Specification: AQA A-level Physics (7408) 3.6.1.2 | Paper: 1 | Data and formulae booklet: given in the exam
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📌 KEY DEFINITIONS – learn them word for word
x = A cos ωt: gives the displacement of an object in simple harmonic motion that is released from maximum displacement (x = +A) at t = 0. The angle ωt is in radians.
Amplitude: the maximum displacement from the equilibrium position, A. The period T is the time for one complete oscillation, and ω = 2π ÷ T = 2πf.
Phase difference: how far one oscillation is ahead of or behind another, as a fraction of a cycle. In radians, phase difference = 2πΔt ÷ T, where Δt is the time between corresponding points.
The equation
displacement = amplitude × cos(angular frequency × time) x = A cos ωt with ω = 2πf = 2π ÷ T
| Quantity | Symbol | Unit |
|---|---|---|
| displacement from equilibrium | x | metres (m) |
| amplitude | A | metres (m) |
| angular frequency | ω | radians per second (rad s⁻¹) |
| time since release from x = +A | t | seconds (s) |
| period | T | seconds (s) |
Rearranged: cos ωt = x ÷ A | t = cos⁻¹(x ÷ A) ÷ ω | A = x ÷ cos ωt | Put your calculator in radians before you use cos or cos⁻¹: ωt is an angle in radians, not degrees.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) An oscillator is released from its amplitude of 0.050 m at t = 0. Its angular frequency is 4.0 rad s⁻¹. Calculate its displacement at t = 0.30 s, to 2 significant figures. (calculator in radians)
Show solution
F: x = A cos ωt
I: x = 0.050 × cos(4.0 × 0.30) = 0.050 × cos(1.2)
F: x is already the subject
A: 0.050 × 0.3623… = 0.01811… m = 0.018 m (2 s.f.)
Watch out: in degrees mode cos(1.2) = 0.9997…, giving 0.04999 m. The angle ωt is always in radians.
Q2 (Core) A pendulum bob is released from a displacement of 12 cm and swings with simple harmonic motion with a period of 2.0 s. Calculate its displacement 0.25 s after release, to 2 significant figures. (calculator in radians)
Show solution
Step 1 – angular frequency
ω = 2π ÷ T = 2π ÷ 2.0 = π rad s⁻¹ = 3.141… rad s⁻¹
Step 2 – displacement
F: x = A cos ωt
I: x = 12 × cos(π × 0.25) = 12 × cos(π/4)
F: x is already the subject
A: 12 × 0.7071… = 8.485… cm = 8.5 cm (2 s.f.)
Q3 (Core) A mass on a spring is pulled down 2.5 cm and released at t = 0. It oscillates with a period of 1.6 s. Calculate its displacement at t = 0.60 s, to 2 significant figures. (calculator in radians)
Show solution
Convert first: 2.5 cm = 0.025 m; ω = 2π ÷ T = 2π ÷ 1.6 = 3.926… rad s⁻¹
F: x = A cos ωt
I: x = 0.025 × cos(3.926… × 0.60) = 0.025 × cos(2.356…)
F: x is already the subject
A: 0.025 × (−0.7071…) = −0.01767… m = −1.8 cm (2 s.f.), so 1.8 cm on the other side of the equilibrium position
Watch out: a negative displacement is correct here. After more than a quarter of a period the mass has passed through equilibrium.
Q4 (Core) An oscillator has amplitude 0.080 m and angular frequency 4.0 rad s⁻¹ and is released from x = +A at t = 0. Calculate the time at which its displacement first becomes 0.040 m, to 2 significant figures. (calculator in radians)
Show solution
F: x = A cos ωt
I: 0.040 = 0.080 × cos(4.0t)
F: cos(4.0t) = 0.040 ÷ 0.080 = 0.50, so 4.0t = cos⁻¹(0.50) = 1.047… rad, t = 1.047… ÷ 4.0
A: 0.2617… s = 0.26 s (2 s.f.)
Watch out: cos⁻¹(0.50) in degrees mode gives 60, and 60 ÷ 4.0 = 15 s, which is far too long.
Q5 (Exam) A mass on a spring is released from a displacement of 40 mm and oscillates at 2.5 Hz. Calculate its displacement 0.13 s after release, to 2 significant figures. (calculator in radians)
Show solution
Step 1 – angular frequency
ω = 2πf = 2π × 2.5 = 15.70… rad s⁻¹
Step 2 – displacement
F: x = A cos ωt
I: x = 40 × cos(15.70… × 0.13) = 40 × cos(2.042…)
F: x is already the subject
A: 40 × (−0.4539…) = −18.15… mm = −18 mm (2 s.f.)
Watch out: in degrees mode this gives 39.97 mm, which would mean the mass had barely moved. Always check your calculator is in radians.
