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 oscilate in simple harmonic motion for which the maximum velocities will be
`v_(1)=a_(1)omega_(1)=a_(1)xx(2pi)/(T_(1))`, `v_(2)=a_(2)omega_(2)=a_(2)xx(2pi)/(T_(2))`
Given that `v_(1)=v_(2)` `a_(1)xx(2pi)/(T_(1))=a_(2)xx(2pi)/(T_(2))` implies `(a_(1))/(a_(2))=(T_(1))/(T_(2))=(2xxsqrt((m)/(k_(1))))/(2xxsqrt((m)/(k_(2))))=sqrt((K_(2))/(K_(1)))`