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

$U = U _{0}(1-\cos \alpha x )$
$\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}$
$T =2 \pi \sqrt{\frac{ m }{ k }}=2 \pi \sqrt{\frac{ m }{ U _{0} \alpha^{2}}}$
$\Rightarrow a =\frac{1}{2}, b =\frac{-1}{2}, c =-1$
$\therefore abc =\left(\frac{1}{2}\right)\left(\frac{-1}{2}\right)(-1)=0.25$