Physics - SHM Question with Solution | TestHub

The amplitude of oscillations will be the maximum compression in the spring. At the time of maximum compression velocities of both the blocks are equal say$v$, then using law of conservation of momentum,

$m_1{v}_0=\left(m_1+m_2\right)v$

or$1\times 12=\left(1+2\right)v \mathrm{o}\mathrm{r} v=4 \mathrm{c}\mathrm{m }{\mathrm{s}}^{-1}$

Using law of conservation of energy, we have

$\frac{1}{2}{m}_1{v}_0^2=\frac{1}{2}k{x}^2+\frac{1}{2}\left(m_1+m_2\right)v^2$

Putting the value and solving we get$x=2$cm