A particle executes simple harmonic motion with a period of `16s`. At time `t=2s`, the particle crosses the mean position while at `t=4s`, its velocity is `4ms^-1` amplitude of motion in metre is

Options

A. `sqrt(2) pi`
B. `16 sqrt(2) pi`
C. `(32 sqrt(2))/(pi)`
D. `(4)/(pi)`

Answer

C

Explanation

`y = A sin omega t" "t_(1) = 2s = 4, t = 2`
`V = A omega cos omega t`
`4 = A xx (2pi)/(16)cos ((2pi)/(16)xx2), 4 = A xx (pi)/(6) xx (1)/(sqrt(2))`
`A = (32 sqrt(2))/(pi)`