Question

The potential energy of a particle in SHM, 0.2 sec after passing the mean position is 1/4 of its total energy. The period of oscillation is:

• A. 3 s
• B. 2.4 s
• C. 1.8 s
• D. 1.2 s

Answer 1

Answer: (B)

2.4 s

$TE=\frac{1}{2}m{\omega }^2{A}^2$
$PE=\frac{1}{2}m{\omega }^2{x}^2$
$PE=\frac{TE}{4}$
$\frac{1}{2}m{\omega }^2{x}^2=\frac{1}{4}\times \frac{1}{2}m{\omega }^2{A}^2$
$x=\frac{A}{2}$
$x=Asin\theta$
$Asin\theta =\frac{A}{2}$
$\theta =\omega t=\frac{\pi }{6}$
or
$\omega =\frac{\pi }{1.2}$
Thus we get $T=\frac{2\pi }{\omega }=\frac{2\pi \times 1.2}{\pi }=2.4sec$