 2.2 m s−1 (2.21) | ✓ 1 | |
| Part (c) | ||
| amax = Aω2 = 0.80 × 10−3 × 27652 | ✓ 1 | |
| = 6.1 × 103 m s−2 | ✓ 1 | |
Common error: Leaving the amplitude in millimetres. Convert 0.80 mm to 0.80 × 10−3 m before substituting.
A simple pendulum consists of a small metal bob on a thread. The length from the point of suspension to the centre of the bob is 0.65 m. g = 9.81 N kg−1.
Calculate the period of the pendulum for small oscillations.
State two conditions needed for the equation T = 2π√(l/g) to give an accurate value of the period.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| T = 2π√(0.65 ÷ 9.81) | ✓ 1 | |
| = 1.6 s (1.62 s) | ✓ 1 | |
| Part (b) | ||
| any two from: the amplitude is small (less than about 10°) | ✓ 1 | |
| the thread is light / inextensible, or the bob is small and dense so it acts as a point mass / air resistance is negligible | ✓ 1 | |
Exam tip: The length of a pendulum is measured to the centre of the bob, not to its top or bottom.
The graphs show how the acceleration a of four different oscillating objects varies with the displacement x from the equilibrium position.
Which graph shows simple harmonic motion?
| Answer | Mark | Notes |
|---|---|---|
| SHM needs a ∝ −x: a straight line through the origin with a negative gradient. Only graph 1 has this shape. Graphs 3 and 4 are restoring (the acceleration is opposite to x) but not proportional to x; graph 2 is not restoring at all. | ✓ 1 | Answer A |
Answer: A
Common error: Accepting any graph where the acceleration points back towards equilibrium. The acceleration must also be proportional to the displacement: a straight line through the origin.
The graph shows how the displacement x of an oscillating object varies with time t. Four points P, Q, R and S are labelled on the curve.
At which point is the velocity of the object negative and greatest in size?
| Answer | Mark | Notes |
|---|---|---|
| Velocity is the gradient of the x–t graph. The gradient is negative and steepest where the curve crosses the time axis going downwards: R. | ✓ 1 | Answer C |
Answer: C
Common error: Linking the largest displacement with the largest velocity. At Q and S the curve is flat, so the velocity there is zero.
A block of mass 0.15 kg is fixed to the end of a horizontal steel strip clamped at its other end. The block is moved sideways by measured amounts and the resultant horizontal force on it is recorded. The graph shows the results.
Explain how the graph shows that the block will move with simple harmonic motion when released.
Use the graph to determine the constant k in F = −kx.
Calculate the period of the oscillation.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the graph is a straight line through the origin, so the force (and acceleration) is proportional to the displacement | ✓ 1 | |
| the gradient is negative, so the force is always opposite to the displacement / towards the equilibrium position | ✓ 1 | |
| Part (b) | ||
| k = 3.6 ÷ 0.060 = 60 N m−1 | ✓ 1 | |
| Part (c) | ||
| T = 2π√(m/k) = 2π√(0.15 ÷ 60) | ✓ 1 | |
| = 0.31 s (0.314 s) | ✓ 1 | |
Exam tip: Any system with F = −kx has T = 2π√(m/k), whether the force comes from a coil spring, a steel strip or a pair of elastic cords.
A hacksaw blade is clamped to a bench with a small mass fixed to its free end. The free end is plucked and vibrates from side to side. The graph shows how the displacement of the free end varies with time.
Use the graph to determine the period and the frequency of the vibration.
Calculate the maximum speed of the free end.
State a time shown on the graph at which the velocity is negative and greatest in size. Explain your answer.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| T = 0.40 s (e.g. peaks at 0.1 s and 0.5 s) | ✓ 1 | |
| f = 1/T = 2.5 Hz | ✓ 1 | |
| Part (b) | ||
| ω = 2π ÷ 0.40 = 15.7 rad s−1; amplitude from the graph 15 mm | ✓ 1 | |
| vmax = Aω = 0.015 × 15.71 = 0.24 m s−1 (0.236) | ✓ 1 | |
| Part (c) | ||
| t = 0.2 s (or 0.6 s or 1.0 s) | ✓ 1 | |
| velocity is the gradient of the displacement–time graph, which is steepest and negative where the curve crosses the time axis going downwards | ✓ 1 | |
Common error: Reading the amplitude as the full peak-to-trough height. The amplitude is measured from the equilibrium line: 15 mm here, not 30 mm.
