The equation of a progressive wave is given b | vedclass

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(C) Comparing the given equation $y = 0.2 \cos \pi (0.04t + 0.02x - \frac{\pi}{6})$ with the standard wave equation $y = a \cos (\omega t + kx - \phi)

Vedclass · Accepted Answer

$25$

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(C) Comparing the given equation $y = 0.2 \cos \pi (0.04t + 0.02x - \frac{\pi}{6})$ with the standard wave equation $y = a \cos (\omega t + kx - \phi)$,we distribute $\pi$ inside the bracket:$y = 0.2 \cos (0.04\pi t + 0.02\pi x - \frac{\pi^2}{6})$.Here,the wave number $k = 0.02\pi$.We know that $k = \frac{2\pi}{\lambda}$,so $0.02\pi = \frac{2\pi}{\lambda}$.Solving for wavelength $\lambda$: $\lambda = \frac{2\pi}{0.02\pi} = 100 \ cm$.The relationship between phase difference $\Delta \phi$ and path difference $\Delta x$ is given by $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$.Given $\Delta \phi = \frac{\pi}{2}$,we have $\frac{\pi}{2} = \frac{2\pi}{100} \Delta x$.$\Delta x = \frac{\pi}{2} 	imes \frac{100}{2\pi} = 25 \ cm$.

The equation of a progressive wave is given by $y = 0.2 \cos \pi (0.04t + 0.02x - \frac{\pi}{6})$. The distance is expressed in $cm$ and time in seconds. What will be the minimum distance between two particles having a phase difference of $\pi / 2$ in $cm$?

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