Two particles P and Q performs S.H.M. of same amplitude 'A' and frequency n along the same straight line the resultant S.H.M. have the amplitude `sqrt(2)` times amplitude of individual. The initial phase difference between two particles will be nearly.

Options

A. zero
B. `(pi)/(6)` rad
C. `(pi)/(2)` rad
D. `(3pi)/(4)` rad

Answer

C

Explanation

`R=sqrt(A^(2)+A^(2)+2A^(2)cos(prop_(1)-prop_(2)))`
`sqrt(2)A=sqrt(2A^(2)+2A^(2)cos(prop_(1)-prop_(2)))`
`1=1+cosprop_(1)-prop_(2)`
`1-1=cosprop_(1)-prop_(2)`
`cosprop_(1)-prop_(2)=0`
`prop_(1)-prop_(2)=cos^(-1)(0)`
`prop_(1)-prop_(2)=(pi)/(2)`.