A small, heavy lantern hangs on a light chain and swings gently, as shown. Treat it as a simple pendulum of length 0.80 m. g = 9.81 N kg−1.
Calculate the period of the swing.
Calculate the amplitude of the motion, measured along the arc.
Calculate the maximum speed of the lantern.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| T = 2π√(l/g) = 2π√(0.80 ÷ 9.81) | ✓ 1 | |
| = 1.8 s (1.79 s) | ✓ 1 | |
| Part (b) | ||
| A = lθ = 0.80 × (6 × π ÷ 180) = 0.084 m (0.0838 m) | ✓ 1 | |
| Part (c) | ||
| ω = 2π ÷ 1.794 = 3.50 rad s−1 | ✓ 1 | |
| vmax = Aω = 0.0838 × 3.502 = 0.29 m s−1 (0.293) | ✓ 1 | |
Exam tip: Arc length = radius × angle only when the angle is in radians. Convert 6° first: 6 × π/180 = 0.105 rad.
The tip of a vibrating ruler moves with simple harmonic motion of amplitude 0.040 m and period 0.50 s.
What is the maximum speed of the tip?
| Answer | Mark | Notes |
|---|---|---|
| ω = 2π/T = 2π ÷ 0.50 = 12.6 rad s−1. vmax = Aω = 0.040 × 12.57 = 0.50 m s−1. | ✓ 1 | Answer B |
Answer: B
Exam tip: Maximum speed is Aω, reached at the equilibrium position. Maximum acceleration, Aω2, has a different unit — check which one the question asks for.
A trolley of mass 0.50 kg is attached to a spring of spring constant 80 N m−1 and oscillates on a smooth horizontal track.
What is the period of the oscillation?
| Answer | Mark | Notes |
|---|---|---|
| T = 2π√(m/k) = 2π√(0.50 ÷ 80) = 0.50 s (0.497 s). | ✓ 1 | Answer C |
Answer: C
Exam tip: Sense-check: a heavier mass oscillates more slowly, so m must be on top inside the square root.
A simple pendulum swinging through a small angle has a period of 1.60 s. Its length is reduced to one quarter of its original value.
What is the new period?
| Answer | Mark | Notes |
|---|---|---|
| T ∝ √l, so a quarter of the length gives √(1/4) = 1/2 of the period: 0.800 s. | ✓ 1 | Answer B |
Answer: B
Exam tip: Write the proportionality before calculating: T ∝ √l. A factor of 4 in length is a factor of 2 in period.
A component on a test rig oscillates with simple harmonic motion of period 0.25 s.
What is the magnitude of its acceleration when its displacement is 0.015 m?
| Answer | Mark | Notes |
|---|---|---|
| ω = 2π ÷ 0.25 = 25.1 rad s−1. |a| = ω2x = 25.132 × 0.015 = 9.5 m s−2 (9.47). | ✓ 1 | Answer D |
Answer: D
Common error: Using ω instead of ω2. Check the unit: ωx has the unit of speed, not acceleration.
A displacement–time graph for an oscillating mass is a sinusoidal curve with peaks at +30 mm and troughs at −30 mm. The graph shows exactly 5 complete cycles in 2.0 s.
What is the maximum speed of the mass?
| Answer | Mark | Notes |
|---|---|---|
| T = 2.0 ÷ 5 = 0.40 s, so ω = 2π ÷ 0.40 = 15.7 rad s−1. vmax = Aω = 0.030 × 15.71 = 0.47 m s−1. | ✓ 1 | Answer D |
Answer: D
Exam tip: From a graph, find the period from several cycles (here 2.0 s ÷ 5) — it is more precise than reading one cycle.
The end of a diving board vibrates vertically with simple harmonic motion of amplitude 0.12 m and frequency 1.25 Hz after a diver leaves it. Take t = 0 when the end of the board is at its highest point, and upwards as positive.
Calculate the angular frequency.
Calculate the displacement of the end of the board at t = 0.30 s.
Calculate the velocity of the end of the board at t = 0.30 s.
