Question

A rod of mass $m$ and length $l$ hinged at one end is connected by two springs of spring constants $k_1$ and $k_2$ so that it is horizontal at equilibrium. The angular frequency of the system is (in $\text{rad/s}$)

[$\text{Take}l=1\text{m},b=\frac{1}{2\sqrt{2}}\text{m},k_1=16\text{N/m},k_2=1\text{N/m},m=\frac{1}{64}\text{kg}$]

Answer 1

At equilibrium, let elongation and compression of spring 1 and 2 be $x_1$ and $x_2$ respectively.
Then, $k_1{x}_{1}b\cos \theta +k_2{x}_{2}l\cos \theta =mg\frac{l}{2}\cos \theta \dots \left(i\right)$
[balancing torques about ${}^{\prime }{\text{O}}^{\prime }$]
Let us rotate the rod by an angle $\theta$. Then, spring 1 gets elongated further by $b\theta$ and spring 2 is compressed further by $l\theta$.


Writing torque about point ${}^{\prime }{\text{O}}^{\prime }$ again,
${\tau }_0=-\left[k_{1}b\cos \theta \right(x_1+b\theta )+k_{2}l\cos \theta (x_2+l\theta \left)\right]+mg\frac{l}{2}\cos \theta$
[taking clockwise positive]
Using (i), restoring torque about point ${}^{\prime }{\text{O}}^{\prime }$ becomes
${\tau }_0=-(k_{1}b\theta \times b\cos \theta +k_{2}l\theta \times l\cos \theta )$
From this, we can say that
$\tau =-(k_1{b}^2\cos \theta +k_2{l}^2\cos \theta )\theta$
But we know that,
$\tau =I\alpha$
Moment of inertia of rod about point ${}^{\prime }{\text{O}}^{\prime }$ is
$I=\frac{m{l}^2}{3}$
When $\theta$ is very small, $\cos \theta \approx 1$
From this, we can say that,
$\frac{m{l}^2}{3}\alpha =-(k_1{b}^2+k_2{l}^2)\theta$
$\Rightarrow \alpha =-\frac{3(k_1{b}^2+k_2{l}^2)}{m{l}^2}\theta$
Comparing this with $\alpha =-{\omega }^2\theta$, we get
$\omega =\sqrt{\frac{3(k_1{b}^2+k_2{l}^2)}{m{l}^2}}$
From the data given in the question,
$\omega =\sqrt{\frac{3\times [16\times {\left(\frac{1}{2\sqrt{2}}\right)}^2+1\times (1{)}^2]}{\frac{1}{64}\times 1^2}}\text{rad/s}$
$\Rightarrow \omega =24\text{rad/s}$