A mass (M) is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes `(5T)/3`. Then the ratio of `m/M` is .

Options

A. `(9)/(16)`
B. `(25)/(9)`
C. `(16)/(9)`
D. `(4)/(3)`

Answer

C

Explanation

`m_(1) = M,T_(1) = T`
`M_(2) = M +m T_(1) = (57)/(3)`
`(T_(1))/(T_(2)) = (2pi sqrt(m_(1)//k))/(2pi sqrt(m_(2)//k))= sqrt((m_(1))/(m_(2))) = sqrt((M)/(M+m))`
`(M +m)/(M) = ((T_(2))/(T_(1)))^(2) = ((5T//3)/(T))^(2) = (25)/(9)`
`(m)/(M) = (16)/(9)`