Question

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 1

Correct Answer - A::B::C
(i) `x_1=v_0t-A(1-cos omega t)`
`tx_(cm)=(m_1x_1+m_2x_2)/(m_1+m_2)=v_0t`
`:. x_2=v_0t+m_1/m_2A(1-cos omegat)`
(ii) `a_1=(d^2x_1)/(dt^2)=-omega^2Acos 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_2A(1-cos omega t)+A(1-cos omegat)`
or `l_0=(m_1/m_2+1)A` `(cos omega=0)`
Thus, the relation between `l_0` and A is,
`l_0(m_1/m_2+1)A`