Time and position in SHM

Choosing sin or cos

• Timing starts at the centre: x = a sin ωt, and v = aω cos ωt.
• Timing starts at an end point: x = a cos ωt, and v = -aω sin ωt.

Your calculator must be in radians.

Time between two positions

Find the time to each position from the same starting point, then subtract. Use symmetry: the motion from O to an end takes a quarter of a period, T/4.

Time spent beyond a point

To find how long in each oscillation the particle is more than d from O, find the time t_1 to go from O to d, then the time beyond d in one quarter is T/4 - t_1. Pearson's examiners advise finding one value of t and thinking about its significance by symmetry, rather than finding many values and trying to combine them.

Worked example

A particle moves with SHM of amplitude 3 m and ω = 2 rad/s. Find the time to go from the centre to a point 1.5 m from the centre, and the total time in each oscillation for which it is more than 1.5 m from the centre.

1. x = 3 sin 2t. Set 1.5 = 3 sin 2t: 2t = π/6, so t_1 = π/12.
2. T = π, so T/4 = π/4. Time beyond 1.5 m in one quarter: π/4 - π/12 = π/6.
3. Each of the four quarters of an oscillation contains π/6 beyond 1.5 m, so the total is 4 × π/6 = 2π/3.

Answer: t_1 = π/12 = 0.262 s; total 2.09 s per oscillation

Check yourself

1. In the worked example, find the time from the centre to the point 1.5 m from the centre, to 3 significant figures.
   Show answer

   0.262 s. sin 2t = 0.5, t = π/12 = 0.262 s.

2. A particle moving with SHM of amplitude 4 m and ω = 0.5 rad/s is released from rest at an end point. Find the time it takes to reach a point 2 m from the centre (on the same side), to 3 significant figures.
   Show answer

   2.09 s. x = 4 cos 0.5t. 2 = 4 cos 0.5t gives 0.5t = π/3, t = 2π/3 = 2.09 s.

3. For the same particle (released from an end point, a = 4 m, ω = 0.5), find its distance from the centre after 1 second, to 3 significant figures.
   Show answer

   3.51 m. x = 4 cos 0.5 = 3.51 m (radians).

4. A particle moves with SHM of amplitude 2 m and ω = 3 rad/s. Find the time taken to travel directly from a point 1 m from the centre to a point 2 m from the centre, to 3 significant figures.
   Show answer

   0.349 s. From O: 1 = 2 sin 3t gives t = π/18; 2 = 2 sin 3t gives t = π/6. Difference = π/9 = 0.349 s.

5. A particle moves with SHM of amplitude 3 m and ω = 2 rad/s. Find the total time in one oscillation for which it is more than 1.5 m from the centre, to 3 significant figures.
   Show answer

   2.09 s. Each quarter has π/4 - π/12 = π/6 beyond 1.5 m, so the total is 4 × π/6 = 2.09 s.

6. A particle moves so that x = 3 sin 2t. Find its speed when t = 0.5, to 3 significant figures.
   Show answer

   3.24 m/s. v = 6 cos 2t, so v(0.5) = 6 cos 1 = 3.24 m/s.

7. A particle in SHM is released from rest at time t = 0. Which expression gives its displacement from the centre?
   • x = a sin ωt
   • x = a tan ωt
   • x = a cos ωt
   • x = aωt
   Show answer

   x = a cos ωt. At t = 0 it is at an end point, where x = a, and cos 0 = 1.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3), Unit M3
• Pearson Edexcel IAL Mechanics M3 (WME03) question paper, June 2025: front-cover instruction on g and accuracy
• Pearson Edexcel IAL Mechanics M3 (WME03) examiners' report, June 2025

Last updated 8 November 2026. Spotted a mistake? Tell us.