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. `32sqrt2pi`

Answer

D

Explanation

`x=asin(omegat+phi)`
`omega=(2pi)/(T)=(2pi)/(16)=pi/8`
`x=asin(pi/8t+phi)`
At `t=2s`, `x=0`
`0=asin(pi/8xx2+phi)`
`sin(pi/4+phi)=0impliespi/4+phi=piimpliesphi=(3pi)/(4)`
`v=(dx)/(dx)=a*pi/8cos(pi/8t+(3pi)/(4))`
`4=(pia)/(8)cos(pi/8xx4+(3pi)/(4))=(-pia)/(8)sin((3pi)/(4))`
`4=-(pia)/(8)sin(pi-pi//4)=-(pia)/(8)sinpi//4`
`=-(pia)/(8sqrt2)implies|a|=(32sqrt2)/(pi)`