A particle move in S.H.M. such that its acceleration is a $=-\mathrm{px}$, where ' $\mathrm{x}$ ' is the…
A particle move in S.H.M. such that its acceleration is a $=-\mathrm{px}$, where ' $\mathrm{x}$ ' is the displacement of particle from equilibrium position and ' $\mathrm{p}$ ' is a constant. The period of oscillation is
$2 \pi \sqrt{p}$
$2 \sqrt{\frac{\pi}{p}}$
$\frac{2 \pi}{p}$
$\frac{2 \pi}{\sqrt{p}}$
Solution
For SHM the acceleration a is proportional to the displacement $x$ and it is directed towards the mean position of the particle:
$\mathrm{a}=-\omega^2 \mathrm{x}=-\mathrm{px}$
where, $\omega$ is the angular frequency of the SHM.
On comparison, $\omega=\sqrt{\mathrm{p}}$
Thus, the period of oscillation can be written as, $T=\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{p}}$