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. `(3)/(5)`
B. `(25)/(9)`
C. `(16)/(9)`
D. `(5)/(3)`

Answer

C

Explanation

`T=2pisqrt((M)/(k))`
`T'=2pisqrt((M+m)/(k))`
`implies" "(5T)/(3)=2pi sqrt((M+m)/(k))`
`implies" "(5T)/(3)=2pisqrt((M+m)/(k))`
Dividing Eq. (i) by Eq. (ii), we have
`(3)/(5)=sqrt((M)/((M+m)))`
`(9)/(25)=(M)/(M+m)`
`implies9M+9m=25M`
`implies" "16M=9m`
`(m)/(M)=(16)/(9)`