Question

Block A of mass m is performing SHM of amplitude a. Another block B of mass m is gently placed on A when it passes through mean position and B sticks to A. Find the amplitude of new S.H.M.

• A. $\frac{a}{\sqrt{2}}$
• B. $\frac{a}{2}$
• C. $\frac{a}{2\sqrt{2}}$
• D. $a\sqrt{2}$

Answer 1

The correct option is A $\frac{a}{\sqrt{2}}$.

Let the amplitude of new S.H.M. = $A$
Let the initial angular frequency be $\omega =\sqrt{\frac{k}{m}}$'

Initial velocity = $u$

Final angular frequency be ${\omega }^{\prime }=\sqrt{\frac{k}{2m}}$

Final velocity of both the blocks = $v$
Now, by conservation of linear momentum we have,
$mu=mv+mv=2mv$
$\Rightarrow v=\frac{u}{2}$
$\Rightarrow {\omega }^{\prime }A=\frac{\omega a}{2}$

Snice, $v=\omega r$
$\Rightarrow \sqrt{\frac{k}{2m}}\times A=\sqrt{\frac{k}{m}}\times \frac{a}{2}$
$\Rightarrow A=\frac{a}{\sqrt{2}}$.