A Level Physics 17.1 Simple Harmonic Oscillations MCQs – 30 Practice Questions with Answers (Cambridge 9702)

These 30 original practice MCQs cover Cambridge AS & A Level Physics (9702) Section 17.1, Simple Harmonic Oscillations – an A Level (A2) topic in Topic 17, Oscillations. They test the defining equation a = −ω2x, the displacement, velocity and acceleration equations for SHM, and how to use them in calculations and with graphs.

Formula recap (full list on the A Level Physics Formula Sheet):

  • a = −ω2x, where ω = 2πf = 2π/T
  • x = x0 cos(ωt) or x = x0 sin(ωt), depending on the starting position
  • vmax = ωx0 and amax = ω2x0
  • v = ω√(x02 − x2)

Defining Simple Harmonic Motion

Q1. Which statement correctly defines simple harmonic motion?

  • (A) The acceleration is constant in magnitude and always directed towards a fixed point.
  • (B) The acceleration is proportional to the displacement from a fixed point and is always directed towards that point.
  • (C) The acceleration is proportional to the displacement from a fixed point and is always directed away from that point.
  • (D) The velocity is proportional to the displacement from a fixed point and is always directed towards that point.
Show Answer

Answer: B – SHM is defined by a = −ω2x: the acceleration is proportional to the displacement x from a fixed (equilibrium) point, and the minus sign shows it is always directed back towards that point.

Q2. A mass on a spring oscillates in SHM because the resultant force on the mass obeys which condition?

  • (A) F is proportional to the velocity of the mass.
  • (B) F is constant in magnitude throughout the motion.
  • (C) F is proportional to the displacement and opposite in direction to it, F = −kx.
  • (D) F is proportional to the square of the displacement.
Show Answer

Answer: C – Combining F = ma with a = −ω2x shows the resultant force must be proportional to the displacement and act towards the equilibrium position, F = −kx. This restoring force is what produces SHM.

Q3. A simple pendulum swings with a small amplitude. Its motion is approximately simple harmonic because

  • (A) for small angles, the component of the weight acting as the restoring force is approximately proportional to the displacement.
  • (B) air resistance on the bob is negligible.
  • (C) the string is inextensible.
  • (D) the period of the pendulum does not depend on its length.
Show Answer

Answer: A – For small angles, sin(θ) is approximately equal to θ, so the restoring component of the weight is approximately proportional to the displacement and directed towards the equilibrium position – the condition for SHM.

Q4. The amplitude of an oscillation is

  • (A) the distance between the two extreme positions.
  • (B) the displacement of the object at a particular instant.
  • (C) the total distance travelled in one complete oscillation.
  • (D) the maximum displacement of the object from its equilibrium position.
Show Answer

Answer: D – The amplitude x0 is the maximum displacement from the equilibrium position. The distance between the two extremes is 2x0, and the distance travelled in one full cycle is 4x0.

Q5. Two oscillators have a phase difference of π rad. This means they are

  • (A) exactly in phase with each other.
  • (B) in antiphase: when one is at its positive extreme, the other is at its negative extreme.
  • (C) a quarter of a period apart in their cycles.
  • (D) three quarters of a period apart in their cycles.
Show Answer

Answer: B – One complete cycle corresponds to a phase angle of 2π rad, so a phase difference of π rad is half a cycle: the oscillators are in antiphase (in opposite states).

Q6. Oscillator B starts its cycle exactly one quarter of a period after oscillator A. The phase difference between them is

  • (A) π/6 rad
  • (B) 2π rad
  • (C) π/2 rad
  • (D) π rad
Show Answer

Answer: C – Phase difference in radians = (fraction of a cycle) x 2π. One quarter of a cycle gives (1/4) x 2π = π/2 rad.

Q7. An oscillation described as isochronous has the property that

  • (A) its period is independent of its amplitude.
  • (B) its period is the same whatever mass is oscillating.
  • (C) its frequency increases as its amplitude decreases.
  • (D) its amplitude remains constant in time.
Show Answer

Answer: A – For true SHM, ω is a constant of the system, so the period T = 2π/ω does not depend on the amplitude. This isochronous property is why pendulums and springs make good timekeepers.

Q8. Which of the following motions is NOT simple harmonic?

  • (A) A mass oscillating on the end of a spring.
  • (B) A pendulum swinging through small angles.
  • (C) A cork bobbing up and down on water.
  • (D) A basketball repeatedly bouncing on the floor.
Show Answer

Answer: D – While the ball is in the air its acceleration is simply g, constant in magnitude and direction, not proportional to its displacement from a fixed point. The other three all have a restoring effect approximately proportional to displacement.

Q9. An object in SHM is at its maximum displacement from equilibrium. At this instant

  • (A) its speed is maximum and its acceleration is zero.
  • (B) its speed is zero and its acceleration is zero.
  • (C) its speed is zero and its acceleration is maximum, directed towards the equilibrium position.
  • (D) its speed and its acceleration are both maximum.
Show Answer

Answer: C – At the extremes the object momentarily stops (v = 0) before returning. Since a = −ω2x, the acceleration has its greatest magnitude, amax = ω2x0, where |x| is greatest.

