Physics - SHM Question with Solution | TestHub

Given:

$m_1=30 \mathrm{gm}$ and $m_2=20 \mathrm{gm}$.

Moment of inertia about the axis of rotation is

$I=m_1{r_1}^2+m_2{r_2}^2$

X-coordinate of the centre of mass can be written as,

$$\begin{gathered} x_{CM}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2} \\ =\frac{0+\left(20\times {10}^{-3}\right)\times \left(10\times {10}^{-2}\right)}{\left(30\times {10}^{-3}\right)+\left(20\times {10}^{-3}\right)} \\ =4\times {10}^{-2} \mathrm{m} \\ =4 \mathrm{cm}=r_1 \end{gathered}$$

Therefore, 

And $r_2=\left(10-4\right) \mathrm{cm}=6 \mathrm{cm}$

$\therefore I=\left(30\times {10}^{-3}\times 16\times {10}^{-4}\right)+\left(20\times {10}^{-3}\times 36\times {10}^{-4}\right)$

$\Rightarrow I=1200\times {10}^{-7} \mathrm{kg} {\mathrm{m}}^2$

If the system is rotated by small angle ‘$\theta$’, the restoring torque is ${\tau }_{\left(R\right)}=-k\theta$

And $\frac{d^2\theta }{d{t}^2}=\frac{-k}{I}·\theta =-{\omega }^2\theta =\frac{-1.2\times {10}^{-8}}{1200\times {10}^{-7}}·\theta$

$\therefore {\omega }^2={10}^{-4}$

$\Rightarrow \omega =10\times {10}^{-3} \mathrm{rad} {\mathrm{s}}^{-1}$

Therefore, $n=10$.