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.

Explanation

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))`