A Level Physics 17.2 Energy in Simple Harmonic Motion MCQs – 30 Practice Questions with Answers (Cambridge 9702)
These 30 original practice MCQs cover Section 17.2, Energy in Simple Harmonic Motion, of the Cambridge International AS & A Level Physics (9702) syllabus. This is an A Level (A2) topic: it explains how kinetic and potential energy interchange during an oscillation, why the total energy depends on the square of the amplitude, and how to calculate the energy of an oscillator at any displacement.
Formula recap: total energy E = ½ m ω2 x02; kinetic energy at displacement x: Ek = ½ m ω2 (x02 − x2); potential energy at displacement x: Ep = ½ m ω2 x2; maximum speed vmax = ω x0. See the full A Level Physics Formula Sheet for every equation used in this topic.
Energy Interchange in SHM
Q1. A mass oscillates in simple harmonic motion. At which position is its kinetic energy maximum?
- (A) At x = +x0
- (B) At the equilibrium position, x = 0
- (C) At x = −x0
- (D) The kinetic energy is the same at every position
Show Answer
Answer: B – The oscillator moves fastest as it passes through the equilibrium position, so its kinetic energy is maximum at x = 0 and zero at the extremes of the motion.
Q2. For an oscillator in SHM, the potential energy is maximum when the displacement is
- (A) at its maximum values, x = +x0 and x = −x0
- (B) zero
- (C) half the amplitude
- (D) The potential energy does not change during SHM
Show Answer
Answer: A – Potential energy is stored when the system is displaced furthest from equilibrium. At x = ±x0 the oscillator momentarily stops, so all of its energy is potential energy.
Q3. A mass on a spring reaches its maximum displacement. At that instant its kinetic energy is
- (A) maximum
- (B) half of the total energy
- (C) zero
- (D) equal to its average kinetic energy
Show Answer
Answer: C – At maximum displacement the mass is momentarily at rest before moving back, so its speed and kinetic energy are both zero; the total energy is entirely potential energy there.
Q4. A pendulum bob swings in SHM. As it passes through the lowest point of its swing,
- (A) its gravitational potential energy is maximum and its kinetic energy is zero
- (B) its kinetic energy and gravitational potential energy are both maximum
- (C) its gravitational potential energy is maximum and its kinetic energy is maximum
- (D) its gravitational potential energy is minimum and its kinetic energy is maximum
Show Answer
Answer: D – At the lowest point the bob is at its smallest height, so its gravitational potential energy is minimum, and it is moving fastest, so its kinetic energy is maximum. Energy continually swaps between the two forms.
Q5. A mass on a horizontal spring passes through the equilibrium position. At that instant,
- (A) its speed and kinetic energy are maximum, and the elastic potential energy of the spring is minimum
- (B) its speed is zero and the spring stores maximum elastic potential energy
- (C) its acceleration and speed are both maximum
- (D) its kinetic energy and elastic potential energy are both zero
Show Answer
Answer: A – At equilibrium the spring is at its natural length (minimum stored elastic energy) and the mass is moving at its maximum speed, so the energy is entirely kinetic.
Q6. An oscillator performs SHM with no friction or resistive forces acting. Its total mechanical energy
- (A) increases steadily
- (B) remains constant
- (C) decreases steadily
- (D) is zero at the equilibrium position
Show Answer
Answer: B – With no damping, no energy is transferred away, so the total of kinetic + potential energy stays constant throughout the oscillation; only the share between the two forms changes.
Q7. A mass in SHM moves from the equilibrium position towards an extreme of the oscillation. During this part of the motion,
- (A) kinetic energy increases and potential energy decreases
- (B) both kinetic and potential energy increase
- (C) kinetic energy decreases and potential energy increases
- (D) both kinetic and potential energy stay constant
Show Answer
Answer: C – The restoring force opposes the motion, so the mass slows down: kinetic energy is transferred into potential energy until the mass stops at the extreme position.
Q8. At an intermediate displacement, halfway through a swing, an undamped oscillator has
- (A) only kinetic energy
- (B) only potential energy
- (C) no energy at all
- (D) both kinetic and potential energy, with their sum equal to the total energy
Show Answer
Answer: D – At any displacement between 0 and x0 the oscillator is moving (so it has kinetic energy) and is displaced (so it has potential energy). The two shares add up to the constant total energy.
Q9. During one complete oscillation of an undamped oscillator, the kinetic energy reaches its maximum value
- (A) twice
- (B) once
- (C) four times
- (D) never, because it is constant
Show Answer
Answer: A – The oscillator passes through the equilibrium position twice in each complete cycle (once in each direction), so its speed and kinetic energy peak twice per oscillation.
Q10. An oscillator performs SHM at a frequency of 3.0 Hz. At what frequency does its potential energy rise to a maximum?
