Question

The potential energy of a particle of mass $1\text{kg}$ moving along the $X$-axis is given by $U=4(1+\cos 2x)\text{J},$ where $x$ is in meters. Find the time period of small oscillation (in second).

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

Answer 1

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

For a conservative force, we can say that
$\overrightarrow{F}=-(\frac{\partial U}{\partial x}\hat{i}+\frac{\partial U}{\partial y}\hat{j}+\frac{\partial U}{\partial z}\hat{k})$
Since the given potential energy is only a function $x$ we can write the above equation as
$F=-\frac{dU}{dx}$
From the data given in the question,
$F=-\frac{d}{dx}\left(4\right(1+\cos 2x)=-8\sin 2x....(1)$
Let $m$ be the mass and $a$ be the acceleration of the particle.
From Newton's second law, $F=ma$
Using$\left(1\right)$ in the above equation we get,
$a=-8\sin 2x$
For small values of $x,\sin 2x$ is approximately equal to $2x$.
Thus,
$a=-16x$
Comparing this with $a=-{\omega }^{2}x$
we get, ${\omega }^2=16\Rightarrow \omega =4$
We know that,
The time period of oscillation is given by $T=\frac{2\pi }{\omega }$
$\Rightarrow T=\frac{2\pi }{4}=\frac{\pi }{2}\text{s}$
Thus, option (c) is the correct answer.