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

`l_(0) = ((m_1)/(m_2) + 1)A`

Explanation

acceleration of mass `m_1, a_1 = (d^2x_1)/(dt^2) = -omega^2 A cos omegat `
The separation `x_2 – x_1` between the two blocks will be equal to `l_0` when `a_1 = 0` or `cos omega t = 0`
`x_2 - x_1 = (m_1)/(m_2) A(1 - cos omegat) + A(1 - cos omegat) " or " l_(0) = ((m_1)/(m_2) + 1) A` .