Practise › Questions › Simple harmonic motion
Simple harmonic motion questions
One differential equation, and every oscillation on the paper is an instance of it. Prove the equation holds, read off ω, and all the standard results follow.
6 original questions · 28 marks · the simple harmonic motion notes · Further Mechanics 2
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
Write down the defining equation of simple harmonic motion and say where the particle moves fastest and where its acceleration is greatest.
Worked 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. The minus sign is part of the definition and dropping it loses the mark.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.
Worked 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.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.
Worked 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.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.
Worked 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.Explain why the period of simple harmonic motion does not depend on the amplitude.
Worked 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. That is what makes a pendulum useful as a clock.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.
Worked 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 particle covers the middle half of its range in a third of the time, so it is far quicker through the centre than the ends. The answer depends on neither a nor ω, and choosing x = a cos ωt instead gives the same third with the intervals shifted.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise simple harmonic motion one question at a time
The player marks nothing for you. It shows one question, waits, then shows the worked answer so you can mark yourself, and brings a question back sooner when it went badly.