Question

The system shown in the figure can move on a smooth surface. The spring is initially compressed by $6cm$ and then released. Find
(a) time period
(b) the amplitude of $3kg$ block and
(c) the maximum momentum of $6kg$ block

Answer 1

The reduced mass of system $\mu =\frac{3\times 6}{3+6}=2kg$

Angular frequency of the system $\omega =\sqrt{\frac{k}{\mu }}=\sqrt{\frac{800}{2}}$

$\Rightarrow \omega =20$

$a)$Time period of oscillation $T=\frac{2\pi }{\omega }=\frac{\pi }{10}{sec}^{-1}$

$b$The spring is initially compressed by $6cm$. Hence, $A_1+A_2=6cm...1$ ($A_1$=amplitude of $3kg$ mass, $A_2$=amplitude of $6kg$mass. )

As, no external force acting on the system we can write$m_1{A}_1=m_2{A}_2$ (From the concept of momentum conservation )

$\Rightarrow A_2=\frac{m_1{A}_1}{m_2}$ ($m_1=3kg$, $m_2=6kg$

Putting this vale in $1$, we get, $A_1+\frac{m_1}{m_2}{A}_1=6$

$\Rightarrow A_1=\frac{m_2}{m_1+m_2}6=\frac{6}{6+3}6$

$\Rightarrow A_1=4cm$, and $A_2=2cm$

$c)$Maximum velocity of mass $6kg$ is $v_{max}=\omega A_2$

So, max momentum, $m_2{v}_{max}=6\omega A_2=6\times 20\times 2=240$