The pahse difference between the instantaneous veliocity and acceleration of a particle executing simple harmonic motion is

Options

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

Answer

B

Explanation

The displacement equation of particle execttion `SHM`
`x = a cos omega t`…(i)
velocity `v = (dx)/(dt6) = - a omega sin omega t`….(ii)
and its acceleration
`a = (dx)/(dt) = - a omega^(2) cos omega t` …(iii)
and its acceleration
fig (i) shown of eq (i) fig(i) eq(ii) and fig (ii) is a plot of eq (ii) it should be reted that in the figure the curve `v` is shiffted by the leftfrom the curve of x by one quarter period `((1)/(4)T)`

similarly the acceleration curve is shiffted (to the left) `(1)/(4)T` relative to the velocity curve `v` this impless that velocity is `90^(@) (0.5pi)` out of phase but dipendent and the velocity but `180^(@)` out of phase with displecement