- (A) 1.5 Hz
- (B) 6.0 Hz
- (C) 3.0 Hz
- (D) 9.0 Hz
Show Answer
Answer: B – Potential energy is maximum at both extremes of the swing, which are reached twice per oscillation. The energy therefore swaps back and forth at twice the frequency of the motion: 2 × 3.0 = 6.0 Hz.
Total Energy & Amplitude
Q11. The total energy of an undamped mass m performing SHM with angular frequency ω and amplitude x0 is
- (A) ½ m ω x02
- (B) ½ m ω2 x02
- (C) m ω2 x02
- (D) ½ m ω2 x0
Show Answer
Answer: B – The total energy equals the maximum kinetic energy, ½ m vmax2, and since vmax = ωx0, this gives E = ½ m ω2 x02.
Q12. The amplitude of an oscillator is doubled while its mass and frequency stay the same. Its total energy becomes
- (A) twice as large
- (B) eight times as large
- (C) four times as large
- (D) unchanged
Show Answer
Answer: C – Total energy is proportional to (amplitude)2. Doubling x0 multiplies E by 22 = 4.
Q13. The frequency of an oscillator is doubled, with the mass and amplitude unchanged. Its total energy becomes
- (A) four times as large
- (B) twice as large
- (C) half as large
- (D) unchanged
Show Answer
Answer: A – Since ω = 2πf, total energy is proportional to f2 as well as x02. Doubling f multiplies E by 22 = 4.
Q14. An oscillator has total energy E. Its amplitude is increased to three times the original value (same mass and frequency). The new total energy is
- (A) 3E
- (B) 6E
- (C) 27E
- (D) 9E
Show Answer
Answer: D – E ∝ x02, so tripling the amplitude multiplies the energy by 32 = 9.
Q15. For SHM of amplitude x0 and angular frequency ω, the kinetic energy of the mass at a displacement x is
- (A) ½ m ω2 (x02 − x2)
- (B) ½ m ω2 x2
- (C) ½ m ω2 x02
- (D) ½ m ω2 (x0 − x)2
Show Answer
Answer: A – Kinetic energy = total energy − potential energy = ½ m ω2 x02 − ½ m ω2 x2 = ½ m ω2 (x02 − x2). This is zero at x = ±x0 and maximum at x = 0, as expected.
Q16. For the same oscillator, the potential energy at a displacement x is
- (A) ½ m ω2 (x02 − x2)
- (B) ½ m ω2 x02
- (C) ½ m ω2 x2
- (D) ½ m ω x2
Show Answer
Answer: C – Potential energy grows with the square of the displacement, Ep = ½ m ω2 x2. It is zero at the equilibrium position and equals the total energy at x = ±x0.
Q17. During SHM, the kinetic energy and potential energy are equal when the displacement is
- (A) x0 / 2
- (B) x0 / √2 ≈ 0.71 x0
- (C) x0
- (D) zero
Show Answer
Answer: B – Setting ½ m ω2 x2 = ½ m ω2 (x02 − x2) gives x2 = x02 / 2, so x = x0 / √2 ≈ 0.71 x0. At x = x0/2 the potential energy is only a quarter of the total.
Q18. Which graph correctly shows how the potential energy of an SHM oscillator varies with displacement x between −x0 and +x0?
- (A) A straight line rising from 0 at x = −x0 to a maximum at x = +x0
- (B) An inverted parabola with its maximum at x = 0
- (C) A horizontal straight line
- (D) A parabola with its minimum (zero) at x = 0 and maxima at x = ±x0
Show Answer
Answer: D – Because Ep = ½ m ω2 x2, the graph is a U-shaped parabola, zero at x = 0 and equal to the total energy at both extremes. The kinetic energy graph is the inverted parabola, and the total energy graph is the horizontal line.
Q19. An oscillator is at a displacement equal to half its amplitude, x = x0 / 2. The potential energy at this point is what fraction of the total energy?
- (A) 1/4
- (B) 1/2
- (C) 3/4
- (D) 1/√2
Show Answer
Answer: A – Ep / E = x2 / x02 = (x0/2)2 / x02 = 1/4. The remaining 3/4 of the energy is kinetic energy at this displacement.
Q20. An oscillator has a total energy of 2.0 J at an amplitude of 0.10 m. The amplitude is reduced to 0.050 m (mass and frequency unchanged). The new total energy is
- (A) 1.0 J
- (B) 1.4 J
- (C) 0.50 J
- (D) 0.25 J
Show Answer
Answer: C – E ∝ x02. Halving the amplitude quarters the energy: 2.0 × (0.050/0.10)2 = 2.0 × 0.25 = 0.50 J.