Q10. A cylindrical block floats vertically in water. It is pushed a short distance down and released. Its bobbing motion is approximately SHM because

  • (A) the block has a very small mass.
  • (B) the extra upthrust when it is pushed down is proportional to the extra depth submerged, so the resultant force towards the equilibrium position is proportional to the displacement.
  • (C) the upthrust on the block is constant.
  • (D) the period of bobbing does not depend on the block.
Show Answer

Answer: B – Pushing the block down by x increases the submerged volume, and hence the upthrust, by an amount proportional to x. The resultant force is therefore a restoring force proportional to the displacement, giving approximately SHM.

Equations of SHM

Q11. The displacement of an oscillator is given by x = x0 cos(ωt), with t = 0 when timing starts. At t = 0 the object is

  • (A) at the equilibrium position, moving in the positive direction.
  • (B) at its negative extreme position.
  • (C) at its positive extreme position (+x0), about to move back towards equilibrium.
  • (D) moving at its maximum speed.
Show Answer

Answer: C – cos(0) = 1, so at t = 0, x = +x0: the object starts from rest at its maximum positive displacement, which is why the cosine form is used for motion released from an extreme.

Q12. An oscillator is timed from the instant it passes through its equilibrium position moving in the positive direction. Its displacement is best described by

  • (A) x = x0 sin(ωt)
  • (B) x = x0 cos(ωt)
  • (C) x = x0 cos(ωt + π)
  • (D) x = −x0 cos(ωt)
Show Answer

Answer: A – sin(0) = 0, so x = 0 at t = 0, and the displacement initially increases in the positive direction, matching an object leaving equilibrium in the +x direction. The cosine forms start at an extreme.

Q13. An oscillator has amplitude 0.050 m and frequency 2.0 Hz. Its maximum speed is

  • (A) 0.10 m/s
  • (B) 0.20 m/s
  • (C) 0.31 m/s
  • (D) 0.63 m/s
Show Answer

Answer: D – ω = 2πf = 2π x 2.0 = 12.6 rad/s. vmax = ωx0 = 12.6 x 0.050 = 0.63 m/s, reached as the object passes through the equilibrium position.

Q14. An object performs SHM with angular frequency 10 rad/s and amplitude 0.040 m. Its maximum acceleration is

  • (A) 0.40 m/s2
  • (B) 4.0 m/s2
  • (C) 25 m/s2
  • (D) 40 m/s2
Show Answer

Answer: B – amax = ω2x0 = 102 x 0.040 = 100 x 0.040 = 4.0 m/s2, occurring at the extreme positions.

Q15. An oscillator has a period of 0.25 s. Its angular frequency ω is

  • (A) 0.79 rad/s
  • (B) 1.6 rad/s
  • (C) 25 rad/s
  • (D) 40 rad/s
Show Answer

Answer: C – ω = 2π/T = 2π/0.25 = 8π = 25 rad/s.

Q16. An object moves with SHM of amplitude 5.0 cm and angular frequency 10 rad/s. When its displacement is 3.0 cm, the magnitude of its velocity is

  • (A) 0.30 m/s
  • (B) 0.40 m/s
  • (C) 0.50 m/s
  • (D) 0.80 m/s
Show Answer

Answer: B – v = ω√(x02 − x2) = 10 x √(0.0502 − 0.0302) = 10 x √(0.0016) = 10 x 0.040 = 0.40 m/s.

Q17. For motion described by x = x0 cos(ωt), one quarter of a period after t = 0 the object is

  • (A) at x = +x0, at rest.
  • (B) at x = x0/2, moving outwards.
  • (C) at an extreme position with zero acceleration.
  • (D) at x = 0, moving at maximum speed through the equilibrium position.
Show Answer

Answer: D – At t = T/4 the phase ωt = (2π/T)(T/4) = π/2 rad, and cos(π/2) = 0. The object is at equilibrium, where its speed is the maximum, ωx0.

Q18. If x = x0 cos(ωt), the velocity of the object is given by

  • (A) v = −ωx0 sin(ωt)
  • (B) v = ωx0 cos(ωt)
  • (C) v = −ωx0 cos(ωt)
  • (D) v = ω2x0 sin(ωt)
Show Answer

Answer: A – Velocity is the rate of change of displacement: v = dx/dt = −ωx0 sin(ωt). Its maximum value, when sin(ωt) = −1, is ωx0.

Q19. An object in SHM is at a displacement equal to half its amplitude, x = x0/2. The magnitude of its acceleration at this point is

  • (A) zero.
  • (B) one quarter of its maximum value.
  • (C) one half of its maximum value.
  • (D) equal to its maximum value.
Show Answer

Answer: C – Since a = −ω2x, the acceleration is proportional to the displacement. At x = x0/2, |a| = ω2x0/2, exactly half of amax = ω2x0.

