A body of mass `m` is attached to the lower end of a spring whose upper end is fixed .The spring has negligible mass .When the mass `m` is slightly pulled down and released it oscillates with a time period of `3 s` when the mass `m` is increased by `1kg`, time period of oscillations becomes `5s`. The value of `m` in `kg` is

Options

A. `(3)/(4)`
B. `(4)/(3)`
C. `(16)/(3)`
D. `(9)/(16)`

Answer

D

Explanation

(d) Time period,
`T=2pisqrt((m)/(k))`
Case I, `T_(1)=2pisqrt((m)/(k))" ...(i)`
Case II when the mass m is increased by
1 kg , then =m+1
`T_(2)=2pisqrt((m+1)/(k))" "…(i)`
From Eqs. (ii) and (i) , we get
`(T_(2))/(T_(1))=sqrt((m+1)/(m))implies(5)/(3)=sqrt((m+1)/(m))`
`implies (25)/(9)=(m+1)/(m)implies(25)/(9)=1+(1)/(m)`
`implies(1)/(m)=(16)/(9)`
`therefore m=(9)/(16) kg`