Two interfering waves have the same wavelengt | vedclass

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(C) Let the equations of the two waves be $y_1 = A \sin(kx - \omega t)$ and $y_2 = A \sin(kx - \omega t + \frac{\pi}{2})$.Since $\sin(kx - \omega t +

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wavelength and velocity but different amplitude

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(C) Let the equations of the two waves be $y_1 = A \sin(kx - \omega t)$ and $y_2 = A \sin(kx - \omega t + \frac{\pi}{2})$.Since $\sin(kx - \omega t + \frac{\pi}{2}) = \cos(kx - \omega t)$,the resultant wave is $y = y_1 + y_2 = A \sin(kx - \omega t) + A \cos(kx - \omega t)$.Using the identity $a \sin 	heta + b \cos 	heta = \sqrt{a^2 + b^2} \sin(	heta + \phi)$,where $\phi = 	an^{-1}(b/a)$,we get:$y = \sqrt{A^2 + A^2} \sin(kx - \omega t + \frac{\pi}{4}) = \sqrt{2}A \sin(kx - \omega t + \frac{\pi}{4})$.The resultant wave has an amplitude of $\sqrt{2}A$,which is different from the individual amplitude $A$.However,the wavelength $\lambda = \frac{2\pi}{k}$ and the frequency $f = \frac{\omega}{2\pi}$ (and thus the velocity $v = f\lambda = \frac{\omega}{k}$) remain the same as those of the individual waves.Therefore,the resultant wave has the same wavelength and velocity as the individual waves,but a different amplitude.

Two interfering waves have the same wavelength,frequency,and amplitude. They are traveling in the same direction but are $90^o$ out of phase. Compared to the individual waves,the resultant wave will have the same.

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