Question

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

• A. $2\pi$
• B. $\pi$
• C. $\frac{\pi }{2}$
• D. $3\pi$

Answer 1

The correct option is C $\frac{\pi }{2}$

The potential energy of particle is $U=4(1-\text{cos}2x)\text{J}$.
The restoring conservative force acting on the particle is given by,
$F=\frac{-dU}{dx}$
$\Rightarrow F=-4\times 2\times [-(-\text{sin}2x\left)\right]=-8\text{sin}2x$
For small oscillations, $\text{sin}2x\approx 2x$
$\Rightarrow F=-16x$
$\Rightarrow ma=-16x$
$\Rightarrow a=\frac{-16x}{1}=-16x$
Time period of oscillation,
$T=2\pi \sqrt{\frac{m}{k}}$
Substituting the data in the equation,
$T=2\pi \sqrt{\frac{1}{16}}$
$\Rightarrow T=2\pi \times \frac{1}{4}=\frac{\pi }{2}\text{s}$