A transverse wave in a medium is described by | vedclass

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Question

$Y=\frac{A}{2}[1-\cos (2 \omega t-2 k x)]$

particle velocity $(\mathrm{VP})_{\max }=\frac{\mathrm{A}}{2} 	imes 2 \omega=\mathrm{A} \omega$

wave

Vedclass · Accepted Answer

$\lambda / \pi$

Vedclass · Answer

$Y=\frac{A}{2}[1-\cos (2 \omega t-2 k x)]$

particle velocity $(\mathrm{VP})_{\max }=\frac{\mathrm{A}}{2} 	imes 2 \omega=\mathrm{A} \omega$

wave velocity $V=\frac{2 \omega}{2 k}=\frac{2 \omega}{2 \pi} \lambda$

Given $\left(\mathrm{V}_{\max }ight) \mathrm{p}=\mathrm{V}$

$\frac{2 \omega \lambda}{2 \pi}=\omega A ightarrow A=\lambda / \pi$

A transverse wave in a medium is described by the equation $y = A \sin^2 \,(\omega t -kx)$. The magnitude of the maximum velocity of particles in the medium will be equal to that of the wave velocity, if the value of $A$ is ($\lambda$ = wavelngth of wave)

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