A particle executes two types of SHM. x_(1) = A_(1) sin omega t " and "x_(2) =A_(2) sin [omega t+(pi)/(3)], then find the displacement at time t=0.

A particle executes two types of SHM. x_(1) = A_(1) sin omega t " and

1.	A particle executes two types of SHM. x_(1) = A_(1) sin omega t " and "x_(2) =A_(2) sin [omega t+(pi)/(3)], then find the displacement at time t=0.
Answer» Solution :Here phase difference `phi = (PI)/(3)rad` At t=0, time, `x_(1)= 0" and "x_(2) = A_(2) sin"" (pi)/(3)` RESULTANT displacement `X= x_(1)+ x_(2)` `=0+ A_(2) sin ""(pi)/(3) = (sqrt(3)A_(2))/(2)` Resultant amplitude `A= sqrt(A_(1)^(2) +A_(2)^(2)+ 2A_(1) A_(2) cos ""(pi)/(3))` `= sqrt(A_(1)^(2) +A_(2)^(2)+ 2A_(1) A_(2))`