Question

A small block is connected to a massless rod placed on a fulcrum, which in turn is attached to a spring of force constant $k$ as shown in the figure. The block is displaced down slightly, and left free. Find time period of small oscillations.

• A. $2\pi \left(\frac{b}{a}\right)\sqrt{\frac{m}{k}}$
• B. $2\pi \left(\frac{a}{b}\right)\sqrt{\frac{m}{k}}$
• C. $2\pi \left(\frac{b}{a}\right)\sqrt{\frac{k}{m}}$
• D. $2\pi \left(\frac{a}{b}\right)\sqrt{\frac{k}{m}}$

Answer 1

The correct option is A $2\pi \left(\frac{b}{a}\right)\sqrt{\frac{m}{k}}$


Let the angular displacement of the block is $\theta$, then the extension of the spring will be $a\theta$. If $F$ is the force in spring, then restoring torque, ${\tau }_{\text{rest}}=-Fa=-k\left(a\theta \right)\times a$
$[\because F=kx=ka\theta ]$
and $\alpha =\frac{{\tau }_{\text{rest}}}{I}=\frac{k{a}^2}{I}(-\theta )$
where
$\alpha =$ Angular acceleration
$I=$ Moment of inertia
Now comparing with $\alpha =-{\omega }^2\theta$, we get
$\omega =\sqrt{\frac{k{a}^2}{I}}$ and $T=2\pi \sqrt{\frac{I}{k{a}^2}}$
Here $I=m{b}^2$
$\therefore T=2\pi \sqrt{\frac{m{b}^2}{k{a}^2}}=2\pi \left(\frac{b}{a}\right)\sqrt{\frac{m}{k}}$

Hence, option (a) is correct.