Which one of the following statement is true for the speed `v` and the acceleration a of a particle executing simple harmonic motion?

Options

A. When `v` is maximum `a` is maximum
B. When `v` is minimum `a` is also minimum
C. When `v` is zero `a` is zero
D. When `v` is maximum `a` is zero

Answer

D

Explanation

In simple harmonic motion the displacement equation is `y = A sin omega t`
Where `A` is amplitude of the motion
Velocity `v = (dv)/(dt) = A cos omega t`
`v = A omega sqrt(1 - sin^(2) omega t)`
`v = omega sqrt(A^(2) - y^(2))` ….(i)
Acceleration `a = (dv)/(dt) = (d)/(dt) (A omega t cos omega t)`
`a = -A omega ^(2) sin omega t`
`a = - omega^(2) t`....(ii)
When `y = 0 , y = A omega = v_(max)`
`a= 0 = a_(min)`
when
`y = A, y = 0 = v_(min)`
`a = omega^(2) A = a_(max)`
Hence it is clean that when `x` is maximum then `a` is minimum (i.e. zero) at vics versa.