A particle moves with simple harmonic motion in a straight line. In first `taus`, after starting form rest it travels a destance a, and in next `tau s` it travels 2a, in same direction, then:

Options

A. amplitude of motion is 4a
B. time period of oscillation is `6tau`
C. amplitude of motion is 3a
D. time period of oscillation is `8tau`

Answer

B

Explanation

As particle starts from rest, we have, `x=Acos omegat `
At `t=0, x=Acosomegaxx0=A`
When `t=tau, x=(A-a)=Acosomegatau` ..(i)
or `cos omegatau=(A-a)/(A)`
When `t=2tau, x=A-3a=Acos2omegatau`
`=2Acos^(2)omegatau-A`
or `A-3a=2A((A-a)/(A))^(2)-A`
or ` (A-3A)/(A)=2((A-a)/(A))^(2)-1`
or `A^(2)-3aA=2A^(2)+2a^(2)-4Aa-A^(2)`
`Aa=2a^(2) or A=2a`
From (i)`A-a=Acosomegatau`
`2a-a=2a cos omegatau or cos omega tau =1//2=cos pi//3`
or `(2pi)/(T)tau=(pi)/(3) or T=6tau`