Q88 / 90 · KVPY 2010 (SB/SX) · Physics
A particle of mass $\mathrm{m}$ under goes oscillations about $\mathrm{x}=0$ in a potential given by $\mathrm{V}(\mathrm{x})=\frac{1}{2} \mathrm{kx}^{2}-\mathrm{V}_{0} \cos \left(\frac{\mathrm{x}}{\mathrm{a}}\right)$, where $V_{0}, k$, a are constants. If the amplitude of oscillation is much smaller than a, the time period is given by-
- Correct answerA$2 \pi \sqrt{\frac{m a^{2}}{k a^{2}+V_{0}}}$
- B$2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$
- C$2 \pi \sqrt{\frac{\mathrm{ma}^{2}}{\mathrm{~V}_{0}}}$
- D$2 \pi \sqrt{\frac{m a^{2}}{k a^{2}-V_{0}}}$
Solution
$$
\begin{array}{l}
V(x)=\frac{1}{2} k x^{2}-V_{0} \cos \left(\frac{x}{a}\right) \\
E=-\frac{d V}{d x}=-\left(k x+V_{0} \sin \left(\frac{x}{a}\right) \times \frac{1}{a}\right)
\end{array}
$$
since $x< < $ a
$$
\therefore \sin \left(\frac{\mathrm{x}}{\mathrm{a}}\right) \simeq \frac{\mathrm{x}}{\mathrm{a}} \text { or } \mathrm{E}=-\left(\mathrm{k}+\frac{\mathrm{V}_{0}}{\mathrm{a}^{2}}\right) \mathrm{x}
$$
This resembles $\mathrm{F}=-\mathrm{kx}$
$$
\begin{array}{l}
\therefore \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}=2 \pi \sqrt{\frac{\mathrm{ma}^{2}}{\mathrm{ka}^{2}+\mathrm{v}_{0}}} \\
\Gamma=2 \pi \sqrt{\frac{\mathrm{m} \cdot \mathrm{a}^{2}}{\mathrm{ka}^{2}+\mathrm{v}_{0}}}
\end{array}
$$
Hence, the correct answer is option: 'A'.