Question

A particle of mass $m$ is bound by a linear potential $U=Kr$. It will have stationery circular motion with angular frequency ${\omega }_o$ with radius $r$ about the origin. If the particle is slightly disturbed from this circular motion, it will have small oscillations. If the angular frequency $\omega$ of the oscillations is $\omega =\sqrt{n}{\omega }_o$ The value of $n$ is :

• A. $2$
• B. $3$
• C. $5$
• D. $6$

Answer 1

The correct option is B $3$

$E_{total}$ = U + (kinetic energy)
$=Kr+\frac{m{v}^2}{2}=Kr+\frac{m{\omega }^2{r}^2}{2}$$=\frac{3Kr}{2}$
($\because$ for circular motion F = m ${\omega }^{2}r$$=\left|\frac{dU}{dr}\right|=K$)
Angular momentum about the origin:
L = $m\omega r^2$$=m{r}^2\sqrt{\frac{K}{mr}}$$=\sqrt{mK{r}^3}$
Angular frequency of circular motion is
$\omega =\sqrt{\frac{K}{mr}}$
Effective potential
$U_{eff}=Kr+\frac{L^2}{2m{r}^2}$
Radius $r_o$ of the stationery circular motion is
${\left(\frac{d{U}_{eff}}{dr}\right)}_{r=r_o}=K-\frac{L^2}{m{r}_0^3}=0$
$\Rightarrow r_o={\left(\frac{L^2}{mK}\right)}^{\frac{1}{3}}$
${\left(\frac{d^2{U}_{eff}}{d{r}^2}\right)}_{r=r_o}=\frac{3L^2}{m{r}^4}{|}_{r=r_o}$$=\frac{3L^2}{m}{\left(\frac{mK}{L^2}\right)}^{\frac{4}{3}}$
The angular frequency of small radial oscillations about $r_o,$ if it is slightly disturbed from the stationary circular motion:
${\omega }_r={\sqrt{\frac{1}{m}\left(\frac{d^2{U}_{eff}}{d{r}^2}\right)}}_{r=r_o}$ = ${\sqrt{\frac{3K}{m}\left(\frac{mK}{L^2}\right)}}^{\frac{1}{3}}$$=\sqrt{\frac{3K}{m{r}_o}}=\sqrt{3}{\omega }_o$, where ${\omega }_o$ is angular frequency of stationary circular motion.

So, according to given condition, $n=3$.