IB DP Physics: Simple harmonic motion: equations, period and energy (DP Physics) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover sub-topic C.1, Simple harmonic motion, in the IB Physics guide (first assessment 2025). Questions and parts marked “HL only” use additional higher level content; everything else is for SL and HL. Take g = 9.81 m s⁻² where it is needed. Learn the content in the study guide and the revision notes. Other links: the course hub, the printable checklist and the free 10-minute diagnostics.

Questions

1. State the two conditions that must be satisfied for an object’s oscillation to be simple harmonic. [2]

2. The defining equation of simple harmonic motion is a = –ω²x. Explain the significance of the minus sign. [3]

3. An oscillating block completes 45 oscillations in 27.0 s. Determine its time period, its frequency and its angular frequency. [3]

4. A block of mass 0.30 kg oscillates on a horizontal spring of spring constant 48 N m⁻¹.

(a) Calculate the time period. [2] (b) The block is replaced by one of a different mass so that the period becomes 1.5 s. Calculate the new mass. [3]

5. A simple pendulum has length 1.20 m.

(a) Calculate its period on Earth. [2] (b) Calculate its period on the Moon, where g = 1.62 m s⁻². [2] (c) State why the formula only gives a good result for small swings. [1]

6. In an experiment, the period T of a simple pendulum is measured for different lengths l. A graph of T² against l is a straight line through the origin. It passes through the point (0.80 m, 3.22 s²).

(a) Show that T² is proportional to l, and state the gradient in terms of g. [2] (b) Determine g from the graph. [3]

7. A block on a horizontal frictionless surface is attached to a spring. It is pulled to its maximum extension and released. Describe the energy changes during one complete cycle. [4]

8. (HL only) A particle moves with x = x₀ sin(ωt + φ), with x₀ = 0.050 m and period 1.2 s. At t = 0 its displacement is +0.030 m and it is moving towards the equilibrium position.

(a) Calculate ω. [1] (b) Determine φ in radians. [3] (c) Calculate the displacement at t = 0.50 s. [2] (d) Calculate the velocity at t = 0. [1]

9. (HL only) A mass on a spring oscillates with amplitude 0.080 m that completes 3.2 oscillations every second.

(a) Determine the angular frequency. [1] (b) Calculate the speed when the displacement is 0.050 m. [2] (c) State the maximum speed. [1] (d) Calculate the maximum acceleration. [1]

10. (HL only) A mass of 0.25 kg oscillates with ω = 14 rad s⁻¹ and amplitude 0.060 m.

(a) Calculate the total energy. [2] (b) Calculate the potential energy at displacement 0.040 m. [2] (c) Determine the kinetic energy and the speed at that displacement. [2]

11. A mass of 0.40 kg hangs at rest from a vertical spring and extends it by 0.056 m. It is then displaced a further 0.035 m below that position, held still, and let go.

(a) Determine the spring constant. [2] (b) Calculate the time period of the oscillation. [2] (c) (HL only) Calculate the maximum speed. [2] (d) (HL only) Calculate the maximum kinetic energy, and show that it equals ½kx₀². [3]

12. (HL only) A pendulum bob swings on a string of length 0.90 m with amplitude 0.050 m measured along the arc.

(a) Treating the motion as SHM, calculate ω and the maximum speed. [3] (b) Find the maximum height the bob rises above its lowest point, and use energy conservation to find the speed at the lowest point. [3]

Answers

1. The net (restoring) force acts towards the equilibrium position [1]. Its magnitude is proportional to the displacement from equilibrium [1]. [2] Examiner insight: Answers that say only “force is proportional to displacement” leave out the direction, which is the other half of the condition.

2. The acceleration is directed towards the equilibrium position [1]. The displacement is measured away from equilibrium, so a and x have opposite signs [1]. This keeps the object returning to equilibrium, so the motion is oscillatory [1]. [3] Examiner insight: A good answer links the sign to direction. Saying “the acceleration is negative” is incomplete, because it is only negative when the displacement is positive.

3. T = 27.0/45 = 0.600 s [1]. f = 1/T = 1.67 Hz [1]. ω = 2πf = 10.5 rad s⁻¹ [1]. [3] Examiner insight: Keep units on all three. Hz and rad s⁻¹ are different and mixing them up is an easy way to lose credit.

4. (a) T = 2π√(m/k) = 2π√(0.30/48) [1] = 0.497 s [1] (b) Rearrange: m = kT²/4π² [1] = 48 × 1.5²/(4π²) [1] = 2.74 kg [1] [5] Examiner insight: Show the rearranged formula before substituting. Squaring both sides is where most slips happen in a rearrangement.

5. (a) T = 2π√(l/g) = 2π√(1.20/9.81) [1] = 2.20 s [1] (b) T = 2π√(1.20/1.62) [1] = 5.41 s [1] (c) The restoring force is only proportional to displacement for small angles [1] [5] Examiner insight: For (c), simply writing “the angle must be small” is a statement of the condition. Linking it to the restoring force gives the physics.

