Questions on simple harmonic motion

Question 1

Write down the defining equation of simple harmonic motion and say where the particle moves fastest and where its acceleration is greatest.

Accepted Answer

ẍ = −ω²x. The acceleration is proportional to the displacement from a fixed point and directed back towards it. The speed is greatest at the centre, where x = 0 and the acceleration vanishes. The acceleration is greatest at the ends, where the particle is momentarily at rest. B1 for the defining equation, B1 for greatest speed at the centre, B1 for greatest acceleration at the ends.

Question 2

A particle moves with simple harmonic motion of amplitude 0.4 m and ω = 5 rad/s. Find its maximum speed, its maximum acceleration and its period.

Accepted Answer

Maximum speed = aω = 2 m/s. Maximum acceleration = aω² = 0.4 × 25 = 10 m/s². Period = 2π/ω = 1.26 s. B1 B1 B1, one for each of the three values.

Question 3

A particle moves with simple harmonic motion with ω = 3 rad/s and amplitude 0.5 m. Find its speed when its displacement from the centre is 0.3 m, and the displacement at which its speed is half its greatest value.

Accepted Answer

Use v² = ω²(a² − x²), which links speed to displacement with no time involved. At x = 0.3: v² = 9(0.25 − 0.09) = 1.44, so v = 1.2 m/s. The greatest speed is aω = 1.5 m/s, so half of it is 0.75: 0.5625 = 9(0.25 − x²), x² = 0.1875 and x = 0.433 m. M1 for v² = ω²(a² − x²), A1 for 1.2 m/s, A1 for 0.433 m. Half the speed is reached at 0.433/0.5 = 87% of the amplitude, not at half of it, because the particle keeps most of its speed until close to the ends.

Question 4

State the two standard solutions of ẍ = −ω²x for the displacement of a particle in simple harmonic motion, and say which starting condition each one describes.

Accepted Answer

x = a sin ωt, for a particle timed from the centre, where x = 0 at t = 0 and the particle is moving at its greatest speed; and x = a cos ωt, for a particle timed from one end, where x = a at t = 0 and the particle is at rest. B1 for the sine form with its condition, B1 for the cosine form with its condition. Any other start needs a phase angle, x = a sin(ωt + α), and both standard forms are cases of it.

Question 5

A particle moving with simple harmonic motion has speed 1.6 m/s at a displacement of 0.3 m and speed 1.2 m/s at a displacement of 0.4 m. Find the amplitude and ω.

Accepted Answer

Using v² = ω²(a² − x²) twice: 2.56 = ω²(a² − 0.09) and 1.44 = ω²(a² − 0.16). Dividing removes ω²: 2.56/1.44 = (a² − 0.09)/(a² − 0.16), so 1.778(a² − 0.16) = a² − 0.09 and 0.778a² = 0.194, giving a² = 0.25 and a = 0.5 m. Then ω² = 2.56/0.16 = 16, so ω = 4 rad/s. M1 for using v² = ω²(a² − x²), A1 A1 for the two equations, M1 for eliminating ω², A1 for a = 0.5 m, A1 for ω = 4 rad/s.

Question 6

For that motion, find the time taken to travel from the centre to a displacement of 0.25 m, and the next time it is at that displacement.

Accepted Answer

Timing from the centre, x = 0.5 sin4t. Setting 0.25 = 0.5 sin4t gives sin4t = 0.5, so 4t = π/6 and t = π/24 = 0.131 s. The particle returns to the same displacement when 4t = 5π/6, that is t = 5π/24 = 0.654 s, on the way back. M1 for x = 0.5 sin4t, M1 for solving sin4t = 0.5, A1 for 0.131 s, M1 for 4t = 5π/6, A1 for 0.654 s.

Question 7

Explain why the period of simple harmonic motion does not depend on the amplitude.

Accepted Answer

A larger amplitude means a longer distance to travel. It also means a proportionally larger restoring force and so a proportionally larger speed at every corresponding point. Because the force is proportional to the displacement, the two effects cancel exactly and T = 2π/ω whatever the amplitude. B1 for the greater distance, B1 for the proportionally greater restoring force and speed, B1 for the two effects cancelling.

Question 8

A particle moving with simple harmonic motion has greatest speed 3 m/s and greatest acceleration 12 m/s². Find ω, the amplitude and the period of the motion, and the speed of the particle at a displacement of 0.45 m.

Accepted Answer

The greatest speed is aω and the greatest acceleration is aω², so dividing one by the other removes the amplitude: ω = 12/3 = 4 rad/s. Then a = 3/ω = 3/4 = 0.75 m, and the period is 2π/4 = 1.57 s.
For the speed, v² = ω²(a² − x²) = 16(0.5625 − 0.2025) = 16 × 0.36 = 5.76, so v = 2.4 m/s.
M1 for dividing aω² by aω to isolate ω, A1 for ω = 4 rad/s, A1 for a = 0.75 m, B1 for the period 1.57 s, M1 for using v² = ω²(a² − x²), A1 for 2.4 m/s.

