OxfordAQA International AS and A-level Physics: Oscillating systems, damping and resonance (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 follow sections 3.5.1 and 3.5.2 of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4. Both sections are shared AS and A-level content, except question 11, which uses the section 3.6.2 results and is International A-level only. Take g = 9.81 m s⁻² and the speed of sound as 340 m s⁻¹. Learn the content from the study guide or the revision notes. The course hub, checklist and free diagnostics cover the rest.

Questions

1. A 0.40 kg mass hangs on a spring of stiffness 64 N m⁻¹. Calculate the period and frequency of its vertical oscillations. [3]

2. A simple pendulum on the Moon (g = 1.62 m s⁻²) must have a period of 3.0 s. Calculate its length. [3]

3. (a) A mass-spring system has a period of 0.90 s. The mass is increased to four times its value. State the new period. [2] (b) A pendulum has a period of 2.4 s. Its length is reduced to one-ninth of the original. State the new period. [2]

4. A pendulum bob of mass 0.080 kg is released from rest with its centre 2.5 cm above its lowest point.

(a) Calculate the maximum kinetic energy and the maximum speed. [3] (b) Describe how the kinetic energy, potential energy and total energy vary with time during one oscillation. [3]

5. The amplitude of a damped oscillator falls from 9.0 cm to 5.0 cm in 8 oscillations. The energy is proportional to the amplitude squared.

(a) Calculate the fraction of the original energy that remains. [2] (b) State where the lost energy goes and the effect of light damping on the period. [2]

6. Describe the motion of a displaced system that is (a) lightly damped, (b) critically damped, (c) heavily damped. [3] (d) Name one device that relies on critical damping. [1]

7. A trolley on springs has a natural frequency of 1.5 Hz. A motor shakes its base at an adjustable frequency.

(a) Once the motion has settled, state the frequency of the trolley when the motor is at 2.5 Hz. [1] (b) Name the effect when the motor is at 1.5 Hz and describe the amplitude. [2] (c) Friction pads are now added to the trolley. Describe two changes to the amplitude-frequency graph. [2]

8. A loose panel in a car has a natural frequency of 25 Hz. The engine shakes the car once per revolution.

(a) Calculate the engine speed, in revolutions per minute, at which the panel rattles most. [2] (b) Explain why the rattling is greatest at this speed. [2] (c) Suggest one way to reduce the shaking. [1]

9. A string of length 0.60 m is fixed at both ends. The speed of waves on it is 48 m s⁻¹. A vibration generator drives it.

(a) Calculate the lowest driving frequency that gives resonance. [2] (b) State the next two resonant frequencies. [1] (c) Explain why the amplitude is small at 60 Hz. [2] (d) A 680 Hz tuning fork is held over a tube closed at the bottom. Calculate the two shortest air column lengths at which resonance occurs. Ignore end effects. [3]

10. In a pendulum investigation, a student measures the time for 10 oscillations as 18.9 s, with an uncertainty of 0.2 s, for a length of 0.85 m measured from the pivot to the top of the bob.

(a) Explain why 10 oscillations are timed, and say where timing should start and why. [2] (b) Calculate T, and the percentage uncertainties in T and T². [3] (c) Calculate the value of g from these results. [2] (d) The percentage difference from 9.81 m s⁻² is larger than the uncertainty in T². Suggest one systematic error that explains this. [1]

11. International A-level only. A 0.40 kg mass oscillates on a spring of stiffness 90 N m⁻¹ with amplitude 0.030 m.

(a) Calculate ω and the period. [2] (b) Calculate the total energy. [2] (c) Calculate the speed at a displacement of 0.018 m. [3] (d) Calculate the maximum speed. [1]

Answers

1. T = 2π√(m/k) [1]. T = 2π√(0.40/64) = 0.50 s [1]. f = 1/T = 2.0 Hz [1] [3] Examiner insight: Writing the formula and the substitution gives a clear route to credit even if the arithmetic slips.

2. Rearrange: l = gT²/4π² [1]. l = 1.62 × 9.0 / 39.48 [1] = 0.37 m [1] [3] Examiner insight: Rearranging before substituting shows the algebra step separately. Using T instead of T² is an easy slip.

3. (a) T ∝ √m [1], so T doubles to 1.8 s [1] (b) T ∝ √l, so T is multiplied by √(1/9) = 1/3 [1], giving 0.80 s [1] [4] Examiner insight: State the proportionality used. A bare number can be right, but the reasoning is what shows understanding.

