OxfordAQA International AS and A-level Physics: Simple harmonic motion: equations, graphs and energy (9630) – 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 – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 3.6.2 (simple harmonic motion) of the OxfordAQA International AS and A-level Physics (9630) specification, version 4.4. The section is International A-level only. Learn the content first from the study guide and the revision notes. The course hub and printable checklist show where this fits. Use g = 9.81 m s⁻² where needed.

Questions

1. State the condition for simple harmonic motion, in words and as an equation, defining each symbol. [2]

2. The acceleration–displacement graph of an oscillator is a straight line through the origin and through the point x = 0.040 m, a = −1.44 m s⁻². The amplitude is 0.055 m.

(a) Show that the motion is simple harmonic and find ω. [2] (b) Calculate the period. [1] (c) Calculate the maximum speed. [1] (d) Calculate the maximum acceleration. [1]

3. An oscillator has amplitude 0.085 m and angular frequency 14 rad s⁻¹.

(a) Calculate its speed when its displacement is 0.050 m. [2] (b) State the magnitude and direction of its acceleration at this displacement, which is on the positive side of equilibrium. [1]

4. A tip of a vibrating ruler moves with amplitude 3.5 cm at frequency 2.5 Hz. Timing starts when it is at maximum positive displacement.

(a) Write down x as a function of t. [1] (b) Calculate x at t = 0.060 s. [2] (c) Calculate the first time at which x = −1.75 cm. [2]

5. A displacement–time graph for an oscillating block is a cosine curve with amplitude 6.0 cm and period 0.80 s.

(a) Calculate ω. [1] (b) Calculate the maximum speed and state the first time it occurs. [2] (c) Describe the acceleration–time graph, giving its maximum value and where it starts. [2]

6. A mass m is attached to a spring of constant k and oscillates horizontally without friction.

(a) Show that the period is T = 2π√(m/k). [3] (b) Calculate T for m = 0.45 kg and k = 28 N m⁻¹. [1] (c) Calculate T if the mass is changed to 1.80 kg. [1]

7. A simple pendulum has a light string of length l and a bob of mass m.

(a) Derive T = 2π√(l/g) for small oscillations. [3] (b) Calculate l for T = 1.50 s. [2] (c) The same pendulum is used where g = 1.62 m s⁻². Calculate its period. [1]

8. A 0.30 kg oscillator has ω = 8.0 rad s⁻¹ and amplitude 0.045 m.

(a) Calculate its total energy. [2] (b) Calculate its kinetic and potential energy at x = 0.027 m. [3]

9. An oscillator of amplitude 0.072 m has period 1.2 s.

(a) Sketch in words how Ek and Ep vary with displacement, stating where the curves meet. [2] (b) Calculate the displacement at which Ek = Ep. [1] (c) Starting at x = A, calculate the time taken to first reach that displacement. [1] (d) State the period of the Ek–time graph. [1]

10. A 0.60 kg mass hangs at rest from a vertical spring and stretches it by 0.12 m. It is then pulled down a further 0.040 m and released.

(a) Calculate the spring constant. [2] (b) Calculate the period of the oscillation. [2] (c) Calculate the maximum speed. [1] (d) Calculate the total energy of the oscillation. [2] (e) Calculate the displacement from equilibrium at which Ek = Ep. [2]

11. A pendulum bob of mass 0.15 kg hangs on a string 0.80 m long. It is released from rest with the bob 0.060 m (measured along the arc) from its lowest point.

(a) Calculate the angle of the string in radians and degrees, and comment on whether the small-angle model is reasonable. [2] (b) Calculate the period. [2] (c) Calculate the maximum speed of the bob. [2] (d) Check this speed using energy conservation, by finding the height gained by the bob. [3] (e) Calculate the maximum acceleration along the arc. [1]

Answers

1. The acceleration is directly proportional to the displacement from equilibrium and is directed towards equilibrium [1]. a = −ω²x, where a is acceleration, x is displacement from equilibrium and ω is the (constant) angular frequency [1]. [2] Examiner insight: A statement of only “a is proportional to x” misses the direction, so include “towards equilibrium” or the minus sign.

2. (a) The graph is a straight line through the origin with negative gradient, so a is proportional to −x [1]. Gradient = −1.44/0.040 = −36 s⁻², so ω² = 36 and ω = 6.0 rad s⁻¹ [1]. (b) T = 2π/6.0 = 1.05 s [1]. (c) vmax = ωA = 6.0 × 0.055 = 0.33 m s⁻¹ [1]. (d) amax = ω²A = 36 × 0.055 = 1.98 m s⁻² [1]. [5] Examiner insight: Explaining why a straight line through the origin means SHM, rather than just quoting ω, is what a “show that” part rewards.

3. (a) v = ω√(A² − x²) = 14 × √(0.085² − 0.050²) [1] = 14 × 0.0687 = 0.96 m s⁻¹ [1]. (b) a = −ω²x = −196 × 0.050 = 9.8 m s⁻² towards equilibrium (negative direction) [1]. [3] Examiner insight: Keep the squares inside the root. Subtracting before squaring is a frequent slip.

4. (a) ω = 2π × 2.5 = 15.7 rad s⁻¹, so x = 3.5 cos(15.7t) cm [1]. (b) ωt = 15.708 × 0.060 = 0.942 rad [1]; x = 3.5 × cos(0.942) = 2.1 cm [1]. (c) cos ωt = −0.5, so ωt = 2π/3 [1]; t = (2π/3)/15.708 = 0.133 s [1]. [5] Examiner insight: Using degrees in cos ωt gives a wrong x, so check the angle unit and keep answers in cm or convert to metres consistently.