Question 9

A particle moving with simple harmonic motion has amplitude 0.6 m and period 3 s. It is released from rest at one end of its path. Find the time it takes to reach a displacement of 0.3 m from the centre, and the time it takes to first pass through the centre.

Accepted Answer

Released from rest at an end means timing starts at x = a, so take the cosine form: with ω = 2π/3, x = 0.6 cos(2πt/3).
Setting 0.3 = 0.6 cos(2πt/3) gives cos(2πt/3) = ½, so 2πt/3 = π/3 and t = 0.5 s.
The centre is x = 0, so 2πt/3 = π/2 and t = 0.75 s, a quarter of the period, as the journey from an end to the centre must take.
B1 for choosing the cosine form with ω = 2π/3, M1 for solving cos(2πt/3) = ½, A1 for 0.5 s, A1 for 0.75 s. Note that the particle covers the first half of the distance in two thirds of the time, because it moves slowest near the end of its path; assuming uniform speed here is the standard error.

Question 10

A particle moves with simple harmonic motion of period 2 s. Find the magnitude of its acceleration when its displacement from the centre is 0.2 m. Given also that its greatest speed is 1.5 m/s, find the amplitude.

Accepted Answer

ω = 2π/T = 2π/2 = π rad/s. The defining equation gives ẍ = −ω²x, so at x = 0.2 the acceleration has magnitude π² × 0.2 = 1.97 m/s², directed towards the centre. The greatest speed is aω, so a = 1.5/π = 0.477 m. B1 for ω = π, M1 for ẍ = −ω²x, A1 for 1.97 m/s², A1 for 0.477 m. The acceleration at 0.2 m does not depend on the amplitude at all; the amplitude only fixes how far out the motion reaches, and so the greatest speed.

Question 11

A particle moves with simple harmonic motion about a centre O with amplitude 2 m and period 4 s. Find its speed when it is 1 m from O, and the time it takes to travel directly from the point 1 m from O to the end of its path on the same side.

Accepted Answer

ω = 2π/4 = π/2 rad/s. Speed at x = 1: v² = ω²(a² − x²) = (π²/4)(4 − 1), so v = (π/2)√3 = 2.72 m/s. For the time, take x = 2 sin(πt/2) with t = 0 at O. The particle is at x = 1 when sin(πt/2) = ½, so πt/2 = π/6 and t = 1/3 s; it reaches the end x = 2 when πt/2 = π/2, so t = 1 s. The journey from 1 m out to the end takes 1 − 1/3 = 2/3 s. B1 for ω = π/2, M1 for v² = ω²(a² − x²), A1 for 2.72 m/s, M1 for timing with the sine form, A1 for 2/3 s.

Question 12

Starting from the equation of motion, derive v² = ω²(a² − x²). Hence show that a particle moving with simple harmonic motion spends one third of each period within half the amplitude of the centre.

Accepted Answer

Write the acceleration as v dv/dx, so v dv/dx = −ω²x. Separating gives ∫v dv = −ω²∫x dx, so ½v² = −½ω²x² + c. At the extreme of the motion x = a and v = 0, so c = ½ω²a². Substituting back and doubling gives v² = ω²(a² − x²). The result contains no time, so it answers position and speed questions in one step.
For the second part, time from the centre with x = a sin ωt. Then |x| ≤ a/2 requires |sin ωt| ≤ ½, that is ωt between 0 and π/6, or between 5π/6 and 7π/6, or between 11π/6 and 2π.
Those stretches measure π/6, π/3 and π/6, totalling 2π/3 out of a full 2π, so the proportion is 1/3.
M1 for writing the acceleration as v dv/dx, M1 for separating and integrating, A1 for ½v² = −½ω²x² + c, A1 for evaluating c and reaching the printed result, M1 for |sin ωt| ≤ ½, A1 for the three intervals, A1 for the total 2π/3, A1 for the proportion 1/3. The answer depends on neither a nor ω, and choosing x = a cos ωt instead gives the same third with the intervals shifted.

Question 13

A particle moves along a straight line so that its displacement from a point O at time t seconds is x = 0.3 sin 2t + 0.4 cos 2t metres. Show that the particle moves with simple harmonic motion about O, and find the amplitude of the motion, its greatest speed, and the first time at which the displacement from O is greatest.

Accepted Answer

Differentiate twice: ẋ = 0.6 cos 2t − 0.8 sin 2t, and ẍ = −1.2 sin 2t − 1.6 cos 2t = −4(0.3 sin 2t + 0.4 cos 2t) = −4x. The acceleration is a negative constant times the displacement from O, which is the defining equation with ω² = 4, so the motion is simple harmonic with ω = 2 rad/s.
For the amplitude, write the sum as a single sinusoid: x = R sin(2t + α) with R = √(0.3² + 0.4²) = 0.5 m and tan α = 0.4/0.3, so α = 0.927.
Greatest speed = aω = 0.5 × 2 = 1 m/s, at O.
The displacement is greatest when sin(2t + α) = 1, so 2t + 0.927 = π/2 and t = 0.322 s.
M1 for differentiating twice, A1 for ẍ = −4x with the conclusion stated and ω = 2 named, M1 for the R sin(2t + α) form, A1 for the amplitude 0.5 m with α = 0.927, B1 for the greatest speed 1 m/s, M1 for setting sin(2t + α) = 1, A1 for 0.322 s. Showing ẍ = −4x must display the −4 multiplying the original x; quoting the R form without the differentiation proves nothing about the acceleration.

