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

Options

A. `(1)/(6) msqrt((m)/(k))`
B. `sqrt((m)/(k))`
C. `(2pi)/(3)sqrt((m)/(k))`
D. `(pi)/(64)sqrt((m)/(k))`

Answer

C

Explanation

According to the question,
Givne, that the amplitude of the motion , `A = 2x_(0)`
WE know that the time needed to cover from compressed position to mean position (normal ) ` = (T)/(4)`.
Where , T is the time period of oscillation for the displacement from the mean position to `x_(0)` , the time taken, t can be obtained from ` y = A sin omega t `
where , y is the displacement from mean position , A is the amplitude
therefore, `x_(0) = 2x_(0) sin omega t rArr x_(0) = 2x_(0) "sin"(2pi)/(T) t`
`rArr "sin "(pi)/(6) = "sin" (2pi)/(T) t rArr t =(T)/(12)`
Therefore , the time taken to hit the wall will be,
`(T)/(4) +(T)/(12) = (T)/(3)`
If a mass m is suspended from a spring of force constant k, then the time period will be
`T = 2pi sqrt((m)/(k))`
Hence, the time taken to hit the wall
` (T)/(3)= (2pi)/(3) sqrt((m)/(k))`