Assertion: In simple harmonic motion the velocity is maximum when the acceleration is minimum
Reason : Displacementand velocity of `SHM`differ in phase by `(pi)/(2)`

Options

A. If both assertion and reason are true and the reasopn is correct explanation of the assertion
B. If both assertion and reason are true and but not the correct explanation of assertion
C. If assertion is true but the reason is false
D. If both assertion and reason are false

Answer

B

Explanation

As velocity
`(dy)/(dt) = omega sqrt(A^(2) - y^(2))`…(1)
and acceleration
`(d^(2)y)/(dt^(2)) = - omega^(2)y`
When `y = 0`……(2)
Velocity`(dy)/(dx) = omega A` (maximum)
and acceleration `(d^(2)y)/(dt^(2)) = 0` (minimum)
When `y = A,v = (dy)/(dt) = 0` (minimum)
and acceleration
`(d^(2)y)/(dt^(2)) = - omega^(2)A`(maximum)
Hence accleration of a particle executing `SHM` is zero (wherevelocity is maximum) at mean position and maximum at exerme position where velocity is minimum Now velocity `(dy)/(dt) = omega cos (omega t + varphi)`
`= a omega sin [(omega t + (varphi)/(2)) + (pi)/(2))` Thus displacement and velocity of `SHM` difference by `(pi)/(2)` in phase