The figure shows the wave y = A sin(omega t - | vedclass

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(C) The given wave equation is $y = A \sin(\omega t - kx)$.To find the slope of the curve at any point $x$ at a given time $t$,we calculate the partia

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$kA$

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(C) The given wave equation is $y = A \sin(\omega t - kx)$.To find the slope of the curve at any point $x$ at a given time $t$,we calculate the partial derivative of $y$ with respect to $x$:$\frac{\partial y}{\partial x} = A \cos(\omega t - kx) \cdot (-k) = -kA \cos(\omega t - kx)$.The magnitude of the slope is $|\frac{\partial y}{\partial x}| = |kA \cos(\omega t - kx)|$.At point $B$,the displacement $y = 0$,which implies $\sin(\omega t - kx) = 0$. This occurs when $(\omega t - kx) = n\pi$ (where $n$ is an integer).At these points,$\cos(\omega t - kx) = \pm 1$.Therefore,the magnitude of the slope at $B$ is $|-kA(\pm 1)| = kA$.

The figure shows the wave $y = A \sin(\omega t - kx)$. What is the magnitude of the slope of the curve at $B$?

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