4. (a) Maximum Ep = mgh = 0.080 × 9.81 × 0.025 = 0.0196 J [1]. Maximum Ek equals this at the lowest point, 0.020 J [1]. v = √(2Ek/m) = 0.70 m s⁻¹ [1] (b) The total energy stays constant (no energy lost) [1]. Ek and Ep each vary between zero and 0.020 J, with Ek maximum when Ep is zero [1]. Each completes two cycles in one oscillation, so the frequency is double that of the motion [1] [6] Examiner insight: A good graph description names the frequency relationship, the maximum value and the phase of the two curves. A sketch with labelled axes supports it.

5. (a) Energy ratio = (5.0/9.0)² [1] = 0.31 [1] (b) Thermal energy of the surroundings (air or the suspension) [1]. The period is almost unchanged [1] [4] Examiner insight: “Lost” is not enough for the energy. Say it is transferred to thermal energy of the surroundings by resistive forces.

6. (a) It oscillates, with the amplitude falling each cycle [1]. (b) It settles at equilibrium in the least possible time, with no oscillation [1]. (c) It returns slowly without oscillating [1]. (d) Car suspension, or a door closer [1] [4] Examiner insight: Critical and heavy damping both avoid oscillation. What separates them is the return time, so say that.

7. (a) 2.5 Hz (the motor’s frequency) [1] (b) Resonance [1]. The amplitude is a maximum [1] (c) The peak is lower [1] and broader (the peak also moves slightly lower in frequency) [1] [5] Examiner insight: For part (c) give two distinct changes. “Lower” and “smaller” count as one idea.

8. (a) 25 revolutions per second [1] × 60 = 1500 rev min⁻¹ [1] (b) At 1500 rev min⁻¹ the engine’s frequency matches the panel’s natural frequency, which is resonance [1]. The energy transferred to the panel is greatest, so the amplitude is large [1] (c) Add damping (a rubber pad), or stiffen the panel to change its natural frequency [1] [5] Examiner insight: Link the two frequencies explicitly. Writing “resonance” without saying what is equal is usually incomplete.

9. (a) λ = 2L = 1.2 m [1]. f = v/λ = 40 Hz [1] (b) 80 Hz and 120 Hz [1] (c) 60 Hz is not a natural frequency of the string, so no stationary wave fits [1] and energy is transferred poorly from the driver [1] (d) λ = 340/680 = 0.50 m [1]. Shortest column: L = λ/4 = 0.125 m [1]. Next: 3λ/4 = 0.375 m [1] [8] Examiner insight: For a closed tube, explain that the closed end is a node and the open end an antinode. Using λ/2 for the tube is an easy slip.

10. (a) Timing 10 oscillations reduces the percentage uncertainty from reaction time and resolution [1]. Start at equilibrium, where the bob moves fastest and the position is easiest to judge (use a fiducial marker) [1] (b) T = 18.9/10 = 1.89 s [1]. Percentage uncertainty in T = 0.2/18.9 = 1.1% [1]; in T² = 2.1% [1] (c) g = 4π²l/T² = 39.48 × 0.85 / 1.89² [1] = 9.4 m s⁻² [1] (d) The length was measured to the top of the bob, so it is too short (it should go to the centre), making g too small [1] [8] Examiner insight: A systematic error shifts every result the same way; compare the size of the discrepancy with the uncertainty before blaming random error.

11. (a) ω = √(k/m) = √(90/0.40) = 15 rad s⁻¹ [1]. T = 2π/ω = 0.42 s [1] (b) E = ½kA² = ½ × 90 × 0.030² [1] = 0.0405 J [1] (c) v = ω√(A² − x²) [1] = 15 × √(0.030² − 0.018²) [1] = 15 × 0.024 = 0.36 m s⁻¹ [1] (d) v max = ωA = 15 × 0.030 = 0.45 m s⁻¹ [1] [8] Examiner insight: Part (c) can also be done by subtracting Ep from the total. Either route is valid when each step is shown.

Where marks are usually lost

• Measuring the pendulum length to the top of the bob, or forgetting the units.
• Failing to rearrange for l or m when squaring, or squaring only one side.
• Saying amplitude or bob mass changes a pendulum period.
• Describing energy against time without the doubled frequency.
• Saying “energy is lost” without stating it becomes thermal energy.
• Confusing critical damping with heavy damping.
• Writing “resonance occurs at high frequency” rather than equal frequencies.
• Using λ/2 rather than λ/4 for a closed tube.
• Giving a graph gradient without the 4π² step.

Next steps

• Revision notes
• Study guide
• OxfordAQA Physics course hub and printable checklist
• Find weak spots with all free 10-minute diagnostics: diagnostics
• Book a free trial class: trial class

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams May/June 2020 onwards, published by OxfordAQA. Sections 3.5.1 and 3.5.2, with the derivation in 3.6.2.