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. `sqrt2pi`
B. `16sqrt2pi`
C. `24sqrt2pi`
D. `(32sqrt2)/(pi)`

Answer

D

Explanation

For simple harmonic motion,
`y=a sin omegat`
`thereforey=a sin ((2pi)/(R))t(at =2s)`
`y_(1)=asin[((2pi)/(16))xx2]`
`=asin((pi)/(4))=(a)/(sqrt2)" "...(i)`
At t =4 s or after 2 s from mean position.
`y_(1)=(a)/(sqrt2), ` velocity =4 m/s
`therefore` Velocity `=omegasqrt(a^(2)-y_(1)^(2))`
or `4=((2pi)/(16))sqrt(a^(2)-(a^(2))/(2))` [From Rq. (ii)]
or `4=pi/8xx(a)/(sqrt2)`
or `a=(32sqrt2)/(pi)m`