Question

A particle of mass $m$ undergoing oscillation about $x=0$ has varying potential energy field given by $U\left(x\right)=\frac{1}{2}k{x}^2-V_0\text{cos}\left(\frac{x}{a}\right)$ where $V_0,k,a$ are constants and $x$ is its displacement. If the amplitude of oscillation is much smaller than $a$, the time period is given by

• A. $2\pi \sqrt{\frac{m{a}^2}{k{a}^2+V_0}}$
• B. $2\pi \sqrt{\frac{m}{k}}$
• C. $2\pi \sqrt{\frac{m{a}^2}{V_0}}$
• D. $2\pi \sqrt{\frac{m{a}^2}{k{a}^2-V_0}}$

Answer 1

The correct option is A $2\pi \sqrt{\frac{m{a}^2}{k{a}^2+V_0}}$

Given Potential energy
$U=\frac{1}{2}k{x}^2-V_{0}cos\left(\frac{x}{a}\right)$
Conservative Field Force$F=-\frac{dU}{dx}=-[kx+\frac{V_0}{a}sin\left(\frac{x}{a}\right)]$
as $x<<a\Rightarrow sin\left(\frac{x}{a}\right)\simeq \frac{x}{a}$
Hence, $F=-x[k+\frac{V_0}{a^2}]$
Comparing $F=-m{\omega }^{2}x$
We get $m{\omega }^2=k+\frac{V_0}{a^2}\Rightarrow \omega =\sqrt{\frac{k{a}^2+V_0}{m{a}^2}}$
$\Rightarrow T=2\pi \sqrt{\frac{m{a}^2}{k{a}^2+V_0}}$