# In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support.

Question:

In the given figure, a mass $M$ is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is $\mathrm{k}$. The mass oscillates on a frictionless surface with time period $T$ and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass $m$ is gently fixed upon it. The new amplitude of oscillation will be -

1. (1) $A \sqrt{\frac{M}{M+m}}$

2. (2) $A \sqrt{\frac{M}{M-m}}$

3. (3) $A \sqrt{\frac{M-m}{M}}$

4. (4) $A \sqrt{\frac{M+m}{M}}$

Correct Option: 1

Solution:

We know that $\omega=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$ and $\omega_{\mathrm{i}}=\sqrt{\frac{\mathrm{k}}{\mathrm{M}}} \quad \mathrm{A}_{\mathrm{i}}=\mathrm{A}$

Also, momentum is conserved just before and just after the block of mass $(\mathrm{m})$ is placed because there is no implusive force. So -

$M A_{i} \omega_{i}=(M+m) v^{\prime}$

$v^{\prime}=\frac{M A_{i} \omega_{i}}{(M+m)} \Rightarrow v^{\prime}=A_{f} \omega_{f}$

$\frac{M A \omega_{i}}{(M+m)}=A_{f} \sqrt{\frac{K}{(M+m)}}$

$\Rightarrow \frac{M A}{M+m} \times \sqrt{\frac{K}{M}} \times \sqrt{\frac{M+m}{K}}=A_{f}$

$\Rightarrow A_{f}=A \sqrt{\frac{M}{(M+m)}}$