Two equal mass hung from two massless springs of spring constants `k_(1)` and `k_(2)`. They have equal maximum velocity when executing simple harmonic motion. The ratio of their amplitudes is

Options

A. `((k_(1))/(k_(2)))^(1//2)`
B. `((k_(1))/(k_(2)))`
C. `((k_(2))/(k_(1)))`
D. `((k_(2))/(k_(1)))^(1//2)`

Answer

D

Explanation

As the masses are executing SHM the velocity, `v = a omega`
`because v_(1) = v_(2)`
`therefore a_(1) omega_(1) = a_(2) omega_(2)`
`(a_(1))/(a_(2)) = (omega_(2))` ………(i)
We know that the time period,
`T = 2pi sqrt((m)/(k))`
`(T)/(2pi) = sqrt((m)/(k)) rArr (1)/(omega) = sqrt((m)/(k))`
`rArr omega = sqrt((k)/(m))`
`rArr omega prop sqrt(k)` ........(ii)
From eqs. (i) and (ii), we get
`(a_(1))/(a_(2)) = sqrt((k_(2))/(k_(1)))`