State and explain the direction of the acceleration of the end of the board at t = 0.30 s.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ω = 2π × 1.25 = 7.85 rad s−1 | ✓ 1 | |
| Part (b) | ||
| x = 0.12 cos(7.854 × 0.30) = 0.12 cos(2.356 rad) | ✓ 1 | |
| = −0.085 m (−0.0849 m), i.e. below equilibrium | ✓ 1 | |
| Part (c) | ||
| v = −0.12 × 7.854 × sin(2.356 rad) | ✓ 1 | |
| = −0.67 m s−1 (−0.666), i.e. moving downwards | ✓ 1 | |
| Part (d) | ||
| upwards, because the displacement is negative (below equilibrium) and a = −ω2x is always opposite to the displacement | ✓ 1 | |
Model sketch
Exam tip: Set the calculator to radians before using x = A cos ωt. In degrees, cos 2.356° ≈ 1 and the board would appear not to have moved.
A displacement–time graph for a small mass on a spring is a sine curve. At t = 0 the mass passes through its equilibrium position moving in the positive direction. The amplitude is 4.0 cm and the period is 0.60 s.
State the times during the first period (from t = 0 to t = 0.60 s) at which the speed of the mass is greatest.
Explain why the velocity of the mass is zero at t = 0.15 s.
Calculate the maximum speed of the mass.
Describe the velocity–time graph for the mass over the first period, and state the phase difference between the acceleration and the displacement.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| t = 0, 0.30 s and 0.60 s — where the displacement–time graph crosses the time axis and its gradient is steepest | ✓ 1 | |
| Part (b) | ||
| the mass is at its maximum displacement / a peak of the graph, where the gradient of the displacement–time graph is zero | ✓ 1 | |
| Part (c) | ||
| ω = 2π ÷ 0.60 = 10.5 rad s−1 | ✓ 1 | |
| vmax = 0.040 × 10.47 = 0.42 m s−1 | ✓ 1 | |
| Part (d) | ||
| a cosine curve starting at its maximum value +0.42 m s−1 at t = 0, with the same period 0.60 s, reaching zero at 0.15 s and −0.42 m s−1 at 0.30 s | ✓ 1 | |
| acceleration and displacement are in antiphase / π rad (180°) apart | ✓ 1 | |
Model sketch
Exam tip: Get the velocity–time graph from the gradient of the displacement–time graph: zero gradient → zero velocity, steepest gradient → maximum speed, with the sign of the slope.
A student investigates how the period T of a mass–spring system depends on the mass m. She hangs slotted masses from 0.100 kg to 0.500 kg on a spring, times 10 vertical oscillations for each and calculates T. She plots a graph of T2 (y-axis) against m (x-axis).
Explain why she plots T2 against m.
The gradient of her line is 0.790 s2 kg−1. Calculate the spring constant.
The line has a small positive intercept on the T2 axis. Suggest a reason for this.
For the 0.300 kg mass, 10 oscillations took 4.9 s with an uncertainty of ±0.1 s. Calculate the percentage uncertainty in her value of T2.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| squaring T = 2π√(m/k) gives T2 = (4π2/k)m | ✓ 1 | |
| so T2 against m is a straight line (through the origin) with gradient 4π2/k | ✓ 1 | |
| Part (b) | ||
| k = 4π2 ÷ 0.790 | ✓ 1 | |
| = 50.0 N m−1 | ✓ 1 | |
| Part (c) | ||
| the spring itself has mass, part of which also oscillates, so the effective oscillating mass is larger than m | ✓ 1 | |
| Part (d) | ||
| % uncertainty in T = 0.1 ÷ 4.9 × 100 = 2.0% | ✓ 1 | |
| % uncertainty in T2 = 2 × 2.0 = 4.1% | ✓ 1 | |
Exam tip: When a measured quantity is squared, double its percentage uncertainty. Timing many oscillations keeps that percentage small.
A motion sensor below a mass oscillating on a vertical spring records its velocity. The velocity–time graph is sinusoidal, with peaks at +0.36 m s−1 and troughs at −0.36 m s−1. The time between successive peaks is 0.90 s.
Calculate the amplitude of the oscillation.
Calculate the maximum acceleration of the mass.
