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

The correct option is A The mean position and extreme positions

We know that, the condition for SHM is
$\overrightarrow{a}=-{\omega }^2\overrightarrow{x}\Rightarrow |a|={\omega }^{2}x$
Rate of change of acceleration
$\frac{da}{dt}={\omega }^2\frac{dx}{dt}={\omega }^{2}v$
But we know that, $v=\omega \sqrt{A^2-x^2}$
Using this in the above equation, we get
$\frac{da}{dt}={\omega }^3\sqrt{A^2-x^2}$
For $\frac{da}{dt}$ to be maximum, $x^2$ should be minimum i.e at $x=0$
For $\frac{da}{dt}$ to be minimum , $x^2$ should be maximum i.e at $x=\pm a$
Hence, option (a) is the correct answer.