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

Answer

D

Explanation

In SHM, a particle starts from rest, we have
i.e. `x = A cos omega t, "at t"=0, x = A`
When `t = tau`, then `x = A - a" "...(i)`
When `t = 2 tau`, then `x = A - 3a" "...(ii)`
On comparing Eqs. (i) and (ii), we get
`A - a = A cos o omega tau`
`A - 3a = A cos 2 omega tau`
As, `cos 2 omega tau = 2 cos^(2)omega tau - 1`
`rArr" "(A-3a)/(A)=2((A-a)/(A))^(2)-1`
`= (2A^(2)+2a^(2)-4Aa-A^(2))/(A^(2))`
`A^(2)-3aA=A^(2)+2a^(2)-4Aa`
`rArr" "2a^(2)=aA`
Now, `A - a = A cos omega tau" "[because tau = t]`
`rArr" "cos omega t = 1//2 rArr (2pi)/(T)tau=(pi)/(3)rArr T = 6 tau`