A transverse wave propagating along the x- ax | vedclass

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Question

(A) The given wave equation is $y(x, t) = 8.0 \sin \left( 0.5 \pi x - 4 \pi t - \frac{\pi}{4} \right)$.Comparing this with the standard wave equation

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) The given wave equation is $y(x, t) = 8.0 \sin \left( 0.5 \pi x - 4 \pi t - \frac{\pi}{4} \right)$.Comparing this with the standard wave equation $y(x, t) = A \sin (kx - \omega t + \phi)$:Here,the wave number $k = 0.5 \pi$ and the angular frequency $\omega = 4 \pi$.The speed of the wave $v$ is given by the ratio of angular frequency to wave number:$v = \frac{\omega}{k}$Substituting the values:$v = \frac{4 \pi}{0.5 \pi} = \frac{4}{0.5} = 8 \ m/s$.Therefore,the speed of the wave is $8 \ m/s$.

$A$ transverse wave propagating along the $x-$ axis is represented by $y(x,t) = 8.0 \sin \left( 0.5\pi x - 4\pi t - \frac{\pi}{4} \right)$,where $x$ is in metres and $t$ is in seconds. The speed of the wave is ..... $m/s$.

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