Question

A particle of mass $m$ is suspended at the lower end of a thin rod of negligible mass. The upper end of the rod is free to rotate in the plane of the page about a horizontal axis passing through the point $\text{O}$. The time period of a system when it is slightly displaced from its mean position is $\pi \sqrt{\frac{L}{n}}$ then $n$ is _____.


Take $k=\frac{9mgL}{l^2}$ and $g=10{\text{m/s}}^2$

Answer 1

When the small angular displacement $\theta$ given to rod, the system will be as shown below


From diagram,
$x=l\sin \theta$
$\therefore kx=kl\sin \theta$

Torque about point $\text{O}$ is

${\tau }_o=I_o\alpha$

So, the equation of torque for the given system will be

$\Rightarrow \left(kx\right)l+mg\sin \theta L+m{L}^2\alpha =0(I_o=m{L}^2)$

$\Rightarrow \alpha =-\frac{(k{l}^2+mgL)\sin \theta }{m{L}^2}$

For small $\theta ,\sin \theta \approx \theta$

$\alpha =-\left[\frac{k{l}^2+mgL}{m{L}^2}\right]\theta$

Time period of system will be,

$T=2\pi \sqrt{\frac{m{L}^2}{k{l}^2+mgL}}=2\pi \sqrt{\frac{m{L}^2}{\left(9mgL/l^2\right)l^2+mgL}}$

$T=\pi \sqrt{\frac{L\times 2\times 2}{10\times 10}}=\pi \sqrt{\frac{L}{25}}$

Hence, correct answer is $25$.