Question

The potential energy of a particle of mass 1 kg in motion along the x -axis is given by $U=4(1-\cos 2x)J$, where $x$ is in metres.
The period of small oscillations (in s) is

• A. $2\pi$
• B. $\pi$
• C. $\pi /2$
• D. $\sqrt{2}\pi$

Answer 1

Answer: (C)

Here, $U=4(1-\cos 2x)J$
$\therefore F=-\frac{dU}{dx}=-8sin2x$
Acceleration, $a=\frac{F}{m}=-8sin2x$
$(\because m=1kg)$
For small oscillations, $\sin 2x\approx 2x$
$\therefore a=-16x\dots (i)$
Since $a\propto -x$, the oscillations are SHM in nature. In SHM, $a=-{\omega }^{2}x\dots (ii)$
Comparing (i) and (ii), we get
${\omega }^2=16$ or $\omega =\sqrt{16}$
$\therefore$ Time period, $T=\frac{2\pi }{\omega }=\frac{2\pi }{\sqrt{16}}=\frac{\pi }{2}s$