The phase difference between the instantaneous Velocity and acceleration of a particle executing simple harmonic motion is

Options

A. `0.5 pi`
B. `pi`
C. `0.707 pi`
D. Zero

Answer

A

Explanation

The displacement equation of particle executing SHM is
`x= a cos(omegat+phi)`…(i)
Velocity, `v=(dx)/(dt)= -a omega sin(omega+phi)` ….(ii)
Acceleration,

`A=(dv)/(dt)= -a omega^(2) cos (omegat+phi)`….(iii)
Fig is a plot of Eq.(i) with `phi=0`. Fig.shows Eq.(ii) also with `phi=0`. Fig(iii) is a plot of Eq.(iii). It should be noted that in the figures the curve of `v` is shifted (to the left) from the curve of x by one-quarter period `(1/4T)`. Similarly, the acceleration curve of A is shifted (to the left) by `1/4T` relative to thevelocity curve `v`. This implies that Velocity is `90^(@)(0.5pi)` out phase with the displacement and the acceleration is `90^(@)(0.5pi)` out phase with the velocity but `180^(@)(pi)` out of phase with displacement.