Phase relationship between displacement, velocity and acceleration of SHM

As we have seen that

x = A sin (ωt + φ)

v = A ω cos (ωt+ φ)

= Aω sin (ωt + φ + π/2)

And a= – Aω^{2} sin (ωt + φ)

= Aω^{2} sin (ω t + φ + π)

Thus, we conclude that in SHM, particle velocity is ahead in phase by π/2 as compared to the displacement and acceleration is further ahead in phase by π/2.

In figure, x, v and a as functions of time are illustrated.**Example: **The equation of motion of a particle is given by dp/ dx + mω^{2}n = 0, where p is the momentum and n is the position. Then, the particle

1. Moves along a straight line

2. Moves along a parabola

3. Execute simple harmonic motion

4. Falls freely under gravity

**Solution: **dp/dt = – mω^{2}n

dp/dt = rate of change of momentum

= F (force)

∴ F = – mω^{2}n

m and ω are constants

∴ F α – n

Force α displacement

So, this is the condition for simple harmonic motion, so the particle will execute simple harmonic motion.

**Example: **A particle executing simple harmonic motion has an amplitude of 6 cm. Its acceleration at a distance from the mean position 2 cm is 8 cm s^{-2}. The maximum speed of the particle

**Solution: **Amplitude, a = 6 cm

Displacement, y = 2 cm

Acceleration, A = 8 cm, s^{-2}

A = ω^{2}y

ω = 2 rads^{-1}

Maximum speed, V_{max} = aω = 6 x 2

= 12 cm s^{-1}