6. (a) T = 2π√(l/g), so T² = 4π²l/g [1]. This is proportional to l, with gradient 4π²/g [1] (b) Gradient = 3.22/0.80 = 4.025 s² m⁻¹ [1]. g = 4π²/gradient [1] = 9.81 m s⁻² [1] [5] Examiner insight: Take the gradient from the line, using the point given. Reading the gradient from a large triangle, or from a point far from the origin as here, reduces the effect of reading errors.

7. At maximum extension the block is momentarily at rest, so the kinetic energy is zero and the elastic potential energy is maximum [1]. As it moves towards equilibrium, elastic potential energy decreases and kinetic energy increases [1]. At equilibrium the kinetic energy is maximum and the elastic potential energy is minimum [1]. The reverse happens on the other side, and the total energy remains constant throughout, with the cycle repeating [1]. [4] Examiner insight: Name the type of potential energy and say what happens to the total. “Energy changes” alone does not describe the cycle.

8. (a) ω = 2π/1.2 = 5.24 rad s⁻¹ [1] (b) 0.030 = 0.050 sin φ gives sin φ = 0.6 [1]. The two solutions are 0.644 rad and π – 0.644 = 2.50 rad [1]. Moving towards equilibrium means negative velocity, so cos φ < 0 and φ = 2.50 rad [1] (c) Argument = 5.236 × 0.50 + 2.498 = 5.116 rad [1]. x = 0.050 sin(5.116) = –0.046 m [1] (d) v = ωx₀ cos φ = 5.236 × 0.050 × cos(2.498) = –0.209 m s⁻¹ [1] [7] Examiner insight: Choosing the correct phase angle depends on the direction of motion, so state your reason for rejecting the other root.

9. (a) ω = 2π × 3.2 = 20.1 rad s⁻¹ [1] (b) v = ω√(x₀² – x²) = 20.1 × √(0.080² – 0.050²) [1] = 1.26 m s⁻¹ [1] (c) v_max = ωx₀ = 1.61 m s⁻¹ [1] (d) a_max = ω²x₀ = 32.3 m s⁻² [1] [5] Examiner insight: Use the unrounded ω in later parts, because small rounding in 20.1 changes the third significant figure of ω².

10. (a) E_T = ½mω²x₀² [1] = ½ × 0.25 × 14² × 0.060² = 0.0882 J [1] (b) E_p = ½mω²x² = ½ × 0.25 × 14² × 0.040² [1] = 0.0392 J [1] (c) E_K = 0.0882 – 0.0392 = 0.0490 J [1]; v = √(2E_K/m) = 0.626 m s⁻¹ [1] [6] Examiner insight: Checking with v = ω√(x₀² – x²) gives the same speed, a useful way to confirm the energy working.

11. (a) At equilibrium kx = mg [1], so k = 0.40 × 9.81/0.056 = 70.1 N m⁻¹ [1] (b) T = 2π√(m/k) = 2π√(0.40/70.07) [1] = 0.475 s [1] (c) The amplitude is 0.035 m, not 0.091 m. ω = 13.2 rad s⁻¹, so v_max = ωx₀ [1] = 0.463 m s⁻¹ [1] (d) E_K,max = ½mv² = ½ × 0.40 × 0.463² [1] = 0.0429 J [1]. ½kx₀² = ½ × 70.07 × 0.035² = 0.0429 J, equal to the kinetic energy [1] [9] Examiner insight: The amplitude is measured from the equilibrium position, not from the natural length of the spring, and this is the usual mistake on a vertical spring.

12. (a) ω = √(g/l) = √(9.81/0.90) = 3.30 rad s⁻¹ [1]. v_max = ωx₀ [1] = 0.165 m s⁻¹ [1] (b) θ = 0.050/0.90 = 0.0556 rad, h = l(1 – cos θ) [1] = 1.39 × 10⁻³ m [1]. v = √(2gh) = 0.165 m s⁻¹, which agrees with part (a) [1] [6] Examiner insight: A comparison of the two methods is worth stating in words, because it shows why the small-angle SHM approximation is acceptable here.

Where marks are usually lost

• Stating the conditions for SHM without mentioning that the force opposes the displacement.
• Reading ω as a speed or mixing up hertz and radians per second.
• Squaring or square-rooting the wrong side when rearranging T = 2π√(m/k) or T = 2π√(l/g).
• Using the spring extension at equilibrium as the amplitude.
• Working in degrees for the phase angle.
• Taking the first solution of sin φ without using the direction of motion.
• Using v = ωx₀ at a general position instead of v = ω√(x₀² – x²).
• Giving energy answers without saying which kind of potential energy it is, or forgetting the total is constant.
• Rounding ω early and then giving a final answer that differs in the third significant figure.

Next steps

• Read the revision notes and the study guide.
• Go to the IB DP Physics hub and the printable checklist.
• Try all free 10-minute diagnostics at /diagnostics/.
• Book a free trial class at /trial/.

Official syllabus

Source: Physics guide, first assessment 2025, International Baccalaureate Organization (Diploma Programme), sub-topic C.1 Simple harmonic motion.