A block A of mass m connected with a spring of force constant k is executing SHM. The displacement time equation of the block is `x =x_(0) + a sin omega t`. An identical block B moving towards negative x-axis with velocity vo collides elastically with block A at time t = 0. Then,

Options

A. displacement time equation of A after collision will be `x=x_(0) -v_(0) sqrt(m/k) sin omegat`
B. displacement time equation of A after collision will be `x=x_(0) + v_(0) sqrt(m/k) sin omega t`
C. velocity of B just after collision will be aw towards positive x-direction.
D. velocity of B just after collision will be vo towards positive x-direction.

Answer

A, C

Explanation

Given: `x=x_(0) +a sin omega t`
`therefore v=(dx)/(dt) = aomega cos omega g` (for block A)
at t=0, `x=x_(0)` and `v= aomega`
i.e., block A is at `x =x_(0)` (mean position) and its velocity is `a omega` in positive x-direction. It collides elastically with an identical block B moving towards negative x-direction with velocity `v_(0)`.
So, the blocks will interchange their velocities i.e.,
`V_(a) =v_(0)` (in negative x-direction) and
`v(B)= aomega` (in positive x-direction) Let A be the new amplitude of block A, then from conservation of mechanical energy
`1/2mv_(A)^(2) =1/2kA^(2)`
`therefore A =v_(A) sqrt(m/k) = v_(0) sqrt(m/k)`
`therefore` New displacement-time equation for block A will be:
`x=x_(0) -v_(0) sqrt(m/k) sin omega t`