The equation of a wave is y = 60 sin (1200 t | vedclass

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Question

(A) The given wave equation is $y = A \sin (\omega t - k x)$,where $A = 60 \ \mu m = 60 	imes 10^{-6} \ m$,$\omega = 1200 \ rad/s$,and $k = 6 \ m^{-1

Vedclass · Accepted Answer

$3.6 	imes 10^{-4}$

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(A) The given wave equation is $y = A \sin (\omega t - k x)$,where $A = 60 \ \mu m = 60 	imes 10^{-6} \ m$,$\omega = 1200 \ rad/s$,and $k = 6 \ m^{-1}$.The maximum particle velocity $(v_p)$ is given by $v_p = A \omega$.$v_p = (60 	imes 10^{-6} \ m) 	imes (1200 \ rad/s) = 72000 	imes 10^{-6} \ m/s = 7.2 	imes 10^{-2} \ m/s$.The wave velocity $(v_w)$ is given by $v_w = \frac{\omega}{k}$.$v_w = \frac{1200}{6} = 200 \ m/s$.The ratio of maximum particle velocity to wave velocity is $\frac{v_p}{v_w} = \frac{A \omega}{\omega/k} = A k$.$\frac{v_p}{v_w} = (60 	imes 10^{-6} \ m) 	imes (6 \ m^{-1}) = 360 	imes 10^{-6} = 3.6 	imes 10^{-4}$.

The equation of a wave is $y = 60 \sin (1200 t - 6 x)$,where '$y$' is in micron,'$t$' is in second,and '$x$' is in metre. The ratio of the maximum particle velocity to the wave velocity of wave propagation is:

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