The equation of a transverse wave travelling | vedclass

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(A) The given equation is $y = 10\sin(0.01\pi x - 2\pi t)$.Comparing this with the standard wave equation $y = A\sin(kx - \omega t)$,we get the amplit

Vedclass · Accepted Answer

$63$

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(A) The given equation is $y = 10\sin(0.01\pi x - 2\pi t)$.Comparing this with the standard wave equation $y = A\sin(kx - \omega t)$,we get the amplitude $A = 10 \ cm$ and the angular frequency $\omega = 2\pi \ rad/s$.The maximum transverse speed of a particle in the rope is given by the formula $v_{\max} = A\omega$.Substituting the values,we get $v_{\max} = 10 	imes 2\pi$.$v_{\max} = 20 	imes 3.14159 = 62.83 \ cm/s$.Rounding to the nearest integer,we get $v_{\max} \approx 63 \ cm/s$.

The equation of a transverse wave travelling on a rope is given by $y = 10\sin \pi (0.01x - 2.00t)$ where $y$ and $x$ are in $cm$ and $t$ is in $seconds$. The maximum transverse speed of a particle in the rope is about .... $cm/s$.

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