Question

A wire is bent at an angle $\theta$. A rod of mass $m$ can slide along the bent wire without friction as shown in the figure. A soap film is maintained in the wire frame kept in a vertical position and the rod is in equilibrium as shown in the figure. If rod is displaced slightly in vertical direction, then the time period of small oscillations of the rod is :-

• A. $2\pi \sqrt{\frac{l}{g}}$
• B. $2\pi \sqrt{\frac{l\cos \theta }{g}}$
• C. $2\pi \sqrt{\frac{l}{g\cos \theta }}$
• D. $2\pi \sqrt{\frac{l}{g\tan \theta }}$

Answer 1

The correct option is A $2\pi \sqrt{\frac{l}{g}}$

Let $S$ be the surface tension of the soap film. For small values of $\theta$, we can say that, $\sin \theta \approx \tan \theta$.

Since soap film has two surfaces, force of surface tension exerted by it on the rod,
$F_s=2S\times 2l\tan \frac{\theta }{2}$
For equilibrium of rod, $F_{net}=0$
i.e $mg=\left(2l\tan \frac{\theta }{2}\right)2S$
or $mg=4Sl\tan \frac{\theta }{2}\dots \left(i\right)$
If the rod is displaced vertically downward from its mean position by a small value $y$, then restoring force on the rod is given by
$F_{rest}=-[F_s^{\prime }-mg]=-\left[4S\right(l+y)\tan \frac{\theta }{2}-mg]$
$=-4Sy\tan \frac{\theta }{2}$ [using (i)]
$\Rightarrow$ Acceleration of rod $a=-\frac{4Sy\tan \frac{\theta }{2}}{m}=\frac{-4Sy\tan \frac{\theta }{2}}{\left(\frac{4Sl\tan \frac{\theta }{2}}{g}\right)}$
[using (i) again]
$\Rightarrow a=-\left(\frac{g}{l}\right)y$
Comparing this with $a=-{\omega }^{2}y$, we get $\omega =\sqrt{\frac{g}{l}}$
$\Rightarrow$ Time period of oscillation $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{l}{g}}$
Thus, option (a) is the correct answer.