A plane wave y=a sin (omega t-k x) propagates | vedclass

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(A) The plane wave is given by $y=a \sin (\omega t-k x)$.At $t=0$,the displacement is $y=a \sin (-k x) = -a \sin (k x)$.The particle velocity $v_{pa}$

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(A) The plane wave is given by $y=a \sin (\omega t-k x)$.At $t=0$,the displacement is $y=a \sin (-k x) = -a \sin (k x)$.The particle velocity $v_{pa}$ is the time derivative of displacement:$v_{pa} = \frac{d y}{d t} = \frac{d}{d t} (-a \sin (k x)) = -a \cos (k x) \cdot \frac{d}{d t} (k x)$.Since $k = \frac{\omega}{v_{wave}}$,this is not quite right. Let's differentiate correctly:$v_{pa} = \frac{\partial y}{\partial t} = a \omega \cos (\omega t - k x)$.At $t=0$,$v_{pa} = a \omega \cos (-k x) = a \omega \cos (k x)$.Using $k = \frac{2 \pi}{\lambda}$,we get $v_{pa} = a \omega \cos \left( \frac{2 \pi x}{\lambda} \right)$.At $x=0$,$v_{pa} = a \omega$.At $x=\frac{\lambda}{4}$,$v_{pa} = a \omega \cos \left( \frac{\pi}{2} \right) = 0$.At $x=\frac{\lambda}{2}$,$v_{pa} = a \omega \cos (\pi) = -a \omega$.This corresponds to a cosine graph starting at a positive maximum. Looking at the provided options,the graph that matches $v_{pa} = a \omega \cos (kx)$ is option $A$.

$A$ plane wave $y=a \sin (\omega t-k x)$ propagates through a stretched string. The particle velocity versus $x$ graph at $t=0$ is

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