Question

A particle of mass $2\text{kg}$ executes simple harmonic motion and its potential energy $U$ varies with position $x$ as shown below. The period of oscillation of the particle is

• A. $\frac{2\pi }{5}\text{s}$
• B. $\frac{2\sqrt{2}\pi }{5}\text{s}$
• C. $\frac{\sqrt{2}\pi }{5}\text{s}$
• D. $\frac{4\pi }{5}\text{s}$

Answer 1

The correct option is D $\frac{4\pi }{5}\text{s}$

Given , mass of the particle $m=2\text{kg}$
Potential energy as a function of the position of particle is given by
$U=\frac{1}{2}m{\omega }^2{x}^2$
At extreme positions, $x=\pm A$. From the data given in the diagram, we can say that
$U=\frac{1}{2}m{\omega }^2{A}^2\Rightarrow \frac{1}{2}m{\omega }^2(0.4{)}^2=1.0\text{J}$
$\Rightarrow \omega =\sqrt{\frac{2\times 1.0}{(0.4{)}^2\times 2}}=2.5\text{rad/s}$
Time period of oscillation of a particle executing SHM is given by,
$T=\frac{2\pi }{\omega }$
$\therefore T=\frac{2\pi }{2.5}=\frac{4\pi }{5}\text{s}$
Thus, option (d) is the correct answer.