In the wave equation y = 0.5 sin frac{2 pi}{l | vedclass

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(C) The standard form of a traveling wave equation is $y = A \sin(\omega t - kx)$.Given the equation: $y = 0.5 \sin \left( \frac{2 \pi}{\lambda} (400

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$400$

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(C) The standard form of a traveling wave equation is $y = A \sin(\omega t - kx)$.Given the equation: $y = 0.5 \sin \left( \frac{2 \pi}{\lambda} (400 t - x) \right) = 0.5 \sin \left( \frac{800 \pi}{\lambda} t - \frac{2 \pi}{\lambda} x \right)$.Comparing this with the standard form,we identify the angular frequency $\omega$ and the wave number $k$:$\omega = \frac{800 \pi}{\lambda}$$k = \frac{2 \pi}{\lambda}$The velocity of the wave $v$ is given by the ratio of angular frequency to wave number:$v = \frac{\omega}{k} = \frac{800 \pi / \lambda}{2 \pi / \lambda} = \frac{800 \pi}{2 \pi} = 400 \, m/s$.Therefore,the velocity of the wave is $400 \, m/s$.

In the wave equation $y = 0.5 \sin \frac{2 \pi}{\lambda}(400 t - x ) \, m$,the velocity of the wave will be ......... $m/s$.

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