Paragraph: When a particle of mass $m$ moves on the $x$-axis in a potential of the form $V(x)=k x^2$, it…
Paragraph:
When a particle of mass $m$ moves on the $x$-axis in a potential of the form $V(x)=k x^2$, it performs simple harmonic motion.
The corresponding time period is proportional to $\sqrt{\frac{m}{k}}$, as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of $x=0$ in a way different from $k x^2$ and its total energy is such that the particle does not escape to infinity. Consider a particle of mass $\mathrm{m}$ moving on the $x$-axis. Its potential energy is $V(x)=\alpha x^4(\alpha>0)$ for $|x|$ near the origin and becomes a constant equal to $V_0$ for $|x| \geq X_0$ (see figure).Question:
For periodic motion of small amplitude $A$, the time period $T$ of this particle is proportional to
$A \sqrt{\frac{m}{\alpha}}$
$\frac{1}{A} \sqrt{\frac{m}{\alpha}}$
$A \sqrt{\frac{\alpha}{m}}$
$\frac{1}{A} \sqrt{\frac{\alpha}{m}}$
Solution
$[\alpha]=\left[\frac{\mathrm{PE}}{x^4}\right]=\left[\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~L}^4}\right]=\left[\mathrm{ML}^{-2} \mathrm{~T}^{-2}\right]$
$
\begin{array}{ll}
\therefore & {\left[\frac{m}{\alpha}\right]=\left[\mathrm{L}^2 \mathrm{~T}^2\right]} \\
\therefore & {\left[\frac{1}{A} \sqrt{\frac{m}{\alpha}}\right]=[\mathrm{T}]}
\end{array}
$
As dimensions of amplitude $A$ is [L].
Hence, the correct option is (b).
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