Question

A particle of mass $m$ is located in a unidimensional potential field, where the potential energy of the particle depends on the coordinated as $U(x) = {U_0}(1 - \sin bx)$; where ${U_0}$ and $b$ are constant. Find the period of small oscillations that the particle performs about the equilibrium position.
(A). $\dfrac{{2\pi }}{{{b^2}}}\sqrt {\dfrac{m}{{{U_0}}}}$
(B). $\dfrac{\pi }{b}\sqrt {\dfrac{m}{{{U_0}}}}$
(C). $\dfrac{{{\pi ^2}}}{{2b}}\sqrt {\dfrac{m}{{{U_0}}}}$
(D). $\dfrac{{2\pi }}{b}\sqrt {\dfrac{m}{{{U_0}}}}$

Answer 1

- Hint: You can start by describing what potential fields are. Then convert the equation $U(x) = {U_0}(1 - \sin bx)$ into a simpler form using the relation $1 - \cos \theta = 2{\cos ^2}\dfrac{\theta }{2}$. Since it is given that the oscillations are small we can say $\cos \dfrac{{bx}}{2} = \dfrac{{bx}}{2}$. Then compare the equations for kinetic energy i.e. $\dfrac{1}{2}k{x^2}$ and potential energy $\dfrac{{{U_0}{b^2}{x^2}}}{2}$ to find the value of $k$. Then put this value in the equation $T = 2\pi \sqrt {\dfrac{m}{k}}$ to reach the solution.

Complete step-by-step answer:
Potential field – It is any field for which Laplace’s equation holds true. Common examples of the potential fields are electric and magnetic fields. In the problem we are not told specifically what the body is, and what the potential field is. But, we are specifically told that the body is undergoing small oscillations that the particles perform about the equilibrium position.
In the problem it is given
$U(x) = {U_0}(1 - \sin bx)$
U(x) = 2{U_0}{\cos ^2}\dfrac{{bx}}{2}$$$(\because 1 - \cos \theta = 2{\cos ^2}\dfrac{\theta }{2})$$ In the problems it mentioned that the particles undergoes small oscillations and for small oscillations $$\cos \dfrac{{bx}}{2} = \dfrac{{bx}}{2}$$ Thus, U(x) = 2{U_0}{\left( {\dfrac{{bx}}{2}} \right)^2}$$U(x) = \dfrac{{{U_0}{b^2}{x^2}}}{2}$The particle is oscillating, at the extreme ends all the energy of the particle is stored in the form of potential energy whose value we have calculated above. At the middle point of oscillations all the energy of the particle is in the form of kinetic energy. So,$PE = KE$$\dfrac{{{U_0}{b^2}{x^2}}}{2} = \dfrac{1}{2}k{x^2}$$k = {U_0}{b^2}$We know that,$T = 2\pi \sqrt {\dfrac{m}{k}} $$ \Rightarrow T = 2\pi \sqrt {\dfrac{m}{{{U_0}{b^2}}}} $$ \Rightarrow T = \dfrac{{2\pi }}{b}\sqrt {\dfrac{m}{{{U_0}}}} $
Hence, option D is the correct choice.

Note: In the solutions we calculated the time period of an oscillating body. The time period of an oscillating body is double the amount of time that the body takes in again reaching the equilibrium position when it starts moving from the equilibrium position. The energy stored in the object may change its form during the oscillations, but overall energy is conserved.