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 .

Explanation

When mass M is suspended from a spring, the time period of oscillation of loaded spring is
`T=2pisqrt((M)/(k))`
When mass m is adde to mass M, then time period of oscillation is
`T^(')=(5T)/(3)=2pisqrt((M+m)/(k))`
or `(5)/(3)xx2pisqrt((M)/(k))=2pisqrt((M+m)/(k))`
or `(25)/(9)M=(M+m)`
or `(16)/(9)M=m or (m)/(M)=(16)/(9)`