The time period of a body of mass M executing SHM, connected to a spring, is 2 s. If the mass of the body is increased by 2 kg, its time period increases by 1 s. Considering that Hooke's law is obeyed, calculate the initial mass M.

Explanation

The time period of SHM executed by the body connected to the spring, `T=2pisqrt(M/k)` , k = force constant of the spring
So, in the first case,
`2=2pisqrt(M/k)" "...(1)`
and in the second case,
`2+1=3=2pisqrt((M+2)/k)" "...(2)`
Dividing (2) by (1) we get,
`3/2=sqrt((M+2)/M)or,9/4=(M+2)/M`
or, 9M = 4M + 8 or, M = 1.6kg.