Question

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

Answer 1

Correct Answer - A
`T = 2pi sqrt((M)/(k))`
`(5T)/(3) = 2pi sqrt ((M + m)/(k))`
Solving these two equation, we get
`(m)/(M) = (16)/(9)`