Find the amplitude of the resultant wave formed by the two waves given as `y_1 = 5 sin(wt-kx), y_2 = 12 sin (wt-kx+2pi/3)`

Find the amplitude of the resultant wave produced due to interference of two waves given as, y_(1)A_(1)sinomegat,y_(2)=A_(2)sin(omegat+phi)

Determine the resultant of two waves given by y_(1) = 4 sin(200 pi t) and y_(2) = 3 sin(200 pi t + pi//2) .

The resultant amplitude due to superposition of two waves y_1 =5 sin (wt -kx) and y_2 =-5 cos (wt -kx-150 ^@)

On the superposition of the two waves given as y_1=A_0 sin( omegat-kx) and y_2=A_0 cos ( omega t -kx+(pi)/6) , the resultant amplitude of oscillations will be

Three one-dimensional mechanical waves in an elastic medium is given as y_1 = 3A sin (omegat - kx), y_2 = A sin (omegat - kx + pi) and y_3 = 2A sin (omegat + kx) are superimposed with each other. The maximum displacement amplitude of the medium particle would be

Two waves are passing through a region in the same direction at the same time . If the equation of these waves are y_(1) = a sin ( 2pi)/(lambda)( v t - x) and y_(2) = b sin ( 2pi)/( lambda) [( vt - x) + x_(0) ] then the amplitude of the resulting wave for x_(0) = (lambda//2) is

Wave equations of two particles are given by y_(1)=a sin (omega t -kx), y_(2)=a sin (kx + omega t) , then

Equation of motion in the same direction is given by y_(1) =A sin (omega t - kx), y_(2) = A sin ( omega t - kx- theta ) . The amplitude of the medium particle will be

In a wave motion y = a sin (kx - omegat) , y can represent

Equation of motion in the same direction are given by ltbygt y_(1)=2a sin (omega t-kx) and y_(2)=2a sin (omega t-kx - theta ) The amplitude of the medium particle will be