Oscillations Test - 59

At the surface of moon, $$g$$ decrease hence time period increase 

On a planet, if a body dropped initial velocity $$(u = 0)$$ from a height $$h$$ and takes time $$t$$ to reach the ground then

The particle will come after a time $$\frac{T} {4}$$ to its mean position.

At equilibrium position

$$\displaystyle \mathrm{F}=-\dfrac{dV}{dx}=0$$

$$\dfrac{dV}{dx }=\dfrac{C(a^{2}-x ^{2})}{(x ^{2}+a^{2})^{2}}=0$$

$$\therefore $$ There are two equilibrium positions

$$x_1 = a$$

$$x_2 = -a$$

Consider $$\dfrac{d^{2}V}{dx ^{2}}=\dfrac{2Cx (x ^{2}-3a^{2})}{(x^{2}+a^{2})^{3}}$$

$$\dfrac{d^{2}V}{dx ^{2}}|_{x_1}< 0$$

$$\dfrac{d^{2}V}{dx ^{2}}|_{x_2}> 0$$

$$\because $$There is a maxima at $$x=a$$ and minima at $$x=-a$$
$$\Rightarrow x _{1}$$ is a position of unstable equilibrium and $$x _{2}$$ is a position of stable equilibrium.

$$E_{total}$$ = U + (kinetic energy)

$$=Kr+\dfrac{mv^{2}}{2}=Kr+\dfrac{m\omega ^{2}r^{2}}{2}$$$$=\dfrac{3Kr}{2}$$

($$\because $$ for circular motion F = m $$\omega ^{2}r$$$$=\left | \dfrac{dU}{dr} \right |=K$$)

Angular momentum about the origin:

L = $$m\omega r^{2}$$$$=mr^{2}\sqrt{\dfrac{K}{mr}}$$$$=\sqrt{mKr^{3}}$$

Angular frequency of circular motion is

$$\omega =\sqrt{\dfrac{K}{mr}}$$

Effective potential

$$U_{eff}=Kr+\dfrac{L^{2}}{2mr^{2}}$$

Radius $$r_{o}$$ of the stationery circular motion is

$$\left ( \dfrac{dU_{eff}}{dr} \right )_{r=r_{o}}=K-\dfrac{L^{2}}{mr_{0}^{3}}=0$$

$$\Rightarrow r_{o}=\left ( \dfrac{L^{2}}{mK} \right )^{\frac{1}{3}}$$

$$\left ( \dfrac{d^{2}U_{eff}}{dr^{2}} \right )_{r=r_{o}}=\dfrac{3L^{2}}{mr^{4}}|_{r=r_{o}}$$$$=\dfrac{3L^{2}}{m}\left ( \dfrac{mK}{L^{2}} \right )^{\frac{4}{3}}$$

The angular frequency of small radial oscillations about $$r_{o},$$ if it is slightly disturbed from the stationary circular motion:
$$\omega _{r}=\sqrt{\dfrac{1}{m}\left ( \dfrac{d^{2}U_{eff}}{dr^{2}} \right )}_{r=r_{o}}$$ = $$\sqrt{\dfrac{3K}{m}\left ( \dfrac{mK}{L^{2}} \right )}^{\frac{1}{3}}$$$$=\sqrt{\dfrac{3K}{mr_{o}}}=\sqrt{3}\omega _{o}$$, where $$\omega _{o}$$ is angular frequency of stationary circular motion.

The displacement of a body executing SHM is given by: