Answer :

Solution :

When we will rotate disc by θ, rod will rotate by angle ϕ as disc is pure rolling on the string

$\therefore L\varphi =r\theta$

$\varphi =\frac{r\theta }{L}$(∵$r=\frac{L}{4}$ as given in question)

$\therefore \varphi =\frac{\theta }{4}$

Moment of Inertia of disc about point D :

$I_D=m{r}^2+\frac{m{r}^2}{2}=\frac{3}{2}m{r}^2$(Using parallel axis theorem)

Now, by energy method :

$\frac{d{E}_{total}}{dt}=0$

E_total = Energy of the disc due to rotation + Energy of the springs

$E_{total}=\frac{1}{2}{I}_{disc}\times {\dot{\theta }}^2+\frac{1}{2}k{\left(2r\theta \right)}^2+\frac{1}{2}2k{\left(\frac{L}{2}\cdot \varphi \right)}^2$

$E_{total}=\frac{1}{2}\left(\frac{3}{2}m{r}^2\right){\dot{\theta }}^2+\frac{1}{2}k{\left(2r\theta \right)}^2+k{\left(\frac{L}{2}\cdot \frac{\theta }{4}\right)}^2$

$\frac{d{E}_{total}}{dt}=0$

$\frac{3}{4}m{r}^2\times 2\dot{\theta }\ddot{\theta }+\frac{1}{2}k\left(4r\right)\times 2\theta \dot{\theta }+\frac{k{L}^2}{64}\times \left(2\theta \dot{\theta }\right)=0$

$\left(\frac{3m}{32}\right)\ddot{\theta }+\left(\frac{9k}{32}\right)\theta =0$

$3m\ddot{\theta }+9k\theta =0$

$\ddot{\theta }+\frac{9k}{3m}\theta =0\Rightarrow \ddot{\theta }+\frac{3k}{m}\theta =0$

Comparing with standard equation

$\because \ddot{\theta }+{\omega }_n^2\theta =0$

${\omega }_n^2=\frac{3k}{m}$

${\omega }_n=\sqrt{\frac{3k}{m}}$