A block of mass m is connected to a spring of spring constant k as shown in Fig. The block is found at its equilibrium position `t=1s` and it has a velocity of `+0.25(m)/(s)` at `t=2s`. The time period of oscillation is 6 s. Based on the given information answer the following question: Q. the amplitude of oscillation is

Options

A. `(3)/(2pi)cm`
B. `3 m`
C. `(1)/(pi)m`
D. `1.5m`

Answer

A

Explanation

Let the equation of simple harmonic motion be
`x=Asin(omegat+delta)`
Then , at `t=1`,`x=0`, so
`0=Asin(delta+omega)=sin(omega+delta)=0`
At `t=2s` ,`v=+0.25(m)/(s)`
So `0.25=Aomegacos(2omega+delta)`
Solving above equation we get `A=(3)/(2pi)m` and `delta=-(pi)/(3)`
Hence equation of SHM will be `x=(3)/(2pi)sin((pi)/(3)t-(pi)/(3))`
For velocity at `t=5s`, i.e., after half a period of `t=2` s the velocity is having same magnitude but opposite direction.