If `i_(1)=3 sin omega t and (i_2) = 4 cos omega t,` then `(i_3)` is

If i_(1) = 3 sin omega t and i_(2) =6 cos omega t , then i_(3) is

In the given figure, if i_(1)=3 sin omegat and i_(2)=4 cos t, then i_(3) is

Two wave are represented by equation y_(1) = a sin omega t and y_(2) = a cos omega t the first wave :-

x_(1) = 3 "sin" omega t , x_(2) = 4 "cos' omega t Find (i) amplitude of resultant SHM. (ii) equation of the resultant SHM.

x_(1) = 3 sin omega t ,x_(2) = 4 cos omega t Find (i) amplitude of resultant SHM, (ii) equation of the resultant SHM.

x_(1) = 3 sin omega t implies x_(2) = 4 cos omega t . Find (i) amplitude of resultant SHm, (ii) equation of the resultant SHm.

The ratio of amplitudes of following SHM is x_(1) = A sin omega t and x_(2) = A sin omega t + A cos omega t

The displacements of two intering lightwaves are y_(1) = 4 sin omega t and y_(2) = 3 cos(omega t) . The amplitude of the resultant wave is ( y_(1) and y_(2) are in CGS system)

Find the resultant amplitude of the following simple harmonic equations : x_(1) = 5sin omega t x_(2) = 5 sin (omega t + 53^(@)) x_(3) = - 10 cos omega t

If i_(1) = i_(0) sin (omega t), i_(2) = i_(0_(2)) sin (omega t + phi) , then i_(3) =