While a particle executes simple harmonic motion, the rate of change of acceleration is maximum and minimum respectively at

Options

A. the mean position and extreme positions
B. the extreme positions and mean position
C. the mean position alternatively
D. the extreme positions alternatively.

Answer

A

Explanation

`a=-omega^2x=omega^2x`(numerically)
Rate of change of acceleration
`(da)/(dt)=omega^2(dx)/(dt)=omega^2v`
`omega^3sqrt(A^2-x^2)`
For `(da)/(dt)` to be minimum `x^2` should be maximum i.e., `x=+-a`.