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. $\sqrt{\frac{2m_2}{m_1+m_2}}$
• D. $d\sqrt{\frac{2m_1}{m_1+m_2}}$

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 $m_2$
$\Rightarrow \frac{1}{2}k{A}^2=\frac{1}{2}{m}_1{v}^2$
$\Rightarrow k{A}^2=\frac{k{m}_1{d}^2}{m_1+m_2}$
$\Rightarrow A=d\sqrt{\frac{m_1}{m_1+m_2}}$