Question

The potential energy of a particle in motion along the x-axis is given by \[U = {U_o} - {U_o}\cos ax\]. The time period of small oscillation is:- (Given mass is \[m\]).
A. \[2m\sqrt {\dfrac{{ma}}{{{U_o}}}} \]
B. \[2\pi \sqrt {\dfrac{{{U_o}}}{{ma}}} \]
C. \[\dfrac{{2\pi }}{a}\sqrt {\dfrac{m}{{{U_o}}}} \]
D. \[2\pi \sqrt {\dfrac{m}{{a{U_o}}}} \]

Answer 1

Hint:Here, we are asked to find the time period of small oscillation of the particle. For this recall the formula for time period of oscillation in S.H.M. Also recall the formula for restoring force and using this formula find the value of force constant. Put this value of force constant in the formula for time period to get the required result.

Complete step by step answer:
Given, the potential energy of the particle, \[U = {U_o} - {U_o}\cos ax\].Mass of the particle is \[m\]. The time period of oscillation of a particle is given by,
\[T = 2\pi \sqrt {\dfrac{m}{k}} \] (i)
where \[m\] is the mass of the particle and \[k\] is the force constant.
The restoring force of a particle using Hookes’ law along x-axis is given by,
\[F = - kx\] (ii)
where \[x\] is the displacement of the particle from its mean position.
Restoring can also be written in terms of potential energy as,
\[F = - \dfrac{{dU}}{{dx}}\] (iii)
Putting the value of \[U\] in equation (iii) we get,
\[F = - \dfrac{d}{{dx}}\left( {{U_o} - {U_o}\cos ax} \right)\]
\[ \Rightarrow F = - {U_o}a\sin ax\] (iv)
We are asked to calculate the time period for small oscillation, which means \[\sin ax\] is very small.

We know if an angle \[\theta \] is small then we can write \[\sin \theta \approx \theta \].Similarly, here as the oscillation is small, we can write, \[\sin ax \approx ax\]. Using this approximation in equation (iv) we get,
\[F = - {U_o}a\left( {ax} \right)\]
\[ \Rightarrow F = - {U_o}{a^2}x\] (v)
Now, equating equations (ii) and (v) we get,
\[ - kx = - {U_o}{a^2}x\]
\[ \Rightarrow kx = {U_o}{a^2}x\]
\[ \Rightarrow k = {U_o}{a^2}\]
Putting this value of force constant \[k\]in equation (i) we get,
\[T = 2\pi \sqrt {\dfrac{m}{{{U_o}{a^2}}}} \]
\[ \therefore T = \dfrac{{2\pi }}{a}\sqrt {\dfrac{m}{{{U_o}}}} \]
Therefore, the time period of small oscillation is \[T = \dfrac{{2\pi }}{a}\sqrt {\dfrac{m}{{{U_o}}}} \].

Hence, the correct answer is option C.

Note: By time period of oscillation we mean the time interval in which a body undergoes oscillation and comes back to its equilibrium or mean position. By the word oscillation we mean equal displacement of a body on both sides of the mean position. And restoring force is the force which brings back the body to its equilibrium or mean position.