A certain ideal spring stretches `20cm` when `m_(1)=40gm` mass is hung from it. If a mass of `m_(2)=205/32gm` gm is hung instead of `40gm` at its end pulled out `20cm` from equilibirum and released at `t=0` sec. Choose the correct option (s) `(g=10m//s^(2))`

Options

A. The magnitude of acceleration of the `m_(2)` mass when it is `2.5cm`from the equilibrium position is `64/45 m//s^(2)`
B. The magnitude of acceleration of the `m_(2)` mass when it si `2.5cm` from the equilibrium position is `32/45m//s^(2)`
C. At `t=(pi)/8` sec `m_(2)` mass is `10cm` from equilibrium position
D. At `t=(pi)/2` sec `m_(2)` is `10cm` from equilibrium position

Answer

B, C, D

Explanation

`k=` stiffness of string `=F/(x_("max"))=(mg)/(x_("max"))=(40xx10^(-3))/(20xx10^(-2))=1.96N//m`
When we replace the mass `40 gm` by the mass `m_(2)=2205/32gm`, we have
`T=2pisqrt(m/k)=(3pi)/8` sec
`omega=(2pi)/T=sqrt(k/m)=16/3rad//sec`
`x=x_(0)cosomegat=0.2 "cos" 16/3 t`
`a=-x_(0)omega^(2)cos omegat=-16/15"sin" 16/3t`
`a=-x_(0)omega^(2)cosomegat=-256/45"cos"16/3t`
If `x=10cmimplies"cos" 16/3 t=1/2=cos((pi)/3)`
`implies16/3t=(2npi+-(pi)/3)`
`impliest=((3npi)/8+-(pi)/8)`sec