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
Simple harmonic motion: an oscillation in which the acceleration is directly proportional to the displacement from the equilibrium position and is always directed towards the equilibrium position, a = −ω²x.
Angular frequency: ω = 2πf = 2π ÷ T, measured in radians per second (rad s⁻¹). One complete oscillation is 2π radians.
Maximum acceleration: amax = ω²A, which occurs at maximum displacement (x = ±A). At the equilibrium position the acceleration is zero.
The equation
acceleration = −(angular frequency)² × displacement a = −ω²x with angular frequency = 2π × frequency ω = 2πf
| Quantity | Symbol | Unit |
|---|---|---|
| acceleration | a | metres per second squared (m s⁻²) |
| angular frequency | ω | radians per second (rad s⁻¹) |
| displacement from equilibrium | x | metres (m) |
| amplitude (maximum displacement) | A | metres (m) |
| frequency | f | hertz (Hz), equivalent to s⁻¹ |
Rearranged: amax = ω²A | ω = √(amax ÷ A) | f = ω ÷ 2π | ω = 2π ÷ T | The minus sign means the acceleration always points back towards equilibrium, so the gradient of an a–x graph is −ω².
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 a frequency of 2.5 Hz. Calculate its angular frequency, to 3 significant figures.
Show solution
F: ω = 2πf
I: ω = 2π × 2.5
F: ω is already the subject
A: 15.70… rad s⁻¹ = 15.7 rad s⁻¹ (3 s.f.)
Q2 (Core) A mass on a spring moves with simple harmonic motion with angular frequency 12 rad s⁻¹. Calculate its acceleration when its displacement is +0.030 m, to 2 significant figures, and state its direction.
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F: a = −ω²x
I: a = −12² × 0.030 = −144 × 0.030
F: a is already the subject
A: −4.32 m s⁻² = −4.3 m s⁻² (2 s.f.), directed towards the equilibrium position (opposite to the displacement)
Q3 (Core) A mass on a spring oscillates with simple harmonic motion with a period of 0.80 s and an amplitude of 2.0 cm. Calculate its maximum acceleration, to 2 significant figures.
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Convert first: 2.0 cm = 0.020 m; ω = 2π ÷ T = 2π ÷ 0.80 = 7.853… rad s⁻¹
F: amax = ω²A
I: amax = 7.853…² × 0.020 = 61.68… × 0.020
F: amax is already the subject
A: 1.233… m s⁻² = 1.2 m s⁻² (2 s.f.)
Q4 (Core) The prong of a tuning fork vibrates with simple harmonic motion at 440 Hz with an amplitude of 0.50 mm. Calculate the maximum acceleration of the tip of the prong, to 2 significant figures.
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Convert first: 0.50 mm = 5.0 × 10−4 m
Step 1 – angular frequency
ω = 2πf = 2π × 440 = 2764.6… rad s⁻¹
Step 2 – maximum acceleration
F: amax = ω²A
I: amax = 2764.6…² × 5.0 × 10−4
F: amax is already the subject
A: 3821.5… m s⁻² = 3.8 × 103 m s⁻² (2 s.f.)
Watch out: forgetting to square ω. Using ωA gives the maximum speed, not the maximum acceleration.
Q5 (Exam) A graph of acceleration a against displacement x for an oscillator is a straight line through the origin. One point on the line is x = 0.040 m, a = −2.5 m s⁻². Calculate the angular frequency and the frequency of the oscillations, to 2 significant figures.
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Step 1 – gradient
gradient = a ÷ x = −2.5 ÷ 0.040 = −62.5 s⁻²
Step 2 – angular frequency
F: a = −ω²x, so gradient = −ω²
I: −ω² = −62.5
F: ω = √62.5 = 7.905… rad s⁻¹
A: 7.9 rad s⁻¹ (2 s.f.)
Step 3 – frequency
f = ω ÷ 2π = 7.905… ÷ 2π = 1.258…
A: 1.3 Hz (2 s.f.)
Watch out: the gradient is −ω², not ω. Square-root the magnitude of the gradient.
Q6 (Exam) A student measures the acceleration of an oscillating trolley at three displacements: x = 0.010 m, a = −0.36 m s⁻²; x = 0.020 m, a = −0.72 m s⁻²; x = 0.030 m, a = −1.08 m s⁻². Show that the motion is simple harmonic and calculate the period, to 2 significant figures.
