Question

The two blocks of masses $m_{1}$ and $m_{2}$ are kept on a smooth horizontal table as shown in the figure. Block of mass $m_{1}$ but not $m_{2}$ is fastened to the spring. If now both the blocks are pushed to the left, so that the spring is compressed at a distance $d$. The amplitude of oscillation of block of mass $m_{1}$ after the system released, is

• A. $d \sqrt{\frac{m_{1}}{m_{1}+m_{2}}}$
• B. $d \sqrt{\frac{m_{2}}{m_{1}+m_{2}}}$
• C. $\sqrt{\frac{2 m_{2}}{m_{1}+m_{2}}}$
• D. $d \sqrt{\frac{2 m_{1}}{m_{1}+m_{2}}}$

Answer 1

Answer: (A)

Block of mass $m_{2}$ shoots off carrying some kinetic energy away from the system.
To find its speed, potential energy of spring = maximum kinetic energy of blocks.
$\frac{k d^{2}}{2}=\left(m_{1}+m_{2}\right) \frac{v^{2}}{2} $
[$k=$ force constant of spring ]
$v^{2}=\frac{k d^{2}}{m_{1}+m_{2}}$ with $m_{1}$ along on the spring.
Maximum potential energy
$=$ Maximum kinetic energy of $m_{2}$
$\Rightarrow \frac{1}{2} k A^{2} =\frac{1}{2} m_{1} v^{2}$
$ \Rightarrow k A^{2}=\frac{k m_{1} d^{2}}{m_{1}+m_{2}}$
$\Rightarrow A =d \sqrt{\frac{m_{1}}{m_{1}+m_{2}}}$