Gujarati
Hindi
14.Waves and Sound
normal

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)

A

$\lambda / 2 \pi$

B

$\lambda / 4 \pi$

C

$\lambda / \pi$

D

$2 \lambda / \pi$

Solution

$Y=\frac{A}{2}[1-\cos (2 \omega t-2 k x)]$

particle velocity $(\mathrm{VP})_{\max }=\frac{\mathrm{A}}{2} \times 2 \omega=\mathrm{A} \omega$

wave velocity $V=\frac{2 \omega}{2 k}=\frac{2 \omega}{2 \pi} \lambda$

Given $\left(\mathrm{V}_{\max }\right) \mathrm{p}=\mathrm{V}$

$\frac{2 \omega \lambda}{2 \pi}=\omega A \rightarrow A=\lambda / \pi$

Standard 11
Physics

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