Question 14

A particle of mass m moves with simple harmonic motion of amplitude a and angular frequency ω about a centre O. Show that its kinetic energy at displacement x is ½mω²(a² − x²), and that the work done against the restoring force in moving it from O to displacement x is ½mω²x². Deduce that the total energy is constant, and find the displacement at which the kinetic and potential energies are equal.

Accepted Answer

Kinetic energy: ½mv² with v² = ω²(a² − x²) gives ½mω²(a² − x²).
The restoring force has magnitude mω²x, so the work done against it from O to x is ∫mω²x dx from 0 to x = ½mω²x², which is the potential energy stored at displacement x.
Adding: ½mω²(a² − x²) + ½mω²x² = ½mω²a², which contains no x and so is the same at every point of the motion. It is the kinetic energy at the centre and equally the potential energy at an end.
Equal energies: ½mω²(a² − x²) = ½mω²x², so a² = 2x² and x = a/√2, about 0.707a, on either side of O.
M1 for ½mv² with the speed-displacement result, A1 for ½mω²(a² − x²), M1 for integrating the restoring force, A1 for ½mω²x², A1 for the constant total ½mω²a², A1 for x = a/√2. The energy is shared equally at 0.707a rather than at a/2 because the potential energy grows with the square of the displacement, slowly at first and steeply near the ends.

Question 15

A particle moves with simple harmonic motion about a centre O with amplitude 5 m and period 6 s. Find its speed when it is 3 m from O. Find the time it takes to travel directly from the point 3 m from O to the point 4 m from O on the same side, and the time it takes to travel from O to the point 1 m from O. State which of these two journeys of 1 m takes longer, and explain why.

Accepted Answer

ω = 2π/6 = π/3 rad/s. Speed at x = 3: v² = (π/3)²(25 − 9), so v = (π/3) × 4 = 4.19 m/s.
Timing from O with x = 5 sin(πt/3): the particle is at x = 3 when sin(πt/3) = 0.6, so t = (3/π) arcsin 0.6 = 0.6145 s, and at x = 4 when sin(πt/3) = 0.8, so t = (3/π) arcsin 0.8 = 0.8855 s. The journey from 3 m to 4 m takes 0.8855 − 0.6145 = 0.271 s.
From O to x = 1: sin(πt/3) = 0.2, so t = (3/π) arcsin 0.2 = 0.192 s.
The outer metre, from 3 m to 4 m, takes longer, because the particle is slowing as it moves away from O: its speed falls from 4.19 m/s at 3 m to (π/3) × 3 = 3.14 m/s at 4 m, whereas near O it is moving at close to its greatest speed of 5π/3 = 5.24 m/s throughout.
B1 for ω = π/3, M1 A1 for the speed 4.19 m/s, M1 for timing with arcsin, A1 for 0.271 s and 0.192 s, B1 for the comparison with the reason. Both times come from the same sine form, since the particle passes through O, 1 m, 3 m and 4 m in that order on a single outward swing.

Question 16

The graph shows the displacement x of a particle that is moving with simple harmonic motion about the origin. Use the graph to write down the amplitude and the period of the motion, and hence find ω, the greatest speed of the particle, and the speed of the particle at the instant its displacement is 0.3 m.

Accepted Answer

The curve rises to 0.6 m above the axis and falls to 0.6 m below it, and those are the extremes of the motion, so the amplitude is a = 0.6 m. B1 for the amplitude. One complete oscillation is the time to return to the same displacement moving the same way: the particle is at the centre and moving in the positive direction at t = 0 and next does so at t = 4, so the period is T = 4 s. B1 for the period. Counting from one peak to the next, t = 1 to t = 5, gives the same 4 s, and the gap between consecutive zeros is only half a period, 2 s, which is the reading to be careful of.
The period and ω are linked by T = 2π/ω. M1 for using T = 2π/ω. So ω = 2π/4 = π/2 = 1.57 rad s⁻¹. A1 for ω.
The greatest speed is aω = 0.6 × π/2 = 0.3π = 0.942 m s⁻¹, and it occurs each time the particle passes through the centre, which is where the graph is steepest. A1 for the greatest speed.
Speed and displacement are linked without any reference to time by v² = ω²(a² − x²). M1 for this formula. At x = 0.3, v² = (π/2)²(0.36 − 0.09) = 2.4674 × 0.27 = 0.6662, so v = 0.816 m s⁻¹. A1 for 0.816 m s⁻¹.
That is about 87 per cent of the greatest speed at half the amplitude, which is the lopsidedness of simple harmonic motion: the particle spends its time near the ends and its speed near the centre.