State the position of the mass at the instants when the gradient of the velocity–time graph is zero.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ω = 2π ÷ 0.90 = 6.98 rad s−1 | ✓ 1 | |
| A = vmax/ω = 0.36 ÷ 6.98 = 0.052 m (0.0516 m) | ✓ 1 | |
| Part (b) | ||
| amax = Aω2 = vmaxω | ✓ 1 | |
| = 0.36 × 6.98 = 2.5 m s−2 (2.51) | ✓ 1 | |
| Part (c) | ||
| at the equilibrium position — the gradient is the acceleration, which is zero only where x = 0 | ✓ 1 | |
Exam tip: A peak of the velocity–time graph gives Aω directly; divide by ω to get the amplitude.
The graph shows how the displacement of a small mass on a spring varies with time.
What is the maximum acceleration of the mass?
| Answer | Mark | Notes |
|---|---|---|
| From the graph, A = 2.0 cm = 0.020 m and T = 0.80 s (peak to peak). ω = 2π ÷ 0.80 = 7.85 rad s−1. amax = Aω2 = 0.020 × 7.8542 = 1.2 m s−2 (1.23). | ✓ 1 | Answer A |
Answer: A
Exam tip: Read the period between two points in the same state, such as two successive peaks. The time from a peak to the next trough is only half a period.
The top graph shows how the displacement of an oscillating object varies with time over one period T.
Which of the graphs 1 to 4 shows how the velocity v of the object varies with time over the same period?
| Answer | Mark | Notes |
|---|---|---|
| Velocity is the gradient of the displacement–time graph. At t = 0 the gradient is zero; between 0 and T/2 it is positive, steepest at T/4; between T/2 and T it is negative, steepest at 3T/4. With x = −A cos ωt, v = +Aω sin ωt: graph 4. | ✓ 1 | Answer D |
Answer: D
Exam tip: Sketch velocity from displacement point by point: zero gradient gives zero velocity, steepest uphill gives maximum positive velocity.
Different masses are hung on the same spring and the period of vertical oscillation is measured for each. The graph shows how the period T varies with the mass m.
What is the spring constant of the spring?
| Answer | Mark | Notes |
|---|---|---|
| Use the labelled point. T = 2π√(m/k) gives k = 4π2m/T2 = 4π2 × 0.40 ÷ 0.502 = 63 N m−1 (63.2). | ✓ 1 | Answer D |
Answer: D
Common error: Squaring only one side when rearranging T = 2π√(m/k). Square the whole equation first: T2 = 4π2m/k.
The graph shows how the acceleration a of a vibrating component varies with its displacement x.
What is the frequency of the vibration?
| Answer | Mark | Notes |
|---|---|---|
| Gradient = −8.0 ÷ 0.0050 = −1600 s−2 = −ω2, so ω = 40 rad s−1. f = ω/2π = 40 ÷ 2π = 6.4 Hz (6.37). | ✓ 1 | Answer B |
Answer: B
Exam tip: The gradient of an a–x graph is −ω2. Take the square root to get ω, then divide by 2π for the frequency.
The graph shows how the velocity of a glider on an air track varies with time. The glider is attached between two springs and moves with simple harmonic motion.
What is the amplitude of the oscillation?
| Answer | Mark | Notes |
|---|---|---|
| From the graph, vmax = 0.30 m s−1 and T = 1.2 s. ω = 2π ÷ 1.2 = 5.24 rad s−1. A = vmax/ω = 0.30 ÷ 5.236 = 0.057 m (0.0573). | ✓ 1 | Answer D |
Answer: D
Exam tip: The peak of a velocity–time graph is Aω, not the amplitude. Find ω from the period on the time axis, then divide.
The graph shows how the displacements of two oscillators, P and Q, vary with time. Both have the same period.
What is the phase difference between P and Q?
| Answer | Mark | Notes |
|---|---|---|
| Period T = 1.6 s (P peaks at 0.4 s and 2.0 s). Q peaks 0.2 s after P. Phase difference = 2π × 0.2 ÷ 1.6 = π/4 = 0.785 rad. | ✓ 1 | Answer C |
Answer: C
Exam tip: Phase difference = 2π × (time shift ÷ period). Measure the time shift between matching points, such as two peaks, and the period on the same graph.
