Assertion: In simple harmonic motion the velocity is maximum when the acceleration is minimum
Reason : Displacement and 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 the assertion is true but reason is false
D. If both the 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 acceleration of a particle executing `SHM` is zero (where velocity is maximum ) at mean position and maximum at exterme position where velocity is minimum .Now let `y = a sin (wt +j)`
Then velocity `(dy)/(dt) = omega cos(omega t + phi)= a omega sin {(omega t + (phi)/(2)) + (pi)/(2)}`
Thus , displecement and velocity of `SHM` difference by `(pi)/(2)` in phase