Question

A small block is connected to a massless rod, which in turn is attached to a spring of force constant k as shown in fig. The block is displaced down slightly, and left free.

• 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 $2\pi \left(\frac{b}{a}\right)\sqrt{\frac{m}{k}}$

Free body diagram:

Let angular displacement of the block is $\theta$, then extension of the spring will be $a\theta$.

If $F$ is the force in spring, then restoring torque,

${\tau }_{rest}=-Fa...\left(1\right)$

Since from figure,

$F=kx=ka\theta$

From $\left(1\right)$,
${\tau }_{rest}=-Fa=-k{a}^2\theta$

Now, we know that angular acceleration of the system,

$\alpha =\frac{{\tau }_{rest}}{I}$

Where, $I=m{b}^2$, is moment of inertia of the block.
Substituting the value in the above equation,

$\alpha =\frac{k{a}^2}{m{b}^2}(-\theta )$

Now comparing with $\alpha =-{\omega }^2\theta$, we get angular velocity of the system, $\omega$ as

$\omega =\sqrt{\frac{k{a}^2}{m{b}^2}}$

and time period $T$,

$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m{b}^2}{k{a}^2}}$

$\therefore T=2\pi \left(\frac{b}{a}\right)\sqrt{\frac{m}{k}}$

Hence, option (b) is correct answer.