Two particles P and Q describe S.H.M. of same amplitue 'A' and frequency n along the same straight line. The resultant displacement amplitude of the two S.H.M.s is `(A//sqrt(2))`. The initial phase difference between the two particles is nearly

Options

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

Answer

D

Explanation

`R=sqrt(A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2)cos(alpha_(1)-alpha_(2)))`
`((A)/(sqrt(2)))^(2)=A^(2)+A^(2)+2A^(2)cos(alpha_(1)-alpha_(2))`
`(1)/(2)=2+2cos(alpha_(1)-alpha_(2))`
`therefore 1+cos(alpha_(1)-alpha_(2))=(1)/(4) therefore cos(alpha_(1)-alpha_(2))=(1)/(4)-1`
`therefore cos(alpha_(1)-alpha_(2))=-(3)/(2xx2)therefore cos(alpha_(1)-alpha_(2))=-(3)/(4)`
`therefore [pi-(alpha_(1)-alpha_(2))]=cos^(-1)((3)/(4))=41.40^(@)`
`therefore alpha_(1)-alpha_(2)=180-41.40^(@)~~(3pi)/(4)`