Springs and SHM - 2

Since the rod is light, net torque on it will be zero. To find net extension in the second spring, we use this equation. First, let $\theta$ be the angle made by the rod at the extreme position. Thus the net extension in the second spring will be $x-\ell\theta$ . So, balancing torques $k_1a^2\theta = k_2\ell(x-\ell\theta)\\\Rightarrow\ell\theta = \dfrac{k_2\ell^2x}{a^2k_1+\ell^2k_2}$

Thus, the force on the block is only due to the second spring, $\begin{aligned}F &= -k_2(x-\ell\theta)\\&=-\dfrac{k_1k_2a^2}{a^2k_1+\ell^2k_2}x \end{aligned}$

Thus, by definition : $\omega = \sqrt{\dfrac{k_1k_2a^2}{m(a^2k_1+\ell^2k_2)}}$

As the rod is light the spring with stiffness k1 can be shifted from distance a to the distance l from the pivot with a equivalent spring of stiffness k1(a/l)^2. Then we see that the springs are in series and we find the net stiffness. This comes out to be 3.7799. Then finding angular frequency is easy.

Let us consider a hypothetical spring of stiffness $k$ at the bottom which is opposite to the spring $k_{2}$ instead of the spring with stiffness $k_{1}$ .

So the elastic potential energy stored in both the springs are equal.

The extension in the springs are $aθ$ & $lθ$ for $k_{1}$ & $k$ respectively.

So by equating their energies we get,

$k$ = $\frac{k_{1}a^{2}}{l^{2}}$

We also know that these two springs are connected in series.So the equivalent stiffness is ,

$\frac{1}{k_{eff}}$ = $\frac{1}{k_{2}}$ + $\frac{1}{k}$

$k_{eff}$ = $\frac{k_{1}k_{2}a^{2}}{k_{1}a^{2} + k_{2}l^{2}}$

On equating the force,

$k_{eff}$ $x$ = $mω^{2}x$

On substituting the values we get

$\boxed{ω = 0.6148}$