Physics - SHM Question with Solution | TestHub

Let the rod be displaced by a small angle $\theta$ from the vertical.

The horizontal displacement of the lower end of the rod is $x=l\theta$.

The restoring force exerted by the two springs is $F=(k_1+k_2)x=(k_1+k_2)l\theta$.

The restoring torque due to the springs about the hinge O is:
${\tau }_s=-F\cdot l=-(k_1+k_2)l^2\theta$

The center of mass of the rod is at a distance $\frac{l}{2}$ from the hinge O. The restoring torque due to gravity about O is:
${\tau }_g=-mg\left(\frac{l}{2}\right)\sin \theta \approx -\frac{mgl}{2}\theta$

The total restoring torque on the rod is:
$\tau ={\tau }_s+{\tau }_g=-\left[(k_1+k_2)l^2+\frac{mgl}{2}\right]\theta$

The moment of inertia of the uniform rod about the hinge O is $I=\frac{m{l}^2}{3}$.

Using the equation of motion $\tau =I\alpha$:
$\frac{m{l}^2}{3}\alpha =-\left[(k_1+k_2)l^2+\frac{mgl}{2}\right]\theta$

$\alpha =-\left[\frac{3(k_1+k_2)}{m}+\frac{3g}{2l}\right]\theta$

Comparing this with the standard equation for simple harmonic motion $\alpha =-{\omega }^2\theta$, the angular frequency is:
$\omega =\sqrt{\frac{3(k_1+k_2)}{m}+\frac{3g}{2l}}$

The frequency of oscillation is:
$f=\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{3(k_1+k_2)}{m}+\frac{3g}{2l}}$

Answer: $\frac{1}{2\pi }\sqrt{\frac{3(k_1+k_2)}{m}+\frac{3g}{2l}}$