Two masses $m_{1}$ and $m_{2}$ are connected to a spring of spring constant $K$ at two ends. The spring is compressed by $y$ and released. The distance moved by $m_{1}$ before it comes to a stop for the first time is

\frac{m_{1} y}{m_{1} + m_{2}}

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\frac{m_{2} y}{m_{1} + m_{2}}

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\frac{2 m_{1} y}{m_{1} + m_{2}}

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\frac{2 m_{2} y}{m_{1} + m_{2}}

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Solution

The correct option is B $\frac{m_{2} y}{m_{1} + m_{2}}$
the spring is compressed by y therefore the force on masses is $F = K y$
let $m_{1}$ traveled distance a and mass $m_{2}$ be b
therefore
$F = m a$
$F (t) = - m_{1} \frac{d a (t)^{2}}{d t^{2}}$
similarly for second mass
$- F (t) = m_{2} \frac{d b (t)^{2}}{d t^{2}}$
Now Assuming simple harmonic motion $a (t) = A s i n ω t$ and $b (t) = B s i n ω t$

as the spring force is equal for both masses
$m_{1} a = m_{2} b$
$m_{1} A s i n ω t = m_{2} B s i n ω t$
$m_{!} A = m_{2} B$

As long as the above amplitude equation will hold center of mass wont move
Eq for center of mass is
$0 = \frac{- B m_{2} + A m_{1}}{m_{1} + m_{2}}$
we know $A + B = y$
$m_{1} A = m_{2} (y - A)$
$A = \frac{m_{2} y}{m_{1} + m_{2}}$

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