Question

A particle of mass $m$ is located in a one dimensional field where potential energy is given by:
$V\left(x\right)=A(1-\cos px)$ where $A$ and $p$ are constants.
The period of small oscillations of the particle is

• A. $2\pi \sqrt{\frac{m}{\left(Ap\right)}}$
• B. $2\pi \sqrt{\frac{m}{\left(A{p}^2\right)}}$
• C. $2\pi \sqrt{\frac{m}{A}}$
• D. $\frac{1}{2\pi }\sqrt{\frac{Ap}{m}}$

Answer 1

Answer: (B)

$2\pi \sqrt{\frac{m}{\left(A{p}^2\right)}}$

We are given that a particle of mass $m$ is located in a one dimensional potential field and the potential energy is given by: $V\left(x\right)=A(1-\cos px)$
So, we can find the force experienced by the particle as
$F=-\frac{dV}{dx}=-Ap\sin px$

For small oscillations, we have, $F\approx -A{p}^{2}x$

Hence, the acceleration would be given by, $\alpha =\frac{F}{m}=-\frac{A{p}^2}{m}x$

Also we know that, $\alpha =\frac{F}{m}=-{\omega }^{2}x$

So, $\omega =\sqrt{\frac{A{p}^2}{m}}$
$\Rightarrow T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{A{p}^2}}$