Two particles perform linear simple harmonic motion along the same path of length 2 A and period T as shown in the graph below. The phase difference between them is

Options

A. Zero rad
B. `(pi)/(4)` rad
C. `(pi)/(2) rad`
D. `(3pi)/(4) rad`

Answer

B

Explanation

From the graph we find that when the particle P is at A or extreme position, particle Q is at a dis"tan"ce of `(A)/(sqrt(2))` from the mean position. U"sin"g `x=A "sin" (omega t) and x=(A)/(sqrt(2))`, we get `(A)/(sqrt(2))=A "sin" (omega t)`
`therefore "sin" (omega t)=(1)/(sqrt(2))="sin" ((pi)/(4)) or (omega t)=(pi)/(4) " for Q while for P, phase " =(pi)/(2)`
`because A= A "sin" (omega t_(1)) therefore omega t_(1)=1`
`therefore 1="sin" ((pi)/(2))="sin" (omega t_(1))`
Phase `(omegat_(1))=(pi)/(2)`
`therefore " Phase difference " =(pi)/(2)-(pi)/(4)=(pi)/(4)` radian