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

Options

A. A)`sqrt(2)pi`
B. B)`16sqrt(2)pi`
C. C)`24sqrt(2)pi`
D. D)`(32sqrt(2))/(pi)`

Answer

D

Explanation

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