Question

A block with mass $M$ is connected by a massless spring with stiffness constant $k$ to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude A about an equilibrium position $x_0$. Consider two cases : (i) when the block is at $x_0$; and (ii) when the block is at $x=x_0+A$. In both the cases, a particle with mass $m(<M)$ is softly placed on the block after which they stick to each other. Which of the following statement(s) is(are) true about the motion after the mass $m$ is placed on the mass $M$ ?

• A. The amplitude of oscillation in the first case changes by a factor of $\sqrt{\frac{M}{m+M}}$, whereas in the second case it remains unchanged
• B. The final time period of oscillation in both the cases is same
• C. The total energy decreases in both the cases
• D. The instantaneous speed at $x_0$ of the combined masses decreases in both the cases

Answer 1

Answer: (A), (B), (D)

$MA\omega =(m+M)A^{\prime }{\omega }^{\prime }$
$A^{\prime }=\frac{M}{m+M}\frac{\omega }{{\omega }^{\prime }}A$
$A^{\prime }=\frac{M}{m+M}\cdot \sqrt{\left(\frac{M+m}{M}\right)A}$
$A^{\prime }=\sqrt{\frac{M}{m+M}}A$
Amplitude in $I^{st}$ Changes by a factor of $\sqrt{\frac{M}{m+M}}$
Time period of $I^{st}$ case $=2\pi \sqrt{\frac{m+M}{K}}$

Amplitude remain same.
But time period because $=2\pi \sqrt{\frac{m+M}{K}}$
Total energy of $I^{\text{st }}$ case decreases due to decrease in amplitude. Instantaneous speed in $I^{\text{st }}$ case decrease
In $I{I}^{nd}$ case $\frac{1}{2}K{A}^2=\frac{1}{2}(m+M)v^{\prime 2}$
So also decrease.