Two harmonic travelling waves are described b | vedclass

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(A) Given equations are:$y_1 = a \sin (kx - \omega t)$$y_2 = a \sin (-(kx - \omega t) + \phi)$Using the identity $\sin(-	heta) = -\sin(	heta)$, we c

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$2a \cos \frac{\phi}{2}$

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(A) Given equations are:$y_1 = a \sin (kx - \omega t)$$y_2 = a \sin (-(kx - \omega t) + \phi)$Using the identity $\sin(-	heta) = -\sin(	heta)$, we can rewrite $y_2$ as:$y_2 = -a \sin (kx - \omega t - \phi)$Using $\sin(	heta - \phi) = \sin 	heta \cos \phi - \cos 	heta \sin \phi$, the superposition $y = y_1 + y_2$ is:$y = a \sin(kx - \omega t) + a \sin(kx - \omega t + \phi)$Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:$y = 2a \sin(\frac{kx - \omega t + kx - \omega t + \phi}{2}) \cos(\frac{kx - \omega t - (kx - \omega t + \phi)}{2})$$y = 2a \sin(kx - \omega t + \frac{\phi}{2}) \cos(-\frac{\phi}{2})$Since $\cos(-	heta) = \cos(	heta)$:$y = [2a \cos(\frac{\phi}{2})] \sin(kx - \omega t + \frac{\phi}{2})$The amplitude of the resultant wave is the coefficient of the sine term, which is $2a \cos(\frac{\phi}{2})$.

Two harmonic travelling waves are described by the equations $y_1 = a \sin (kx - \omega t)$ and $y_2 = a \sin (-kx + \omega t + \phi)$. The amplitude of the superposed wave is:

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