Show solution
Step 1 – test for SHM
a ÷ x = −0.36 ÷ 0.010 = −0.72 ÷ 0.020 = −1.08 ÷ 0.030 = −36 s⁻² for every reading. a ∝ x and the minus sign shows a is opposite to x (towards equilibrium), so the motion is SHM
Step 2 – angular frequency
F: a = −ω²x
I: ω² = 36
F: ω = √36 = 6.0 rad s⁻¹
Step 3 – period
T = 2π ÷ ω = 2π ÷ 6.0 = 1.047… s
A: 1.0 s (2 s.f.)
Q7 (Exam) A piston in an engine moves with approximately simple harmonic motion. The stroke (the total distance between the two extreme positions) is 8.0 cm and the engine runs at 3000 revolutions per minute. Calculate the maximum acceleration of the piston, to 2 significant figures.
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Convert first: amplitude = 8.0 ÷ 2 = 4.0 cm = 0.040 m; f = 3000 ÷ 60 = 50 Hz
Step 1 – angular frequency
ω = 2πf = 2π × 50 = 314.1… rad s⁻¹
Step 2 – maximum acceleration
F: amax = ω²A
I: amax = 314.1…² × 0.040 = 98 696… × 0.040
F: amax is already the subject
A: 3947.8… m s⁻² = 3.9 × 103 m s⁻² (2 s.f.)
Watch out: the amplitude is half the total distance between the extremes. Using 8.0 cm doubles the answer.
Q8 (Exam) A 0.25 kg mass oscillates with simple harmonic motion at 1.5 Hz with an amplitude of 6.0 cm. Calculate the maximum resultant force on the mass, to 2 significant figures.
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Convert first: 6.0 cm = 0.060 m
Step 1 – maximum acceleration
ω = 2πf = 2π × 1.5 = 9.424… rad s⁻¹
F: amax = ω²A
I: amax = 9.424…² × 0.060 = 88.82… × 0.060
A: amax = 5.329… m s⁻²
Step 2 – maximum resultant force
F = mamax = 0.25 × 5.329…
A: 1.332… N = 1.3 N (2 s.f.), at the extremes of the motion, directed towards equilibrium
Q9 (Stretch) A horizontal platform oscillates vertically with simple harmonic motion at a frequency of 3.0 Hz. A small coin rests on the platform. The amplitude is slowly increased. Calculate the amplitude at which the coin just loses contact with the platform, to 2 significant figures. (g = 9.81 m s⁻²)
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Step 1 – where contact is lost
At the top of the motion the acceleration is downwards with magnitude ω²A. The only downward force on the coin is its weight, so the coin can accelerate downwards at no more than g. Contact is lost when amax = g
Step 2 – angular frequency
ω = 2πf = 2π × 3.0 = 18.84… rad s⁻¹, so ω² = 355.3… s⁻²
Step 3 – amplitude
F: amax = ω²A = g
I: 355.3… × A = 9.81
F: A = 9.81 ÷ 355.3…
A: 0.02761… m = 2.8 cm (2 s.f.)
Watch out: contact is lost at the top, not the bottom. At the bottom the platform accelerates upwards and pushes harder on the coin.
Q10 (Stretch) The platform in Q9 oscillates at 3.0 Hz with an amplitude of 1.5 cm, carrying a 0.20 kg block. (a) Calculate the contact force on the block at the top and at the bottom of the motion, to 2 significant figures. (b) The amplitude is kept at 1.5 cm. Calculate the highest frequency at which the block stays in contact throughout, to 2 significant figures. (g = 9.81 m s⁻²)
Show solution
Convert first: 1.5 cm = 0.015 m
Step 1 – (a) maximum acceleration
amax = ω²A = 355.3… × 0.015 = 5.329… m s⁻²
Step 2 – (a) top: acceleration downwards
mg − R = mamax, so R = m(g − amax) = 0.20 × (9.81 − 5.329…) = 0.896… N
A: 0.90 N at the top (2 s.f.)
Step 3 – (a) bottom: acceleration upwards
R − mg = mamax, so R = m(g + amax) = 0.20 × (9.81 + 5.329…) = 3.027… N
A: 3.0 N at the bottom (2 s.f.)
Step 4 – (b) highest frequency
ω²A = g, so ω = √(9.81 ÷ 0.015) = 25.57… rad s⁻¹; f = ω ÷ 2π = 25.57… ÷ 2π = 4.070…
A: 4.1 Hz (2 s.f.)
Watch out: the weight is still 1.962 N throughout. Only the contact force changes as the platform accelerates.
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