A particle executes simple harmonic motion with frequqncy `2.5Hz` and amplitude `2 m`. The speed of the particle `0.3 s` after crossing, the equilibrium position is

Options

A. zero
B. `2pi m//s`
C. `4pi m//s`
D. `pi m//s`

Answer

A

Explanation

`T = (1)/(F) = (1)/(2.5) = 0.4 s`
`t = 0.3s = (3T)/(4)`
At the given time, particle will be in its extreme position if at `t = 0` it crosses the mean position.