Velocity & Acceleration of SHM

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= - Aω2 sin (ωt + φ)

= Aω2 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 + mω2n = 0, where p is the momentum and n is the position. Then, the particle

• Moves along a straight line
• Moves along a parabola
• Execute simple harmonic motion
• Falls freely under gravity

Solution: dp/dt = - mω2n

dp/dt = rate of change of momentum

= F (force)

∴ F = - mω2n

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 = ω2y

ω = 2 rads-1

Maximum speed, Vmax = aω = 6 x 2

= 12 cm s-1