Q6 (Exam) A mass on a spring is released from its amplitude of 5.0 cm and oscillates with a period of 1.2 s. Calculate how long after release its displacement first equals 2.0 cm, to 2 significant figures. (calculator in radians)
Show solution
Step 1 – angular frequency
ω = 2π ÷ T = 2π ÷ 1.2 = 5.235… rad s⁻¹
Step 2 – time
F: x = A cos ωt
I: 2.0 = 5.0 × cos(5.235…t)
F: cos(5.235…t) = 2.0 ÷ 5.0 = 0.40, so 5.235…t = cos⁻¹(0.40) = 1.159… rad, t = 1.159… ÷ 5.235…
A: 0.2214… s = 0.22 s (2 s.f.)
Note: the cm cancel in x ÷ A, so there is no need to convert to metres here.
Q7 (Exam) Two identical mass–spring oscillators each have a period of 0.80 s and an amplitude of 6.0 cm. Oscillator P is released from x = +A at t = 0; oscillator Q is released from x = +A at t = 0.10 s. (a) Calculate the phase difference between them in radians, to 2 significant figures. (b) Calculate the displacement of each oscillator at t = 0.30 s, to 2 significant figures. (calculator in radians)
Show solution
Step 1 – (a) phase difference
F: phase difference = 2πΔt ÷ T
I: phase difference = 2π × 0.10 ÷ 0.80
A: π/4 = 0.785… rad = 0.79 rad (2 s.f.), an eighth of a cycle (45°), with Q lagging behind P
Step 2 – (b) displacements
ω = 2π ÷ 0.80 = 7.853… rad s⁻¹
P: x = 6.0 × cos(7.853… × 0.30) = 6.0 × cos(2.356…) = −4.242… cm = −4.2 cm (2 s.f.)
Q has been moving for 0.30 − 0.10 = 0.20 s: x = 6.0 × cos(7.853… × 0.20) = 6.0 × cos(π/2) = 0 cm, so Q is passing through equilibrium
Q8 (Exam) An object in simple harmonic motion is released from rest at its amplitude at t = 0. Its angular frequency is 8.0 rad s⁻¹. At t = 0.50 s its displacement is −0.026 m. Calculate the amplitude, to 2 significant figures. (calculator in radians)
Show solution
F: x = A cos ωt
I: −0.026 = A × cos(8.0 × 0.50) = A × cos(4.0)
F: A = −0.026 ÷ cos(4.0) = −0.026 ÷ (−0.6536…)
A: 0.03977… m = 0.040 m (2 s.f.)
Watch out: the amplitude must come out positive. If you get a negative amplitude, check the sign of cos ωt and your calculator mode.
Q9 (Stretch) A mass on a spring has an amplitude of 10 cm and a period of 2.4 s. It is released from x = +10 cm at t = 0. Calculate the time it takes to move from x = +8.0 cm to x = −5.0 cm for the first time, to 2 significant figures. (calculator in radians)
Show solution
Step 1 – angular frequency
ω = 2π ÷ T = 2π ÷ 2.4 = 2.617… rad s⁻¹
Step 2 – time to reach +8.0 cm
cos ωt₁ = 8.0 ÷ 10 = 0.80, so ωt₁ = cos⁻¹(0.80) = 0.6435… rad; t₁ = 0.6435… ÷ 2.617… = 0.2457… s
Step 3 – time to reach −5.0 cm
cos ωt₂ = −5.0 ÷ 10 = −0.50, so ωt₂ = cos⁻¹(−0.50) = 2.094… rad; t₂ = 2.094… ÷ 2.617… = 0.80 s
Step 4 – time between
Δt = t₂ − t₁ = 0.80 − 0.2457…
A: 0.5542… s = 0.55 s (2 s.f.)
Watch out: the time is not proportional to the distance moved. The mass moves fastest near equilibrium, so do not use speed = distance ÷ time.
Q10 (Stretch) A mass on a spring oscillates with simple harmonic motion at 2.0 Hz. A sensor records only while the magnitude of the displacement is at least 80% of the amplitude. Calculate the percentage of each cycle during which the sensor records, to 2 significant figures. (calculator in radians)
Show solution
Step 1 – time from x = +A to x = 0.80A
ω = 2πf = 2π × 2.0 = 12.56… rad s⁻¹; cos ωt = 0.80, so ωt = cos⁻¹(0.80) = 0.6435… rad; t = 0.6435… ÷ 12.56… = 0.05120… s
Step 2 – total time per cycle
Near each extreme the mass spends t arriving and t leaving, and there are two extremes per cycle, so total = 4t = 4 × 0.05120… = 0.2048… s
Step 3 – percentage
T = 1 ÷ f = 0.50 s; percentage = 0.2048… ÷ 0.50 × 100
A: 40.96…% = 41% (2 s.f.)
Note: the answer does not depend on the frequency. It is 4cos⁻¹(0.80) ÷ 2π, so an oscillator spends a large fraction of its time near the extremes, where it moves slowly.
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