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Two waves represented by y= a sin(wt - kx) and y= a cos(wt - kx) are superposed The resultant wave will have an amplitude?
the 2nd wave can be written as y2=Asin(wt-kx+pi/2)Now by superposition principle, the resultant wave is y=y1+y2y=A[sin(wt-kx)+sin(wt-kx+pi/2)] =2Acos(pi/4)sin(wt-kx+pi/4) (Transformation formula used)This wave equation is telling us that the amplitude of the resultant wave is 2Acos(pi/4) and the the phase angle of the wave is pi/4 , while frequency remaining constant.
Dear Student,Please find below the solution to your problem.Given, y1 = asin(ωt-kx)y2 = acos(ωt-kx)or y2 = asin(ωt-kx+2π)Phase difference of two waves = 2π∵ Resultant amplitudeR = a2 + a2 + 2aacosϕ= a2 + a2 + 2a2^cos2π= 2a^2 (∵cos2π=0)R = 2a.y1= asin(ωt-kx)Thanks and Regards
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