Two particles execute simple harmonic motion of same amplitude and frequency along the same straight line. They pass on another, when going in opposite directions, each time their displacement is half of their amplitude. What is the phase difference between them?

Options

A. `30^(@)`
B. `60^(@)`
C. `90^(@)`
D. `120^(@)`

Answer

D

Explanation

Let two simple harmonic motions are y = a sin `omega`t and y = a sin `(omega t + phi)`.
In the first case, `(a)/(2)=a sin omega t`
`rArr" "sin omega t = (1)/(2)`
`therefore" "cos omega t=(sqrt(3))/(2)`
In the second case, `(a)/(2)=a sin (omega t + phi)`
`rArr" "(1)/(2)=[sin omegat.cos phi+cos omega t sin phi]`
`rArr" "(1)/(2)=[(1)/(2)cos phi + (sqrt(3))/(2)sin phi]`
`rArr" "1 - cos phi = sqrt(3) sin phi`
`rArr" "(1-cos phi)^(2)=3 sin^(2)phi`
`rArr" "(1-cos phi)^(2)=3(1-cos^(2)phi)`
On solving above equation, we get
`cos phi = + 1`
or `cos phi = (-1)/(2)`
i.e. `phi = 0`
or `phi = 120^(@)`