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

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

Phase difference between two waves y _(1) = A sin (omega t - kx) and y _(2)= A cos (omega t - kx) is ……

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)

The resultant amplitude of a vibrating particle by the superposition of the two waves y_(1)=asin[omegat+(pi)/(3)] and y_(2)=asinomegat is

The resultant amplitude of a vibrating particle by the superposition of the two waves y_(1)=asin[omegat+(pi)/(3)] and y_(2)=asinomegat is

The resultant amplitude of a vibrating particle by the superposition of the two waves y_(1)=asin[omegat+(pi)/(3)] and y_(2)=asinomegat is

The resultant displacement due to superposition of two identical progressive waves is y = 5cos(0.2pix)sin(64pit) , where x, y are in cm and t is in sec. Find the equations of the two superposing waves.

Statement-1 : The superposition of the waves y_(1) = A sin(kx - omega t) and y_(2) = 3A sin (kx + omega t) is a pure standing wave plus a travelling wave moving in the negative direction along X-axis Statement-2 : The resultant of y_(1) & y_(2) is y=y_(1) + y_(2) = 2A sin kx cos omega t + 2A sin (kx + omega t) .