Two particles P and Q describes SHM of same amplitude a and frequency v along the same straight line. The maximum distance between the two particles is `sqrt2a`. The initial phase difference between them is

Options

A. zero
B. `(pi)/(2)`
C. `(pi)/(6)`
D. `(pi)/(3)`

Answer

B

Explanation

`x_1=Asin(omegat+phi_1)`
`x_2=Asin(omegat+phi_2)`
`implies|x_1-x_2|=2Asin(2omegat+(phi_1+phi_2)/(2))cos((phi_1-phi_2)/(2))`
To maximize `|x_1-x_2|`
`sin(2omegat+(phi_1+phi_2)/(2))=1`
`impliesasqrt2=2axx1xxcos((phi_1-phi_2)/(2))`
`implies(1)/(sqrt2)=cos((phi_1-phi_2)/(2))`
`implies((phi_1-phi_2)/(2))=(phi_1-phi_2)/(2)impliesphi_1-phi_2=(pi)/(2)`.
Hence (B)