Question

A simple pendulum made up of light rod of length L and bob of mass m has a spring of force constant k connected to it at a distance x below the point of suspension. Find the frequency of vibration of the system.

Answer 1

Summation of torque about pivot point O

$\sum \tau =mgL\sin \theta +F_{s}H=I_o\alpha =I_o{\theta }^{\prime }$

Since restoring torque $\tau {\theta }^{\prime }=-[mgL\sin \theta +F_{s}H]$

$I_o=m{L}^2,Fs=xk=H\sin \theta k$

Now, $\theta <<1$ $\therefore \sin \theta \approx \theta$

$m{L}^2{\theta }^{\prime }=-[mgL+H^{2}k]\theta$

${\theta }^{\prime }=\frac{-[mgL+H^{2}k]}{m{L}^2}=-[\frac{g}{L}+\frac{H^{2}k}{m{L}^2}]\theta$

${\omega }^2=[\frac{g}{L}+\frac{H^{2}k}{m{L}^2}]$

$f=\frac{1}{2\pi }\sqrt{\frac{g}{l}+\frac{H^{2}k}{m{L}^2}}$