For a wave equation y = 8 sin 2pi (0.1x - 2t) | vedclass

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Question

(D) The given wave equation is $y = 8 \sin 2\pi (0.1x - 2t) \, cm$.Comparing this with the standard wave equation $y = a \sin 2\pi (\frac{x}{\lambda}

Vedclass · Accepted Answer

$72$

Vedclass · Answer

(D) The given wave equation is $y = 8 \sin 2\pi (0.1x - 2t) \, cm$.Comparing this with the standard wave equation $y = a \sin 2\pi (\frac{x}{\lambda} - \frac{t}{T})$,we get the wave number term $k = \frac{2\pi}{\lambda} = 2\pi(0.1) = 0.2\pi$.Thus,$\lambda = \frac{2\pi}{0.2\pi} = 10 \, cm$.The phase difference $\Delta \phi$ is given by the formula $\Delta \phi = \frac{2\pi}{\lambda} 	imes \Delta x$.Given path difference $\Delta x = 2 \, cm$.Substituting the values: $\Delta \phi = \frac{2\pi}{10} 	imes 2 = \frac{4\pi}{10} = 0.4\pi \, radians$.To convert radians to degrees,multiply by $\frac{180^\circ}{\pi}$:$\Delta \phi = 0.4\pi 	imes \frac{180^\circ}{\pi} = 0.4 	imes 180^\circ = 72^\circ$.

For a wave equation $y = 8 \sin 2\pi (0.1x - 2t) \, cm$,find the phase difference in $^\circ$ between two particles separated by a distance of $2 \, cm$.

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