A particle is oscillating in a stright line about a centre of force `O`, towards which when at a distance `x` the force is `mn^(2)x` where m is the mass, n a constant. The amplitude is `a = 15 cm`. When a distance `(asqrt(3))/(2)` from O, find the new amplitude.

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`