If `y=5(mm) sin pi t` is equation of oscillation of source `S_(1)` and `y_(2) = 5 (mm) sin pi//6)` be that of `S_(2)` and it takes `1 sec` and `^(1//_(2 sec)` for the transverse waves to reach point `A` from source `S_(1)` and `S_(2)` respectively then the resulting amplitude at point `A`, is
.

Options

A. `5 sqrt(2 +sqrt(3)) mm`
B. `5 sqrt(3) mm`
C. `5 mm`
D. `5 sqrt(2)mm`

Answer

C

Explanation

Wave originating at `t =0` from `S_(1)` reaches point `A` at `t =1`
Wave originating at `t = (1)/(2)` from `S_(2)` reaches point `A` at `t =1`
So phase difference in these waves `= (pi)/(2)+(pi)/(6),A=sqrt(A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2)cos phi)=5`.