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 position
• B. The extreme position and mean position
• C. The mean position alternatively
• D. The extreme position alternatively.

Answer 1

The correct option is A The mean position and extreme position

Acceleration of SHM particle $a=-{\omega }^{2}x={\omega }^{2}x$ (numerically)
Rate of change of acceleration
$\frac{da}{dt}={\omega }^2\frac{dx}{dt}={\omega }^{2}v$

$\frac{da}{dt}={\omega }^3\sqrt{A^2-x^2}$

For $da/dt$ to be maximum, $x^2$ should be minimum i.e., $x=0$ and For $da/dt$ to be mimum $x^2$ should be maximum i.e., $x=\pm a$.

Hence option A