Question

The potential energy of a particle of mass 'm' situated in a uni-dimensional potential field varies as $U\left(x\right)=U_0[1-\cos ax]$ where $U_0$ and a are constants. The time period of small oscillations of the particle about the mean position:

• A. $2\pi \sqrt{\frac{m}{a{U}_0}}$
• B. $2\pi \sqrt{\frac{am}{U_0}}$
• C. $2\pi \sqrt{\frac{m}{a^2{U}_0}}$
• D. $2\pi \sqrt{\frac{a^{2m}}{U_0}}$

Answer 1

The correct option is C $2\pi \sqrt{\frac{m}{a^2{U}_0}}$

Force and potential energy are related by the equation.

$F=-\frac{dx}{dy}F=-\frac{d\left[V_0(1-cosx)\right]}{dx}F=-\alpha V_{0}sin\alpha x\alpha =-\frac{\alpha V_0}{m}sin\alpha x$

Since $\alpha x$ is very small so $sin\alpha x\approx \alpha x$

Then,

$a=\frac{-{\alpha }^2{V}_{0}x}{m}$

[- indicates restoring nature]

$a={\omega }^{2}x$

$\omega =\sqrt{\frac{{\alpha }^2{V}_0}{m}}T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{{\alpha }^2{V}_0}}$