Specification: AQA A-level Physics (7408) 3.6.1.2 | Paper: 1 | Data and formulae booklet: given in the exam
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Speed in SHM: v = ±ω√(A² − x²). The speed is maximum at the equilibrium position (x = 0) and zero at maximum displacement (x = ±A). The ± shows the object can be moving in either direction at a given displacement.
Maximum speed: vmax = ωA, where ω = 2πf is the angular frequency and A is the amplitude.
Energy in SHM: energy is continually exchanged between kinetic and potential energy. Eₖ = ½mω²(A² − x²) and the total energy ½mω²A² stays constant if there is no damping.
The equation
speed = angular frequency × √(amplitude² − displacement²) v = ±ω√(A² − x²) and maximum speed vmax = ωA
| Quantity | Symbol | Unit |
|---|---|---|
| velocity | v | metres per second (m s⁻¹) |
| angular frequency | ω | radians per second (rad s⁻¹) |
| amplitude | A | metres (m) |
| displacement from equilibrium | x | metres (m) |
| kinetic energy | Eₖ | joules (J) |
Rearranged: A² − x² = v² ÷ ω² | A = √(x² + v² ÷ ω²) | ω = vmax ÷ A | Square A and x separately and subtract before taking the square root: √(A² − x²) is not A − x.
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 object oscillates with simple harmonic motion at 2.0 Hz with an amplitude of 0.035 m. Calculate its maximum speed, to 2 significant figures.
Show solution
Step 1 – angular frequency
ω = 2πf = 2π × 2.0 = 12.56… rad s⁻¹
F: vmax = ωA
I: vmax = 12.56… × 0.035
F: vmax is already the subject
A: 0.4398… m s⁻¹ = 0.44 m s⁻¹ (2 s.f.), as it passes through the equilibrium position
Q2 (Core) An oscillator has angular frequency 5.0 rad s⁻¹ and amplitude 0.10 m. Calculate its speed when its displacement is 0.060 m.
Show solution
F: v = ±ω√(A² − x²)
I: v = ±5.0 × √(0.10² − 0.060²) = ±5.0 × √(0.010 − 0.0036) = ±5.0 × √0.0064
F: v is already the subject
A: ±5.0 × 0.080 = 0.40 m s⁻¹ (in either direction)
Q3 (Core) A mass on a spring oscillates at 1.5 Hz with an amplitude of 4.0 cm. Calculate its speed when its displacement is 2.0 cm, to 2 significant figures.
Show solution
Convert first: 4.0 cm = 0.040 m; 2.0 cm = 0.020 m; ω = 2πf = 2π × 1.5 = 9.424… rad s⁻¹
F: v = ±ω√(A² − x²)
I: v = ±9.424… × √(0.040² − 0.020²) = ±9.424… × √0.0012 = ±9.424… × 0.03464…
F: v is already the subject
A: 0.3264… m s⁻¹ = 0.33 m s⁻¹ (2 s.f.)
Watch out: halfway out the speed is not half the maximum. It is √3 ÷ 2 ≈ 87% of vmax.
Q4 (Core) A mass on a spring passes through its equilibrium position at 1.2 m s⁻¹. The amplitude of its motion is 0.15 m. Calculate the angular frequency and the frequency of the oscillations, to 2 significant figures.
Show solution
F: vmax = ωA
I: 1.2 = ω × 0.15
F: ω = 1.2 ÷ 0.15 = 8.0 rad s⁻¹
A: 8.0 rad s⁻¹; f = ω ÷ 2π = 8.0 ÷ 2π = 1.273… = 1.3 Hz (2 s.f.)
Q5 (Exam) The tip of a tuning fork prong vibrates with simple harmonic motion at 500 Hz with an amplitude of 0.20 mm. Calculate (a) its maximum speed and (b) its speed when the displacement is 0.10 mm, both to 2 significant figures.
Show solution
Convert first: 0.20 mm = 2.0 × 10−4 m; 0.10 mm = 1.0 × 10−4 m; ω = 2π × 500 = 3141.5… rad s⁻¹
Step 1 – (a) maximum speed
vmax = ωA = 3141.5… × 2.0 × 10−4 = 0.6283… m s⁻¹
A: 0.63 m s⁻¹ (2 s.f.)
Step 2 – (b) speed at x = 0.10 mm
F: v = ±ω√(A² − x²)
I: v = ±3141.5… × √(4.0 × 10−8 − 1.0 × 10−8) = ±3141.5… × 1.732… × 10−4
A: 0.5441… m s⁻¹ = 0.54 m s⁻¹ (2 s.f.)
Watch out: (2.0 × 10−4)² = 4.0 × 10−8. Put brackets round standard-form numbers when you square them on a calculator.
Q6 (Exam) An oscillator has amplitude 6.0 cm. Calculate the displacement at which its speed is half its maximum speed, to 2 significant figures.
Show solution
F: v = ω√(A² − x²), with v = ½vmax = ½ωA
I: ½ωA = ω√(A² − x²)
F: ω cancels: ¼A² = A² − x², so x² = ¾A² and x = (√3 ÷ 2)A = 0.8660… × 6.0
A: 5.196… cm = ±5.2 cm (2 s.f.)
Watch out: the speed halves well beyond halfway out, at 87% of the amplitude, because v depends on √(A² − x²), not on x.
Q7 (Exam) A trolley tethered between two springs oscillates with simple harmonic motion with angular frequency 10 rad s⁻¹. When its displacement is 0.040 m its speed is 0.30 m s⁻¹. Calculate the amplitude and the maximum speed.
Show solution
F: v = ±ω√(A² − x²)
I: 0.30 = 10 × √(A² − 0.040²)
F: A² − 0.0016 = (0.30 ÷ 10)² = 0.0009, so A² = 0.0025 and A = √0.0025
A: A = 0.050 m; vmax = ωA = 10 × 0.050 = 0.50 m s⁻¹
Q8 (Exam) A mass on a spring oscillates with simple harmonic motion with a period of 1.25 s and an amplitude of 8.0 cm. Calculate (a) its maximum speed and (b) its speed when it is 5.0 cm from the equilibrium position, both to 2 significant figures.
Show solution
Convert first: 8.0 cm = 0.080 m; 5.0 cm = 0.050 m
Step 1 – angular frequency
ω = 2π ÷ T = 2π ÷ 1.25 = 5.026… rad s⁻¹
Step 2 – (a) maximum speed
vmax = ωA = 5.026… × 0.080 = 0.4021… m s⁻¹
A: 0.40 m s⁻¹ (2 s.f.)
Step 3 – (b) speed at x = 0.050 m
F: v = ±ω√(A² − x²)
I: v = ±5.026… × √(0.0064 − 0.0025) = ±5.026… × 0.06244…
A: 0.3139… m s⁻¹ = 0.31 m s⁻¹ (2 s.f.)
Q9 (Stretch) A 0.40 kg mass on a spring oscillates with simple harmonic motion at 1.5 Hz with an amplitude of 5.0 cm, with no damping. Calculate (a) the total energy of the oscillation and (b) the kinetic energy and the potential energy when the displacement is 3.0 cm, all to 2 significant figures.
Show solution
Convert first: 5.0 cm = 0.050 m; 3.0 cm = 0.030 m; ω = 2π × 1.5 = 9.424… rad s⁻¹, so ω² = 88.82… s⁻²
Step 1 – (a) total energy
F: Etotal = ½mω²A²
I: Etotal = 0.5 × 0.40 × 88.82… × 0.050²
A: 0.04441… J = 0.044 J (2 s.f.)
Step 2 – (b) kinetic energy
F: Eₖ = ½mω²(A² − x²)
I: Eₖ = 0.5 × 0.40 × 88.82… × (0.0025 − 0.0009)
A: 0.02842… J = 0.028 J (2 s.f.)
Step 3 – (b) potential energy
Ep = Etotal − Eₖ = 0.04441… − 0.02842… = 0.01598… J
A: 0.016 J (2 s.f.)
Watch out: round only at the end. Subtracting the rounded values 0.044 − 0.028 still gives 0.016 J here, but it often does not.
Q10 (Stretch) A 0.25 kg mass oscillates on a spring with simple harmonic motion with a period of 1.25 s and an amplitude of 8.0 cm, with no damping. (a) Show that the kinetic energy equals the potential energy when x = A ÷ √2, and calculate this displacement to 2 significant figures. (b) Calculate the speed of the mass at this displacement and show that its kinetic energy is half the total energy.
Show solution
Step 1 – (a) condition
Eₖ = Ep means Eₖ = ½Etotal: ½mω²(A² − x²) = ½ × ½mω²A², so A² − x² = ½A², x² = ½A² and x = A ÷ √2
A: x = 8.0 ÷ √2 = 5.656… cm = ±5.7 cm (2 s.f.)
Step 2 – (b) speed
ω = 2π ÷ 1.25 = 5.026… rad s⁻¹; v = ω√(A² − x²) = ω × A ÷ √2 = 5.026… × 0.080 ÷ √2 = 0.2843… m s⁻¹
A: 0.28 m s⁻¹ (2 s.f.)
Step 3 – (b) energies
Eₖ = ½mv² = 0.5 × 0.25 × 0.2843…² = 0.01010… J; Etotal = ½mω²A² = 0.5 × 0.25 × 25.26… × 0.080² = 0.02021… J
A: 0.01010… ÷ 0.02021… = 0.5, so Eₖ is half the total energy (0.010 J of 0.020 J)
Watch out: Eₖ = Ep is not at x = A ÷ 2. Energy depends on x², so the crossover is at 71% of the amplitude.
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