An accelerometer is fixed to the platform of a laboratory shaker table, which moves up and down with simple harmonic motion. The graph shows how the acceleration of the platform varies with time. Upwards is positive.
Use the graph to determine the angular frequency of the motion.
Calculate the amplitude of the motion.
Calculate the maximum speed of the platform.
State the displacement of the platform at t = 0, and state the two times between 0 and 0.40 s at which its speed is greatest.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| T = 0.40 s | ✓ 1 | |
| ω = 2π ÷ 0.40 = 15.7 rad s−1 | ✓ 1 | |
| Part (b) | ||
| amax = Aω2, with amax = 2.4 m s−2 from the graph | ✓ 1 | |
| A = 2.4 ÷ 15.712 = 9.7 × 10−3 m (9.7 mm) | ✓ 1 | |
| Part (c) | ||
| vmax = amax/ω = 2.4 ÷ 15.71 = 0.15 m s−1 (0.153) | ✓ 1 | |
| Part (d) | ||
| +9.7 mm, i.e. at the top of its motion, because a = −ω2x and the acceleration has its most negative value | ✓ 1 | |
| t = 0.10 s and 0.30 s, where the acceleration is zero so the platform is at the equilibrium position | ✓ 1 | |
Exam tip: Acceleration and displacement are in antiphase: a trough on the a–t graph is a peak of displacement, and zero acceleration means the equilibrium position, where the speed is greatest.
A peg is fixed to the rim of a turntable that rotates at a constant 45 revolutions per minute. Parallel light casts a shadow of the peg on a screen, as shown in the top view. The position x of the shadow is measured from the point on the screen directly opposite the centre of the turntable.
Explain why the shadow moves with simple harmonic motion.
Calculate the angular frequency and the period of the motion of the shadow.
Calculate the maximum speed of the shadow.
For the position shown, state the displacement of the shadow and calculate the size of its acceleration.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the radius to the peg turns through angle θ = ωt, so the shadow position is x = r cos ωt | ✓ 1 | |
| this is the SHM equation x = A cos ωt, with amplitude A = r and the same ω as the turntable | ✓ 1 | |
| Part (b) | ||
| ω = 45 × 2π ÷ 60 = 4.71 rad s−1 | ✓ 1 | |
| T = 60 ÷ 45 = 1.33 s | ✓ 1 | |
| Part (c) | ||
| vmax = Aω = 0.150 × 4.712 = 0.71 m s−1 (0.707) | ✓ 1 | |
| Part (d) | ||
| x = 0.150 cos 60° = 0.075 m (7.5 cm, as on the scale) | ✓ 1 | |
| |a| = ω2x = 4.7122 × 0.075 = 1.7 m s−2 (1.67) | ✓ 1 | |
| (the acceleration is directed towards the zero of the scale) | — | |
Exam tip: Uniform circular motion seen side-on is SHM. The amplitude equals the radius and ω of the oscillation equals the angular speed of the rotation.
A mass of 0.200 kg is hung on a spring beside a vertical ruler. The diagram shows the ruler readings for the end of the spring when it is unstretched, when the mass hangs at rest, and when the mass has been pulled down ready to be released. The mass then oscillates vertically with simple harmonic motion. g = 9.81 N kg−1.
Calculate the spring constant.
Calculate the period of the oscillations.
Calculate the maximum speed of the mass.
State the ruler reading for the end of the spring when the mass is at the top of its motion.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| extension at equilibrium = 24.0 − 15.0 = 9.0 cm = 0.090 m | ✓ 1 | |
| k = mg/Δl = 0.200 × 9.81 ÷ 0.090 = 22 N m−1 (21.8) | ✓ 1 | |
| Part (b) | ||
| T = 2π√(m/k) = 2π√(0.200 ÷ 21.8) | ✓ 1 | |
| = 0.60 s (0.602 s) | ✓ 1 | |
| Part (c) | ||
| amplitude = 28.0 − 24.0 = 4.0 cm = 0.040 m | ✓ 1 | |
| vmax = Aω = 0.040 × 2π ÷ 0.602 = 0.42 m s−1 (0.418) | ✓ 1 | |
| Part (d) | ||
| 24.0 − 4.0 = 20.0 cm | ✓ 1 | |
Common error: Measuring the amplitude from the unstretched position. The mass oscillates about the equilibrium position with the mass attached (24.0 cm here).
