A particle executes simple harmonic motion of period `16 s`. Two seconds later after it passes through the center of oscillation its velocity is found to be `2 m//s`. Find the amplitude.

Answer

B

Explanation

`omega = (2pi)/(T) = (2pi)/(16) = (pi)/(8) rad//s`
If at `t = 0`, particle passes through its mean position `(x = A sin omega t)` with maximum speed its `v - t` equation can be written as
`v = omegaA cos omegat`
Substituting the given values, we have
`2pi = ((pi)/(8)) Acos ((pi)/(8))(2)`
`:. A = (16sqrt(2))/(pi)m`
`= 7.2 m`