Two masses $m_{1}$ and $m_{2}$ are connected by a spring of spring constant k and are placed on a horizontal surface. Initially the spring is stretched through a distance 'd' when the system is move from rest. Find the distance moved by the two masses when before they again come to rest.

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Solution

Mass of $2$ Blocks are $m_{1}$ and $m_{2}$ Initially spring in stretched by $d$ spring constant $K$
For the blocks to come to rest again,
Let the distance travelled by $m_{1}$ and $m_{2}$ be $d_{1}$ and $d_{2}$ towards right and left respectively external
A no Horizontal force act
So,
$m_{1} d_{1} = m_{2} d_{2}$
Again the energy would be conserved in spring

$\frac{1}{2} K \times d^{2} = \frac{1}{2} K (d_{1} + d_{2} - d)^{2}$

$\Rightarrow d = d_{1} + d_{2} - d \Rightarrow d_{1} + d_{2} = 2 d$

$\Rightarrow d_{1} = 2 d - d_{2}$ similarly $d_{1} = (\frac{2 m_{2}}{m_{1} + m_{2}}) d$

$\Rightarrow m_{1} (2 d - d_{2}) = m_{2} d_{2} \Rightarrow 2 m_{1} d - m_{1} d_{2} = m_{2} d_{2}$

$d_{2} = (\frac{2 m_{1}}{m_{1} + m_{2}}) d$

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