Physics - SHM Question with Solution | TestHub

Given,

Potential energy,$U=a+b{x}^2$

Force,$F=-\frac{dU}{dx}$

$=-\frac{d.(a+b{x}^2)}{dx}$

$=-\left[\frac{d}{dx}\left(a\right)+\frac{d}{dx}\left(b{x}^2\right)\right]$

$\Rightarrow F=0-b.\frac{d}{dx}\left(x^2\right)$

$\Rightarrow F=0-b.2x=-2bx$

$\Rightarrow a=\frac{F}{m}=-\frac{2b}{m}x$.......(i)

Compare equation. (i) with the standard relation of displacement in case of simple harmonic motion.

i.e.,$a=\frac{F}{2m}=-{\omega }^{2}x$

i.e.,${\omega }^2=\frac{2b}{m}$

$\omega =\sqrt{\frac{2b}{m}}$

i.e., the frequency of the simple harmonic motion depends on$b$and$m$alone.