Two bodies M and N of equal masses are suspended from two separate massless springs of spring constants `k_(1)` and `k_(2)` respectively . If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of vibration of M to that on N is

Options

A. `(k_1)/(k_2)`
B. `sqrt(k_1//k_2)`
C. `(k_2)/(k_1)`
D. `sqrt(k_2//k_1)`

Answer

D

Explanation

Both the bodies oscillate in simple harmonic motion for which the maximum velocities will be
`v_(1) = a_(1) omega_(1) = a_(1) xx (2 pi)/(T_(1))`
`v_(2) = a_(2) omega_(2) = a_(2) xx (2 pi)/(T_(2))`
Given that `v_(1) = v_(2)`
`a_(1) xx (2pi)/(T_(1)) = a_(2) xx (2pi)/(T_(2))`
`rArr (a_(1))/(a_(2)) = (T_(1))/(T_(2)) = (2 pi sqrt((m)/(k_(1))))/(2pi sqrt((m)/(k_(2)))) = sqrt((k_(2))/(k_(1)))`.