A block `A` of mass m is placed on a smooth horizontal platform P and between two elastic massless springs `S_1` and `S_2` fixed horizontally to two fixed vertical walls. The elastic constants of the two springs are equal to k and the equilibrium distance between the two springs both in relaxed states is d. The block is given a velocity `v_0` initially towards one of the springs and it then oscillated between the springs. The time period `T` of oscillations and the minimum separation `d_m` of the spring will be

Options

A. `T=2((d)/(v)+pisqrt((m)/(k)))`,`d_m=d`
B. `T=2((d)/(v)+2pisqrt((m)/(k)))`,`d_m=d-vsqrt((m)/(k))`
C. `T=2((d)/(v)+2pisqrt((m)/(k)))`,`d_m=d-2vsqrt((m)/(k))`
D. `T=2pisqrt((m)/(k))`,`d_m=d`

Answer

A

Explanation

The oscillation of the block will be periodic but not simple harmonic. The springs are only comressed but not extended since the block loses contact with either spring just in its relaxed state. The minimum distance between the springs will therefore be when both the springs are relaxed i.e., during the interval when the block moves between the springs.
Time taken by the block to move from one either spring is
`t_2=(1)/(2)[2pisqrt((m)/(K))]=pisqrt((m)/(K))`
Hence the periodic time `T` of the oscillation of the block is `T=2(t_1+t_2)`
`=2[(d)/(v)+pisqrt((m)/(K))]`
Also minimum distance between the springs`=d`.