Q20. An oscillator has angular frequency ω = 20π rad/s. Its frequency is

  • (A) 3.2 Hz
  • (B) 10 Hz
  • (C) 20 Hz
  • (D) 63 Hz
Show Answer

Answer: B – ω = 2πf, so f = ω/(2π) = 20π/(2π) = 10 Hz.


SHM Calculations & Graphs

Q21. A displacement-time graph for an oscillator is a cosine curve with a peak displacement of 4.0 cm, and one complete cycle takes 0.80 s. The frequency of the oscillation is

  • (A) 1.25 Hz
  • (B) 0.80 Hz
  • (C) 1.6 Hz
  • (D) 5.0 Hz
Show Answer

Answer: A – The period T = 0.80 s, so f = 1/T = 1/0.80 = 1.25 Hz. (The peak value on the graph gives the amplitude, 4.0 cm.)

Q22. For the oscillator in Q21 (amplitude 4.0 cm, period 0.80 s), the maximum speed is

  • (A) 0.050 m/s
  • (B) 0.10 m/s
  • (C) 0.20 m/s
  • (D) 0.31 m/s
Show Answer

Answer: D – ω = 2π/T = 2π/0.80 = 7.85 rad/s. vmax = ωx0 = 7.85 x 0.040 = 0.31 m/s.

Q23. A mass of 0.50 kg hangs on a spring of spring constant 200 N/m and oscillates vertically. Using T = 2π√(m/k), its period is

  • (A) 0.050 s
  • (B) 0.31 s
  • (C) 0.63 s
  • (D) 3.1 s
Show Answer

Answer: B – T = 2π√(0.50/200) = 2π√(0.0025) = 2π x 0.050 = 0.31 s.

Q24. The mass on the spring in Q23 is replaced by one of four times the mass, on the same spring. The new period is

  • (A) one quarter of the original period.
  • (B) one half of the original period.
  • (C) twice the original period.
  • (D) unchanged.
Show Answer

Answer: C – T = 2π√(m/k) means T is proportional to √m. Increasing m by a factor of 4 increases T by √4 = 2, to about 0.63 s.

Q25. A simple pendulum has length 0.25 m. Using T = 2π√(l/g) with g = 9.81 m/s2, its period is approximately

  • (A) 0.16 s
  • (B) 0.50 s
  • (C) 0.63 s
  • (D) 1.0 s
Show Answer

Answer: D – T = 2π√(0.25/9.81) = 2π x 0.160 = 1.0 s.

Q26. The pendulum in Q25 is taken to the Moon, where the gravitational field strength is about one sixth of that on Earth. Its period there

  • (A) decreases by a factor of about 6.
  • (B) increases by a factor of about 2.4.
  • (C) is unchanged.
  • (D) decreases by a factor of about 2.4.
Show Answer

Answer: B – T = 2π√(l/g), so T is proportional to 1/√g. With g reduced to g/6, T increases by √6, which is about 2.4, to roughly 2.4 s.

Q27. An oscillator has displacement x = 6.0 cos(ωt) cm and period 1.2 s. Its displacement at t = 0.20 s is

  • (A) 6.0 cm
  • (B) 5.2 cm
  • (C) 3.0 cm
  • (D) 0 cm
Show Answer

Answer: C – The phase at t = 0.20 s is ωt = 2π x 0.20/1.2 = π/3 rad. x = 6.0 cos(π/3) = 6.0 x 0.50 = 3.0 cm.

Q28. An object moves with SHM of amplitude 0.13 m and angular frequency 6.0 rad/s. Its speed when its displacement is 0.05 m is

  • (A) 0.72 m/s
  • (B) 0.30 m/s
  • (C) 0.48 m/s
  • (D) 0.78 m/s
Show Answer

Answer: A – v = ω√(x02 − x2) = 6.0 x √(0.132 − 0.052) = 6.0 x √(0.0144) = 6.0 x 0.12 = 0.72 m/s.

Q29. An object performs SHM with angular frequency 8.0 rad/s. When its displacement from equilibrium is 0.030 m, the magnitude of its acceleration is

  • (A) 0.24 m/s2
  • (B) 0.75 m/s2
  • (C) 1.2 m/s2
  • (D) 1.9 m/s2
Show Answer

Answer: D – a = −ω2x, so |a| = 8.02 x 0.030 = 64 x 0.030 = 1.92, about 1.9 m/s2, directed towards the equilibrium position.

Q30. A “seconds pendulum” has a period of exactly 2.0 s. Using T = 2π√(l/g) with g = 9.81 m/s2, its length is approximately

  • (A) 0.99 m
  • (B) 0.25 m
  • (C) 1.6 m
  • (D) 4.0 m
Show Answer

Answer: A – Rearranging, l = gT2/(4π2) = 9.81 x 2.02/(4π2) = 39.2/39.5 = 0.99 m, roughly one metre – which is why seconds pendulums were about a metre long.

Keep practising: work through more free physics MCQs on the Physics MCQs hub, and keep the key equations at hand with the A Level Physics Formula Sheet.

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