Fig. below shows the wave $y = A\sin (\omega t - kx)$ at any instant travelling in the $+ve$ $x-$direction. What is the slope of the curve at $B$
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(c) The particle velocity is maximum at $B$ and is given by
$\frac{{dy}}{{dt}} = {({v_p})_{\max }} = \omega A$
Also wave velocity is $\frac{{dx}}{{dt}} = v = \frac{\omega }{k}$
So slope $\frac{{dy}}{{dx}} = \frac{{{{({v_p})}_{\max }}}}{v} = kA$
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