Question

A wire is bent at an angle $\theta$. A rod of mass $m$ canslide along the bended wire without friction as film is maintained in the 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 oscillation 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

Answer: (A)

Let $S$ be the surface tension of the soap film. For equilibrium of rod
$2(l+y)\tan \frac{\theta }{2}$
$mg={\left(F_{\text{surface }}\right)}_1$

$mg=\left(2l\tan \frac{\theta }{2}\right)S\times 2;mg=4Sl\tan \frac{\theta }{2}$
If the rod is displaced from its mean position by small displacement $y$, then restoring force on the rod is
$F_{\text{rest }}=-\left[{\left(F_{\text{surface }}\right)}_2-mg\right]-{\left(F_{\text{surface }}\right)}_1$
$=-\left[4S(l+y)\tan \frac{\theta }{2}-mg\right]-\left[4S\tan \frac{\theta }{2}y\right]$
$a=-\frac{4S\tan \frac{\theta }{2}}{m}y=\frac{-4S\tan \frac{\theta }{2}}{\left(\frac{4Sl\tan \frac{\theta }{2}}{g}\right)}y$
$\frac{d^{2}y}{d{t}^2}=-\left(\frac{g}{l}\right)y$
$\therefore T=2\pi \sqrt{\frac{l}{g}}$