Question

A rod of mass $m$ and length $l$ hinged at one end is connected by two springs contants $k_1$ and $k_2$ so that it is horizontal at equilibrium. What us the angular frequency of the system ? ( in rad/s) ( Take $\ell =1m,b=1/4m,K_1=16N/m,K_2=61N/m)$.

Answer 1

Applying torque equation about
${\tau }_0=t_0\alpha$
$k_{1}b\theta \times b\cos \theta +\frac{k_2{l}_{\theta }}{\theta }\times l\cos \theta =-\frac{I{d}^2\theta }{d{t}^2}$
Here, $I=\frac{m{l}^2}{3}$, and as $\theta$ is small, $\cos \theta =1$
$\frac{m{l}^2{d}^2\theta }{3d{t}^2}+(k_1{b}^2+k_2{l}^2)\theta =0$
Hence, $\omega =\sqrt{\frac{3k_1{b}^2+k_2{l}^2}{ml}}$
On substituting the value we get $\omega =8rad/s$