Two masses $m_{1}$ and $m_{2}$ are connected by a spring of spring constant $k$ and are placed on a frictionless horizontal surface. Initially the spring is stretched through a distance $x_{0}$ when the system is released from rest. Find the distance moved by the two massed before they again come to rest.

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Solution

Both the block will produce $S H M$ have equilibrium position at natural length of spring $.$
Now moment change in left $=$ moment charge in right
$\Rightarrow m_{2} x_{2} = m_{1} x_{1}$
$\Rightarrow m_{2} (x_{0} - x_{1}) = m_{1} x_{1}$
$\Rightarrow x_{1} = \frac{m_{2} x_{0}}{m_{1} + m_{2}}$ $,$ $x_{2} = \frac{m_{1} x_{0}}{m_{1} + m_{2}}$
$∴$ distance covedered by $m_{1}$ ius $2 x_{1}$ $+$ $m_{2}$ is $2 x_{2}$
$∴$ $m_{1}$ distance $= \frac{2 m_{2} x_{0}}{m_{1} + m_{2}}$
and $m_{2}$ distance $= \frac{2 m_{1} x_{0}}{m_{1} + m_{2}}$

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Two masses m1 and m2 are connected by a spring of spring constant k and are placed on a frictionless horizontal surface. Initially the spring is stretched through a distance x0 when the system is released from rest. Find the distance moved by the two massed before they again come to rest.

Both the block will produce SHM have equilibrium position at natural length of spring.Now moment change in left= moment charge in right⇒m2x2=m1x1 ⇒m2(x0−x1)=m1x1⇒x1=m2x0m1+m2 , x2=m1x0m1+m2∴ distance covedered by m1 ius 2x1+ m2 is 2x2∴ m1 distance =2m2x0m1+m2and m2 distance =2m1x0m1+m2