Following an oscillator through equilibrium

7 marks · A-level · original practice case

Write your working and explanations below or on paper. Your responses stay in this page; PhysicsHelp does not submit or save them. Copy anything you want to keep before closing it. Check calculations separately from method marks. Use unrounded working between parts.

The situation

An undamped oscillator has amplitude 0.040 m, period 0.80 s. It starts at maximum positive displacement. Positive is right. The table gives exact idealised phases; use ω = 2π/T and a = −ω²x.

Positions in one cycle
t / s	x / m	Motion
0.00	+0.040	Momentarily at rest
0.20	0	Moving left
0.40	−0.040	Momentarily at rest
0.60	0	Moving right

Part (a)

Calculate the angular frequency. [2 marks]

Give the numerical answer to 2 significant figures.

Final value / rad s⁻¹

Working

Part (b)

Calculate the signed velocity at t = 0.20 s. [2 marks]

Give the numerical answer to 2 significant figures.

Final value / m s⁻¹

Working

Part (c)

Compare kinetic energy and acceleration at 0.20 s with those at 0.40 s. [2 marks]

Your explanation

Part (d)

Select the effect of doubling oscillation amplitude while period remains unchanged. [1 marks]

Choose one answer Maximum speed doubles and maximum kinetic energy doubles. Maximum speed doubles and maximum kinetic energy quadruples. Maximum speed quadruples and maximum kinetic energy doubles. Maximum speed stays the same and maximum kinetic energy quadruples.

Marking rubric and worked answers

Indicative practice marks. Method marks assess valid working; accuracy marks require the associated valid method. Independent points stand on their own. Tick only points supported by your response. Do not award the same explanation twice. These ticks record your judgement, not an examiner’s result.

Part (a) marking points

Final answer: 7.9 rad s⁻¹. Numerical checks assess the final value only; they do not check working or the number of digits typed.

Use ω = 2π/T. (1 method mark)
Obtain 7.9 rad s⁻¹. (1 accuracy mark)

No earlier numerical result is required. Award method credit for valid working and accuracy credit for the stated answer. Accept a defensible equivalent method; wrong physics cannot earn dependent accuracy credit.

Common errors: This is frequency in hertz, not angular frequency.

Part (b) marking points

Final answer: -0.31 m s⁻¹. Numerical checks assess the final value only; they do not check working or the number of digits typed.

At equilibrium speed magnitude is ωA. (1 method mark)
Motion is left, so v = −0.31 m s⁻¹. (1 accuracy mark)

Use angular frequency from part (a) in v = −ωA at the stated leftward equilibrium crossing. Carry forward arithmetic errors; a positive velocity is inconsistent with the stated direction.

Common errors: The speed magnitude is right, but velocity is left/negative.

Part (c) marking points

At 0.20 s kinetic energy is maximum and acceleration is zero. (1 independent mark)
At 0.40 s kinetic energy is zero and acceleration has maximum positive magnitude. (1 independent mark)

Award each distinct physical point once. Equivalent scientifically correct wording is creditable; a contradictory statement cannot earn the same point.

Part (d) marking points

Final answer: Maximum speed doubles and maximum kinetic energy quadruples..

Maximum speed doubles and maximum kinetic energy quadruples. (1 independent mark)

One correct choice earns the mark; no partial credit.

Common errors: Speed scales with amplitude, but kinetic energy scales with speed squared. Maximum speed depends linearly on amplitude, not amplitude squared. Maximum speed increases when amplitude increases at fixed period.

Maximum speed doubles and maximum kinetic energy doubles. — Speed scales with amplitude, but kinetic energy scales with speed squared.
Maximum speed doubles and maximum kinetic energy quadruples. — At fixed period, maximum speed is proportional to amplitude and kinetic energy to amplitude squared.
Maximum speed quadruples and maximum kinetic energy doubles. — Maximum speed depends linearly on amplitude, not amplitude squared.
Maximum speed stays the same and maximum kinetic energy quadruples. — Maximum speed increases when amplitude increases at fixed period.

Report a problem with this case

Related skills

simple harmonic motion practice

Scope and reference sources

AQA 3.3 and selected 3.6.1.2; OCR A 4.4 and selected 5.3. These are shared original exercises; wording, indicative marks and difficulty have not been endorsed by either board.

AQA specification · OCR A specification

Case reference: oscillator-phases · version 1.1.0 · seed 0. Choose another case