Two particles move parallel to x - axis about the origin with the same amplitude and frequency. At a certain instant they are found at distance `(A)/(3)` from the origin on opposite sides but their velocities are found to be in the same direction. What is the phase difference between the two ?

Options

A. `cos^-1((7)/(9))`
B. `cos^-1((5)/(9))`
C. `cos^-1((4)/(9))`
D. `cos^-1((1)/(9))`

Answer

A

Explanation

Let `x_1=asinomegat` and `x_2=asin(omegat+delta)` be two SHMs.
`(a)/(3)=asinomegat` and `-(a)/(3)=asin(omegat+delta)`
`sinomegat=(1)/(3)` and `sin(omegat+delta)=(-1)/(3)`
Eliminating t, `(1)/(3)cosdelta+sqrt(1-(1)/(9))sindelta=-(1)/(3)`
`9cos^2delta+2cosdelta-7=0`
`cosdelta=-1` or `(7)/(9)`
`delta=180^@` or `cos^-1((7)/(9))`
If put `180^@`, we find that `v_1` and `v_2` are of opposite signs Hence `delta=180^2` is not applicable
`delta=cos^-1((7)/(9))`