Question

A particle of mass $m$ is located in a unidimensional potential field, where the potential energy of the particle depends on the coordinates as $U\left(x\right)=U_0(1-\sin bx)$; where $U_0$ and $b$ are constant. find the period of small oscillations that the particle performs about the equilibrium position.

• A. $\frac{2\pi }{b^2}\sqrt{\frac{m}{U_0}}$
• B. $\frac{\pi }{b}\sqrt{\frac{m}{U_0}}$
• C. $\frac{{\pi }^2}{2b}\sqrt{\frac{m}{U_0}}$
• D. $\frac{2\pi }{b}\sqrt{\frac{m}{U_0}}$

Answer 1

The correct option is D $\frac{2\pi }{b}\sqrt{\frac{m}{U_0}}$

According to the question, given that
$U\left(x\right)=U_0(1-\cos bx)=2U_0{\sin }^2\frac{bx}{2}$
For small oscillations,
$\sin \frac{bx}{2}=\frac{bx}{2}$
Thus, $U\left(x\right)=2U_0{\left(\frac{bx}{2}\right)}^2=\frac{U_0{b}^2{x}^2}{2}$
Comparing with $U=\frac{1}{2}k{x}^2$, we get
$k=U_0{b}^2$
$\because$ Time period, $T=2\pi \sqrt{\frac{m}{k}}$
$=2\pi \sqrt{\frac{m}{U_0{b}^2}}=\frac{2\pi }{b}\sqrt{\frac{m}{U_0}}$.