A data logger records the displacement x and velocity v of a glider oscillating between two springs on an air track. The graph shows v plotted against x for one complete oscillation.
Use the graph to state the amplitude and the maximum speed of the glider.
Calculate the period of the oscillation.
Calculate the maximum acceleration of the glider.
State the displacements at which the acceleration of the glider is greatest, and explain your answer.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| amplitude = 0.080 m (where v = 0) | ✓ 1 | |
| maximum speed = 0.40 m s−1 (at x = 0) | ✓ 1 | |
| Part (b) | ||
| ω = vmax/A = 0.40 ÷ 0.080 = 5.0 rad s−1 | ✓ 1 | |
| T = 2π ÷ 5.0 = 1.3 s (1.26 s) | ✓ 1 | |
| Part (c) | ||
| amax = Aω2 = 0.080 × 5.02 = 2.0 m s−2 | ✓ 1 | |
| Part (d) | ||
| at x = ±0.080 m (where the loop crosses the x-axis), because a = −ω2x is largest in size where the displacement is largest | ✓ 1 | |
Exam tip: On a v–x graph the speed is zero at the ends of the motion and greatest at x = 0. The ratio of the two intercepts gives ω directly.
An object moving with simple harmonic motion is at its maximum positive displacement at time t = 0. The period is T.
Which describes the object at t = 3T/4?
| Answer | Mark | Notes |
|---|---|---|
| With x = A cos ωt and ωt = 3π/2: x = A cos(3π/2) = 0 and v = −Aω sin(3π/2) = +Aω. The object is at equilibrium, moving in the positive direction at maximum speed. | ✓ 1 | Answer A |
Answer: A
Exam tip: Step through the cycle in quarters from the start: +A (rest) → 0 (moving −) → −A (rest) → 0 (moving +) → +A.
A mass hangs at rest from a light spring, which it extends by 0.12 m. The mass is pulled down slightly and released, so that it oscillates vertically. g = 9.81 N kg−1.
What is the period of the oscillation?
| Answer | Mark | Notes |
|---|---|---|
| At equilibrium kΔl = mg, so m/k = Δl/g. T = 2π√(0.12 ÷ 9.81) = 0.69 s (0.695 s). | ✓ 1 | Answer C |
Answer: C
Exam tip: When neither m nor k is given, look for the static extension: m/k = Δl/g.
The body of a small car, of mass 1200 kg, is supported on four identical springs, one at each wheel. When disturbed, the body bounces vertically with simple harmonic motion. The designer wants the frequency of this motion to be 1.4 Hz. Treat the four springs as one combined spring.
Calculate the combined spring constant needed.
Calculate the spring constant of each spring.
Four passengers of total mass 320 kg get into the car. Calculate the new frequency of the bouncing motion.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| f = 1/T and T = 2π√(m/k), so k = 4π2f2m | ✓ 1 | |
| = 4π2 × 1.42 × 1200 | ✓ 1 | |
| = 9.3 × 104 N m−1 | ✓ 1 | |
| Part (b) | ||
| 9.29 × 104 ÷ 4 = 2.3 × 104 N m−1 | ✓ 1 | |
| Part (c) | ||
| f ∝ 1/√m: f = 1.4 × √(1200 ÷ 1520) | ✓ 1 | |
| = 1.2 Hz (1.24 Hz) | ✓ 1 | |
Exam tip: A ratio avoids recalculating k: the springs are unchanged, so f ∝ 1/√m.
A simple pendulum consists of a bob of mass m on a light, inextensible string of length l. When the string makes an angle θ with the vertical, the bob is a distance x, measured along its arc, from the equilibrium position.
State the equation that defines simple harmonic motion, identifying each symbol.
