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Question

(A) The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ and the propagation constant $k$ by the formula: $\Delta \phi = k

Vedclass · Accepted Answer

$5$

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(A) The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ and the propagation constant $k$ by the formula: $\Delta \phi = k \cdot \Delta x$. Given: Propagation constant $k = \frac{5 \pi}{7} \, rad/m$. Path difference $\Delta x = \frac{49}{22} \, m$. Using $\pi = \frac{22}{7}$,we substitute the values: $\Delta \phi = \left( \frac{5}{7} 	imes \frac{22}{7} ight) 	imes \frac{49}{22}$. $\Delta \phi = \left( \frac{5 	imes 22}{49} ight) 	imes \frac{49}{22}$. $\Delta \phi = 5 \, rad$.

If a given wave has a propagation constant of $\frac{5 \pi}{7} \, rad/m$,then the phase difference between two particles having a distance difference of $\frac{49}{22} \, m$ is ..... $rad.$ (Take $\pi = \frac{22}{7}$)

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