A particle is oscillating in a straight line about a centre `O`, with a force directed towards `O`. When at a distance `'x'` from `O`, the force is `mn^(2)x` where `'m'` is the mass and `'n'` is a constant. The amplitude is `a = 15 cm`. When at a distance `sqrt(3)(a)/(2)` from `O` the particle receives blow in the direction of motion which generates an extra velocity `na`. If the velocikty is away from `O` at the time of blow and the new amplitude become `ksqrt(3)cm`, then find `k`.

Answer

A, C

Explanation

As `v = omegasqrt(a^(2) - x^(2))` so `v_(i) = nsqrt(a^(2) - (3a^(2))/(4)) = (na)/(2)`
and `v_(2) = (3)/(2) na =- nsqrt(A^(2) - (3a^(2))/(4)`
`rArr A = sqrt(3a) = 15sqrt(3) cm`