Explain why the pendulum moves with simple harmonic motion only when the amplitude is small, and why its period does not depend on the mass of the bob.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a = −ω2x, where a is the acceleration, x the displacement from equilibrium and ω the angular frequency | ✓ 1 | |
| Part (b) | ||
| Indicative content, one mark per point: | — | |
| the restoring force is the component of the weight at right angles to the string, mg sin θ; tension has no component along the arc | ✓ 1 | |
| the displacement along the arc is x = lθ, with θ in radians | ✓ 1 | |
| for small angles sin θ ≈ θ, so the restoring force ≈ mgθ = (mg/l)x, proportional to x and opposite to it | ✓ 1 | |
| so a = −(g/l)x, which is SHM with ω2 = g/l | ✓ 1 | |
| at larger angles sin θ is noticeably less than θ, so the restoring force is not proportional to x and the motion is not SHM | ✓ 1 | |
| the mass appears in both the restoring force and in a = F/m, so it cancels and T = 2π√(l/g) does not contain m | ✓ 1 | |
Exam tip: In a long explanation, write a chain of linked statements: force → approximation → proportionality → SHM. Each link is a separate point.
A sorting screen in a quarry vibrates vertically with simple harmonic motion of amplitude 3.0 mm. Small stones rest on the screen without being attached to it. As the frequency of the vibration is slowly increased, the stones begin to bounce. g = 9.81 m s−2.
Explain why the stones lose contact with the screen when its maximum acceleration exceeds g.
Calculate the frequency at which the stones first begin to lose contact.
The frequency is increased to 12 Hz with the same amplitude. Calculate the displacement of the screen at which a stone first loses contact, and state whether this is above or below the equilibrium position.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the only downward force on a stone is its weight, so its downward acceleration cannot exceed g (the screen can push up but cannot pull down) | ✓ 1 | |
| when the screen accelerates downwards at more than g, the contact force would have to be negative, so it becomes zero and the stone leaves the screen | ✓ 1 | |
| Part (b) | ||
| Aω2 = g → ω = √(9.81 ÷ 0.0030) = 57.2 rad s−1 | ✓ 1 | |
| f = ω/2π = 9.1 Hz | ✓ 1 | |
| Part (c) | ||
| ω = 2π × 12 = 75.4 rad s−1 | ✓ 1 | |
| ω2x = g → x = 9.81 ÷ 75.42 = 1.7 × 10−3 m (1.73 mm) | ✓ 1 | |
| above the equilibrium position (where the acceleration of the screen is downwards) | ✓ 1 | |
Model sketch
Common error: Placing the loss of contact at the bottom of the motion. There the screen's acceleration is upwards, pressing the stone on; the downward acceleration is largest at the top.
A small heavy bob on a string of length 1.20 m swings with a small amplitude. A peg is fixed 0.90 m vertically below the point of suspension O, as shown. Whenever the bob is on the right of the vertical, the string catches on the peg. g = 9.81 N kg−1.
What is the time for one complete oscillation?
| Answer | Mark | Notes |
|---|---|---|
| Each side takes half a period of the pendulum that is swinging there. Left: ½ × 2π√(1.20 ÷ 9.81) = 1.099 s. Right: ½ × 2π√(0.30 ÷ 9.81) = 0.549 s. Total = 1.65 s (1.648). | ✓ 1 | Answer C |
Answer: C
Exam tip: When the length changes part-way through a swing, add half-periods: each side of the vertical takes half the period of the pendulum that is actually swinging there.
The graph shows how the displacement of a mooring buoy bobbing in a harbour varies with time. Point X is labelled on the graph.
What is the acceleration of the buoy at X?
| Answer | Mark | Notes |
|---|---|---|
| T = 2.0 s, so ω = 2π ÷ 2.0 = π rad s−1. At X, x = +3.0 cm = +0.030 m. a = −ω2x = −π2 × 0.030 = −0.30 m s−2 (−0.296). | ✓ 1 | Answer B |
Answer: B
Common error: Dropping the minus sign in a = −ω2x. When the displacement is positive the acceleration is negative, directed back towards equilibrium.
A student uses a simple pendulum to measure g. For each length she times 20 small oscillations and calculates the period T. She measures the length ℓ of the string from the point of suspension to the top of the spherical bob, not to its centre. The graph shows ℓ against T2 with her best-fit line.
The length to the centre of the bob is L = ℓ + r, where r is the radius of the bob. Use T = 2π√(L/g) to explain why the graph of ℓ against T2 is a straight line.
Use the graph to determine a value for g.