5. (a) ω = 2π/0.80 = 7.85 rad s⁻¹ [1]. (b) vmax = ωA = 7.85 × 0.060 = 0.47 m s⁻¹ [1]; first at equilibrium, t = T/4 = 0.20 s [1]. (c) A negative cosine curve with the same period [1], starting at −ω²A = −61.7 × 0.060 = −3.7 m s⁻² and reaching +3.7 m s⁻² at 0.40 s [1]. [5] Examiner insight: Descriptions of graphs gain credit for the starting value, the sign and the period, not just “it is a cosine curve”.

6. (a) The spring force gives ma = −kx [1], so a = −(k/m)x [1], which is a = −ω²x with ω² = k/m, so T = 2π/ω = 2π√(m/k) [1]. (b) T = 2π√(0.45/28) = 0.80 s [1]. (c) The mass is four times larger, so T doubles: 1.59 s [1]. [5] Examiner insight: In a derivation, state the comparison with a = −ω²x explicitly, as that links the algebra to the definition.

7. (a) Restoring force along the arc F = −mg sin θ [1]. For small θ, sin θ ≈ θ = x/l, so ma = −mg x/l and a = −(g/l)x [1]. So ω² = g/l and T = 2π/ω = 2π√(l/g) [1]. (b) l = gT²/(4π²) = 9.81 × 2.25/39.48 [1] = 0.559 m [1]. (c) T = 2π√(0.559/1.62) = 3.69 s [1]. [6] Examiner insight: Mention that the small-angle approximation is used and that the mass cancels, since both points show you understand the result.

8. (a) E = ½mω²A² = ½ × 0.30 × 64 × 0.045² [1] = 0.0194 J [1]. (b) Ek = ½mω²(A² − x²) = ½ × 0.30 × 64 × (0.002025 − 0.000729) [1] = 0.0124 J [1]. Ep = E − Ek = 0.0070 J [1]. [5] Examiner insight: Use a quick check that Ek + Ep equals the total. It catches slips and shows the energy link.

9. (a) Ep against x is an upright parabola, zero at x = 0 [1]. Ek is an inverted parabola, maximum at x = 0 and zero at ±A, and the two curves cross at x = ±A/√2 [1]. (b) x = 0.072/√2 = 0.051 m [1]. (c) cos ωt = 1/√2, so ωt = π/4, which is T/8 = 0.15 s [1]. (d) 0.60 s (T/2) [1]. [5] Examiner insight: A common slip is to say the energy graphs have the same period as the motion. Count how many times energy peaks per cycle.

10. (a) At rest, kx = mg [1], so k = 0.60 × 9.81/0.12 = 49 N m⁻¹ [1]. (b) ω = √(k/m) = √(49.05/0.60) = 9.04 rad s⁻¹ [1]; T = 2π/9.04 = 0.69 s [1]. (c) The amplitude is 0.040 m, so vmax = ωA = 9.04 × 0.040 = 0.36 m s⁻¹ [1]. (d) E = ½kA² = ½ × 49.05 × 0.040² [1] = 0.039 J [1]. (e) Ek = Ep = E/2 at x = A/√2 [1] = 0.040/√2 = 0.028 m from equilibrium [1]. [9] Examiner insight: The amplitude is 0.040 m, not 0.16 m. Using the total extension as the amplitude is a typical way to lose most of the later marks.

11. (a) θ = x/l = 0.060/0.80 = 0.075 rad [1] = 4.3°, which is small, so sin θ ≈ θ is reasonable [1]. (b) ω = √(g/l) = √(9.81/0.80) = 3.50 rad s⁻¹ [1]; T = 2π/3.50 = 1.79 s [1]. (c) vmax = ωA = 3.50 × 0.060 [1] = 0.21 m s⁻¹ [1]. (d) h = l(1 − cos θ) = 0.80 × (1 − cos 0.075) = 0.00225 m [1]. Gain in gravitational potential energy mgh = 0.15 × 9.81 × 0.00225 = 3.31 × 10⁻³ J [1]. Kinetic energy ½mv² = ½ × 0.15 × 0.2101² = 3.31 × 10⁻³ J, so the two agree [1]. (e) amax = ω²A = 12.26 × 0.060 = 0.74 m s⁻² [1]. [10] Examiner insight: Show the unrounded intermediate values for h, because the difference between 1 and cos θ is tiny and early rounding makes the energy check fail.

Where marks are usually lost

• Leaving out the minus sign or the words “towards equilibrium” when stating the SHM condition.
• Using degrees when evaluating cos ωt.
• Taking the amplitude as the total extension of a hanging spring instead of the extra displacement.
• Using x instead of A in total energy.
• Giving maximum speed at an extreme or maximum acceleration at equilibrium.
• Forgetting that the pendulum result holds only for small angles in radians.
• Drawing Ek and Ep against time with period T rather than T/2.
• Reading a gradient off a chord instead of a tangent.

Next steps

• Review the revision notes and the study guide.
• Go to the OxfordAQA A-level Physics hub and tick off the printable checklist.
• Try all free 10-minute diagnostics: diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), specification version 4.4, for International AS and A-level exams May/June 2020 onwards, published by OxfordAQA. Section 3.6.2 is International A-level only.