A particle excuting `S.H.M.` of amplitude `4 cm `and `T = 4` sec .The time take by it to move position extreme position to half the amplitude is

Options

A. `1/3s`
B. `2/3s`
C. `3/4s`
D. `4/3s`

Answer

B

Explanation

Let particle is at O at `t=0`
`x=A sin (omega t+phi)`
`x=Asin (omegaxx0+phi)impliesphi=0`
`x=Asinomegat`
O to P: `A/2=A sin omegatimpliessin ((2pi)/(T))t=1/2=sinpi/6`
`(2pi)/(T)t=pi/6impliesT/12`
O to A: time `t^'=T//4`
P to A: time `=t^'-t=T/4-(T)/(12)=T/6=4/6=2/3s`
Time from A to P = Time from P to A `=2/3s`