A block of mass `m` moves with a speed `v` towards the right block which is in equilibrium with a spring attached to rigid wall. If the surface is frictionless and collisions are elastic, the frequency of collisions between the masses will be:

Options

A. `(v)/(2L)+(1)/(pi)sqrt((K)/(m))`
B. `2[(v)/(2L)+(1)/(2pi)sqrt((K)/(m))]`
C. `(2)/([(2L)/(v)+pisqrt((m)/(K))])`
D. `(v)/(2l)+(1)/(pi)sqrt((m)/(K))`

Answer

C

Explanation

Time taken to collide on left wall and get back to the mass attached with spring is `t_(1) = (2L)/(v)`. Time to get the spring compressed once and to come back is, `t_(2) = (T)/(2) =(2pi)/(2) sqrt((m)/(K)) =pi sqrt((m)/(K)) :.` Average time between two successive collisions, `t = (t_(1)+t_(2))/(2)` collision freqency `=(1)/(t)`