Question

A particle executes simple harmonic motion. Its instantaneous acceleration is given by $a=-px,$ where $p$ is a positive constant and $x$ is the displacement from the mean position. Angular frequency of the particle is given by:

• A. $\frac{1}{p}$
• B. $p$
• C. $\sqrt{p}$
• D. $\frac{1}{\sqrt{p}}$

Answer 1

Answer: (C)

$\sqrt{p}$

Comparing with $a=-{\omega }^{2}x$
${\omega }^2=p$
$\therefore$ $\omega =\sqrt{p}$