One end of a spring of force constant k is fixed to a vertical wall and the other to a blcok of mass m resting on a smooth horizontal surface. There is another wall at distance `x_(0)` from the block. The spring is then compressed by `2x_(0)` and released. The time taken to strike the wall is

Options

A. `(1)/(6)pisqrt(k/m)`
B. `sqrt(k/m)`
C. `(2pi)/(3)sqrt(m/k)`
D. `(pi)/(4)sqrt(k/m)`

Answer

C

Explanation

`t_(AC)=t_(AB)+t_(BC)`
`implies t_(AC)=(T)/(4)+t_(BC)`
Now,
`BC=AB sin((2pi)/(T))t_(BC)" "implies (BC)/(AB)=sin((2pi)/(T))t_(BC)`
`implies (1)/(2)=sin((2pi)/(T))t_(BC)" "["as "(BC)/(AB)=(1)/(2)]`
`implies "sin"(pi)/(6)=sin((2pi)/(T))t_(BC)implies (pi)/(6)=(2pi)/(T)t_(BC)`
`hArr t_(BC)=(T)/(12)`
So, `t_(AC)=(T)/(4)+(T)/(12)=(T)/(3) implies t_(AC)=(2pi)/(3)sqrt(m/k)`.