Question

A rod of mass $M$ and length $L$ is hinged at its one end and carries a particle of mass $m$ at its lower end. A spring of force constant $k_1$ is installed at distance $a$ from the hinge and another of force constant, $k_2$ at a distance $b$ as shown in the figure. The whole arrangement rests on a smooth horizontal table top. Find the frequency of vibration.

Answer 1

$A$ is the point where the rod is hinged. Let $\theta$ be the angular displacement from mean position. The extension of spring connected at point $B$ is

$=b\theta$ ($\theta$ is so small that we take it as linear displacement)

So, restoring force at $B$ $=k_{2}b\theta$

Moment of this force about point $A$ $=k_{2}b\theta .b=k_2{b}^2\theta$($b$ is the distance of force from $A$)

The contraction of spring connected at point $c$ is $=a\theta$

So, restoring force is$=k_{1}a\theta$

And, the moment of force about $A$ is $=k_{1}a\theta .a=k_1{a}^2\theta$ ($a$ is the distance of point $A$ of restoring force)

So, total restoring force$=(k_1{a}^2+k_1{b}^2)\theta$

Now the total moment of inertia of the system about point $A$ is

$I$= Moment of inertiaabout point $A$ $+$ Moment of inertial of mass $m$ about $A$

$=\frac{m{l}^2}{3}+m{l}^2$

So, the angular acceleration, $\frac{d^2\theta }{{dt}^2}=-\frac{(k_1{a}^2+k_1{b}^2)\theta }{I}=\frac{(k_1{a}^2+k_1{b}^2)\theta }{l^2(\frac{m}{3}+m)}$

$\Rightarrow \frac{d^2\theta }{{dt}^2}+\frac{(k_1{a}^2+k_1{b}^2)\theta }{l^2(\frac{m}{3}+m)}=0$

So, agular frequency, $\omega =\sqrt{\frac{(k_1{a}^2+k_1{b}^2)}{l^2(\frac{m}{3}+m)}}$

and frequency, $f=\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{(k_1{a}^2+k_1{b}^2)}{l^2(\frac{m}{3}+m)}}$