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. Determine the velocity of particle at `t=5s`.

Options

A. `-0.4(m)/(s)`
B. `0.5(m)/(s)`
C. `-0.25(m)/(s)`
D. none of these

Answer

C

Explanation

the equation of simple harmonic motion is
`x=(3)/(2pi)sin((pi)/(3))(t-1)`
and velocity `v=(1)/(2)cos((pi)/(3))(t-1)` Hence at `t=5` s `v=-(1)/(4)(m)/(s)`.