Question

A particle of mass 4 kg moves in simple harmonic motion and its potential energy U varies with position x as shown in the figure. The time period of oscillation of the particle is $\frac{2\pi }{\alpha }$ second. Find the value of $\alpha$.___

Answer 1

In SHM, U-x graph is a parabola.
$\therefore U=C_1{x}^2$, where $C_1$ is a constant.
From figure whe, x = 0.2m, U = 2J
$\therefore 2=C_1(0.2{)}^2$
$\Rightarrow C_1=\frac{2}{0.04}=50$
$\therefore U=50x^2$
Also, in SHM $U=\frac{1}{2}m{\omega }^2{x}^2$
$\therefore \frac{1}{2}m{\omega }^2=50$
$\Rightarrow {\omega }^2=\frac{100}{4}=25$
$\Rightarrow \omega =5s^{-1}$
$\therefore T=\frac{2\pi }{\omega }=\frac{2\pi }{5}$
$T=\frac{2\pi }{5}s$