Energy Calculations
Q21. A mass of 0.50 kg oscillates in SHM with frequency 2.0 Hz and amplitude 0.040 m. The total energy of the oscillation is
- (A) 0.032 J
- (B) 0.063 J
- (C) 0.13 J
- (D) 0.25 J
Show Answer
Answer: B – ω = 2πf = 2π × 2.0 = 12.6 rad s−1. E = ½ m ω2 x02 = 0.5 × 0.50 × (12.6)2 × (0.040)2 = 0.063 J.
Q22. For the oscillator in Q21, the maximum speed of the mass is
- (A) 0.13 m s−1
- (B) 0.25 m s−1
- (C) 0.50 m s−1
- (D) 1.0 m s−1
Show Answer
Answer: C – vmax = ω x0 = 12.6 × 0.040 = 0.50 m s−1. Check with energy: ½ m vmax2 = 0.5 × 0.50 × (0.50)2 = 0.063 J, matching the total energy in Q21.
Q23. A 0.20 kg mass performs SHM with period 0.50 s and amplitude 0.10 m. Its total energy is
- (A) 0.16 J
- (B) 0.32 J
- (C) 0.079 J
- (D) 1.6 J
Show Answer
Answer: A – ω = 2π / T = 2π / 0.50 = 12.6 rad s−1. E = ½ m ω2 x02 = 0.5 × 0.20 × (12.6)2 × (0.10)2 = 0.16 J.
Q24. An oscillator has total energy 0.40 J and amplitude 0.20 m. Its kinetic energy when the displacement is 0.10 m is
- (A) 0.10 J
- (B) 0.20 J
- (C) 0.40 J
- (D) 0.30 J
Show Answer
Answer: D – Potential energy at x = 0.10 m: Ep = E (x/x0)2 = 0.40 × (0.10/0.20)2 = 0.40 × 0.25 = 0.10 J. So Ek = E − Ep = 0.40 − 0.10 = 0.30 J.
Q25. A pendulum bob of mass 0.30 kg is released from rest at a point 0.050 m above its lowest position. The maximum speed of the bob is (g = 9.81 m s−2)
- (A) 0.50 m s−1
- (B) 0.70 m s−1
- (C) 1.4 m s−1
- (D) 0.99 m s−1
Show Answer
Answer: D – Loss of gravitational potential energy = gain of kinetic energy: mgh = ½ m v2, so v = √(2gh) = √(2 × 9.81 × 0.050) = √(0.981) = 0.99 m s−1. The mass cancels, so it is not needed.
Q26. A mass of 0.40 kg oscillates with angular frequency 10 rad s−1 and amplitude 0.060 m. The total energy is
- (A) 0.036 J
- (B) 0.072 J
- (C) 0.14 J
- (D) 0.29 J
Show Answer
Answer: B – E = ½ m ω2 x02 = 0.5 × 0.40 × (10)2 × (0.060)2 = 0.5 × 0.40 × 100 × 0.0036 = 0.072 J.
Q27. For the oscillator in Q26, the kinetic energy at a displacement of 0.030 m is
- (A) 0.018 J
- (B) 0.036 J
- (C) 0.072 J
- (D) 0.054 J
Show Answer
Answer: D – Ek = ½ m ω2 (x02 − x2) = 0.5 × 0.40 × 100 × (0.0602 − 0.0302) = 20 × (0.0036 − 0.0009) = 20 × 0.0027 = 0.054 J. (The potential energy there is 0.018 J, and 0.054 + 0.018 = 0.072 J, the total.)
Q28. A lightly damped oscillator has its amplitude reduced from 0.10 m to 0.070 m as energy is lost. The total energy becomes what fraction of its original value?
- (A) 0.49
- (B) 0.70
- (C) 0.30
- (D) 0.90
Show Answer
Answer: A – Because E ∝ x02, the energy fraction is (0.070/0.10)2 = 0.49. The amplitude has fallen to 70% of its value, but just under half of the energy remains. (Damping is treated fully in Section 17.3.)
Q29. An oscillator has amplitude 0.12 m. At what displacement are its kinetic and potential energies equal?
- (A) 0.060 m
- (B) 0.085 m
- (C) 0.12 m
- (D) 0.10 m
Show Answer
Answer: B – The energies are equal at x = x0 / √2 = 0.12 / 1.41 = 0.085 m from the equilibrium position.
Q30. A pendulum bob of mass 0.50 kg swings so that at the extremes of its motion it is 0.20 m higher than at the lowest point. The total energy of the oscillation is (g = 9.81 m s−2)
- (A) 0.20 J
- (B) 0.49 J
- (C) 0.98 J
- (D) 2.0 J
Show Answer
Answer: C – At the extremes the energy is entirely gravitational potential energy relative to the lowest point: E = mgh = 0.50 × 9.81 × 0.20 = 0.98 J. This same 0.98 J appears entirely as kinetic energy at the lowest point.
Keep practising: revise the whole topic on the Physics MCQs hub and keep the key equations beside you with the A Level Physics Formula Sheet.
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