Two point masses `m_1` and `m_2` are connected by a spring of natural length `l_0`. The spring is compressed such that the two point masses touch each other and then they are fastened by a string. Then the system is moved with a velocity `v_0` along positive x-axis. When the system reached the origin, the string breaks `(t=0)`. The position of the point mass `m_1` is given by `x_1=v_0t-A(1-cos omegat)` where A and `omega` are constants. Find the position of the second block as a function of time. Also, find the relation between A and `l_0`.

Answer

`v_0t + (m_1)/(m_2) A (1 - cos omega t)`

Explanation

The position of the point mass `m_1 : x_1 = v_0 t - A(1 - cos omega t)`
Position of centre of mass of `m_1 and m_2`
`X_(CM) = (m_1x_1 + m_2x_2)/(m_1 + m_2) = v_0t`
Position of mass `m_2 , x_2 = v_0 t + (m_1)/(m_2) A(1 - cos omegat)` .