A transverse sinusoidal wave of amplitude 'a' | vedclass

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Question

(A) The maximum particle velocity $V_{p_{\max}}$ is given by $a\omega = \frac{v}{10}$.Given $a = 10^{-3} \ m$ and $v = 10 \ ms^{-1}$,we have $10^{-3}

Vedclass · Accepted Answer

$\lambda = 2\pi 	imes 10^{-2} \ m, f = \frac{10^3}{2\pi} \ Hz$

Vedclass · Answer

(A) The maximum particle velocity $V_{p_{\max}}$ is given by $a\omega = \frac{v}{10}$.Given $a = 10^{-3} \ m$ and $v = 10 \ ms^{-1}$,we have $10^{-3} 	imes \omega = \frac{10}{10} = 1$.Thus,$\omega = 1000 \ rad/s$.The frequency $f$ is given by $f = \frac{\omega}{2\pi} = \frac{1000}{2\pi} = \frac{10^3}{2\pi} \ Hz$.The wave speed $v$ is related to wavelength $\lambda$ by $v = f\lambda = \frac{\omega}{2\pi} \lambda$.Rearranging for $\lambda$,we get $\lambda = \frac{2\pi v}{\omega} = \frac{2\pi 	imes 10}{1000} = 2\pi 	imes 10^{-2} \ m$.

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