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(D) For a traveling wave represented by $y = A \sin(kx - \omega t)$,the instantaneous velocity of a particle $u$ is given by the partial derivative of

Vedclass · Accepted Answer

$|\vec{u}| = (	ext{slope of wave form}) 	imes |\vec{v}|$

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(D) For a traveling wave represented by $y = A \sin(kx - \omega t)$,the instantaneous velocity of a particle $u$ is given by the partial derivative of displacement with respect to time:$u = \frac{\partial y}{\partial t} = -A\omega \cos(kx - \omega t)$The slope of the waveform is given by the partial derivative of displacement with respect to position:$	ext{Slope} = \frac{\partial y}{\partial x} = Ak \cos(kx - \omega t)$The wave velocity $v$ is given by $v = \frac{\omega}{k}$,which implies $\omega = vk$.Substituting $\omega = vk$ into the expression for $u$:$u = -A(vk) \cos(kx - \omega t)$$u = -v 	imes [Ak \cos(kx - \omega t)]$$u = -v 	imes (	ext{slope of waveform})$Taking the magnitude on both sides:$|u| = |v| 	imes |	ext{slope of waveform}|$Thus,option $D$ is correct.

If $\vec{u}$ is the instantaneous velocity of a particle and $\vec{v}$ is the velocity of the wave,then:

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