Two blocks of masses `3kg` block is attached to a spring with a force constant, `k k = 900N//ma` which is compressed `2m` initially from its equilibrium position. When `3kg` mass is released, it strikes the `6kg` mass and the two stick togther in an inelastic collision.

The amplitude of resulting oscillation after the collision is:

Options

A. `(1)/(sqrt(2))m`
B. `(1)/(sqrt(3))m`
C. `sqrt(2)m`
D. `sqrt(3)m`

Answer

C

Explanation

After collision angular frequency of oscillation has changed however the instantaneous position of block does not change. Also equilibrium position is still ar relaxed length of spring.

Hence `x = 1m`
Now angular velocity `omega' = sqrt((k)/((m_(1)+m_(2))))`,
`omega' = sqrt((900)/(9)) = 10`
Velocity `=omega' sqrt(A'^(2) -x^(2)) or 10 = 10 sqrt(A'^(2) -1^(2))`
Hence `A' = sqrt(2)m`