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. `24 sqrt(2) pi`
D. `32 sqrt(2)//pi`

Answer

D

Explanation

For simple harmonic motion, y = a sin `omega`t
`therefore" "y = a sin ((2pi)/(T))t" "("at t"=2s)`
`y_(1)=a sin [((2pi)/(16))xx2]=a sin((pi)/(4))=(a)/(sqrt(2))" "...(i)`
At t = 4 s or after 2 s from mean position.
`y_(1)=(a)/(sqrt(2))"and velocity"=4 ms^(-1)`
`therefore` Velocity `= omega sqrt(a^(2)-y_(1)^(2))`
or `4 = ((2pi)/(16))sqrt(a^(2)-(a^(2))/(2))" "["From Eq. (i)"]`
or `4 = (pi)/(8) xx (a)/(sqrt(2))or a = (32 sqrt(2))/(pi)m`