Question

A body of mass ' $m$ ' is situated in a potential field $U(x)=U_0(1-\cos \alpha x)$ where $U_0$ and $\alpha$ are constants. The time period of oscillation is found to be proportional to $m^a{U}_0^b{\alpha }^c$. The value $abc$ is _______.

Answer 1

Answer: 0.25

$U=U_0(1-\cos \alpha x)$
$\because F=\frac{-dU}{dx}$
$\therefore F=\frac{-d}{dx}\left[U_0(1-\cos \alpha x)\right]=-U_0\alpha \sin \alpha x$
As $\alpha x$ is small, $\sin \alpha x\approx \alpha x$
$\therefore F=-U_0{\alpha }^{2}x$
$\Rightarrow k=U_0{\alpha }^2$
$T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{m}{U_0{\alpha }^2}}$
$\Rightarrow a=\frac{1}{2},b=\frac{-1}{2},c=-1$
$\therefore abc=\left(\frac{1}{2}\right)\left(\frac{-1}{2}\right)(-1)=0.25$