A progressive wave travelling along the posit | vedclass

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(A) The given wave equation is $y(x, t) = A \sin(kx - \omega t + \phi)$.At $t = 0$,the equation becomes $y(x, 0) = A \sin(kx + \phi)$.From the given g

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$\pi$

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(A) The given wave equation is $y(x, t) = A \sin(kx - \omega t + \phi)$.At $t = 0$,the equation becomes $y(x, 0) = A \sin(kx + \phi)$.From the given graph at $t = 0$,we observe that at $x = 0$,the displacement $y(0, 0) = 0$.Substituting these values into the equation: $0 = A \sin(k(0) + \phi) \Rightarrow \sin(\phi) = 0$.This implies $\phi = 0$ or $\phi = \pi$.To determine the correct value,we check the slope of the wave at $x = 0$. The slope is given by $\frac{\partial y}{\partial x} = Ak \cos(kx + \phi)$.At $x = 0$,the slope is $Ak \cos(\phi)$.From the graph,at $x = 0$,the wave is moving downwards as $x$ increases,meaning the slope is negative.If $\phi = 0$,the slope is $Ak \cos(0) = Ak$,which is positive.If $\phi = \pi$,the slope is $Ak \cos(\pi) = -Ak$,which is negative.Therefore,the phase $\phi$ must be $\pi$.

$A$ progressive wave travelling along the positive $x-$ direction is represented by $y(x, t) = A \sin(kx - \omega t + \phi)$. Its snapshot at $t = 0$ is given in the figure. For this wave,the phase $\phi$ is

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