The stationary wave y = 2a sin kx cos omega t | vedclass

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(B) stationary wave is formed by the superposition of two waves of the same frequency and amplitude traveling in opposite directions.Given the inciden

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$y = -a \sin (\omega t + kx)$

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(B) stationary wave is formed by the superposition of two waves of the same frequency and amplitude traveling in opposite directions.Given the incident wave is $y_1 = a \sin (\omega t - kx)$.For a closed organ pipe,the reflection at the closed end introduces a phase change of $\pi$.Thus,the reflected wave is $y_2 = a \sin (\omega t + kx + \pi)$.Using the trigonometric identity $\sin (	heta + \pi) = -\sin 	heta$,we get $y_2 = -a \sin (\omega t + kx)$.The superposition of these two waves is $y = y_1 + y_2 = a \sin (\omega t - kx) - a \sin (\omega t + kx)$.Using the identity $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$,we get $y = 2a \cos \omega t \sin (-kx) = -2a \sin kx \cos \omega t$.Note: The sign depends on the coordinate system origin; the wave $y = -a \sin (\omega t + kx)$ is the correct reflected component.

The stationary wave $y = 2a \sin kx \cos \omega t$ in a closed organ pipe is the result of the superposition of $y = a \sin (\omega t - kx)$ and

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