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 velocities of a particle executing `S.H.M.` are `30cm//s` and `16 cm//s` when its displacements are `8cm` and `15cm` from the equilibrium position. then its amplitude of oscillation in cm is:

Options

A. `25`
B. `21`
C. `17`
D. `13`

Answer

C

Explanation

`v^(2) = omega^(2) (A^(2)-x^(2))`
For `x = 8cm, 30^(2) = omega^(2) (A^(2) - 8^(2)) …………..(1)`
For `x = 15 cm, 16^(2) = omega^(2) (A^(2) -15^(2)) …………(2)`
Solving (1) and (2),
`16^(2)A^(2) - 16^(2) (8)^(2) = 30^(2)A^(2) - (30)^(2) - (15)^(2)`
`A^(2) (30^(2)-16^(2)) = (30)^(2) (15)^(2) - (16)^(2) (8)^(2) rArr A = 17`