Use the graph to determine the diameter of the bob.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| T2 = 4π2(ℓ + r)/g, so ℓ = (g/4π2)T2 − r | ✓ 1 | |
| this has the form y = mx + c, with gradient g/4π2 and intercept −r (both constant) | ✓ 1 | |
| Part (b) | ||
| gradient = (0.980 − (−0.020)) ÷ 4.00 = 0.250 m s−2 | ✓ 1 | |
| g = 4π2 × gradient | ✓ 1 | |
| = 9.87 m s−2 (allow 9.7–10.0 from a large triangle) | ✓ 1 | |
| Part (c) | ||
| intercept on the ℓ axis = −0.020 m = −r, so r = 0.020 m | ✓ 1 | |
| diameter = 2r = 0.040 m (4.0 cm) | ✓ 1 | |
Exam tip: A systematic error in a length shifts the line up or down but leaves the gradient unchanged, so a value found from the gradient is not affected by it.
A test tube of external diameter 2.0 cm is weighted with sand so that it floats upright in water, as shown. The total mass of the tube and sand is 0.040 kg. The tube is pushed down a small distance and released, and it bobs up and down. Density of water = 1000 kg m−3; g = 9.81 N kg−1. Assume the water level in the beaker does not change, and ignore drag.
Show that the depth d of the base of the tube below the surface at equilibrium is about 0.13 m.
Explain why the tube moves with simple harmonic motion when released.
Show that ω2 = g/d, and calculate the period of the oscillation.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| at equilibrium upthrust = weight: ρAdg = mg, with A = π × 0.0102 = 3.14 × 10−4 m2 | ✓ 1 | |
| d = 0.040 ÷ (1000 × 3.14 × 10−4) = 0.127 m | ✓ 1 | |
| Part (b) | ||
| Indicative content, one mark per point: | — | |
| when the tube is pushed down a further distance x, an extra volume Ax of water is displaced | ✓ 1 | |
| the upthrust increases by ρAgx while the weight is unchanged, so the resultant force is ρAgx upwards | ✓ 1 | |
| this is proportional to x and directed towards the equilibrium position: F = −ρAgx (the same holds above equilibrium, with the force downwards) | ✓ 1 | |
| so a = −(ρAg/m)x, which is SHM with ω2 = ρAg/m | ✓ 1 | |
| Part (c) | ||
| from equilibrium ρA/m = 1/d, so ω2 = ρAg/m = g/d | ✓ 1 | |
| T = 2π√(d/g) = 2π√(0.127 ÷ 9.81) = 0.72 s (0.716 s) | ✓ 1 | |
Exam tip: To prove SHM, find the resultant force at a general displacement x and show it has the form F = −(constant) × x. Then ω2 = constant ÷ mass.
A fishing float bobs up and down with simple harmonic motion on a calm lake. The graph shows how its displacement x varies with time t. A tangent has been drawn to the curve at point P, where t = 0.25 s.
Use the graph to determine the amplitude and the angular frequency of the motion.
Use the tangent to determine the velocity of the float at P.
Use the equation v = −Aω sin ωt to calculate the velocity at P, and comment on your answer to (b).
Calculate the acceleration of the float at P.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A = 50 mm = 0.050 m | ✓ 1 | |
| T = 2.0 s, so ω = 2π ÷ 2.0 = 3.1 rad s−1 (π rad s−1) | ✓ 1 | |
| Part (b) | ||
| gradient = (2 − 63) mm ÷ (0.55 − 0.00) s = −111 mm s−1 | ✓ 1 | |
| v = −0.11 m s−1 (allow −0.10 to −0.12) | ✓ 1 | |
| Part (c) | ||
| v = −0.050 × π × sin(π × 0.25) = −0.050 × 3.142 × 0.707 = −0.11 m s−1 (−0.111) | ✓ 1 | |
| this agrees with the value from the tangent (to within the precision of drawing and reading the tangent) | ✓ 1 | |
| Part (d) | ||
| x = 0.050 cos(π × 0.25) = 0.0354 m | ✓ 1 | |
| a = −ω2x = −π2 × 0.0354 = −0.35 m s−2 (−0.349) | ✓ 1 | |
Exam tip: Work out ωt in radians: here ωt = π × 0.25 = π/4 rad. Using degrees gives sin 0.785° and a velocity about 50 times too small.
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