Question

While a particle executes simple harmonic motion, the rate of change of acceleration is maximum and minimum respectively at
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 1

Correct Answer - A
`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`.