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

Initially, `T=2pisqrt(M//k)`
Finally, `(5T)/(3)=2pisqrt((M+m)/(k))`
`therefore (5)/(3)xx2pisqrt((M)/(k))=2pisqrt((M+m)/(k))or(25)/(9)(M)/(k)=(M+m)/(k)`
or `9m+9M=25Mor(m)/(M)=(16)/(9)`