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(C) Given: Path difference $\Delta x = 50 \ cm = 0.5 \ m$, Phase difference $\Delta \phi = 1.8 \pi$, Speed of sound $v = 340 \ ms^{-1}$.We know the re

Vedclass · Accepted Answer

$612$

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(C) Given: Path difference $\Delta x = 50 \ cm = 0.5 \ m$, Phase difference $\Delta \phi = 1.8 \pi$, Speed of sound $v = 340 \ ms^{-1}$.We know the relation between path difference and phase difference is $\Delta x = \frac{\lambda}{2 \pi} \cdot \Delta \phi$.Substituting the values: $0.5 = \frac{\lambda}{2 \pi} 	imes 1.8 \pi$.$0.5 = \lambda 	imes 0.9 \Rightarrow \lambda = \frac{0.5}{0.9} = \frac{5}{9} \ m$.The frequency $f$ is given by $f = \frac{v}{\lambda}$.$f = \frac{340}{5/9} = \frac{340 	imes 9}{5} = 68 	imes 9 = 612 \ Hz$.

The path difference between two particles of a sound wave is $50 \ cm$ and the phase difference between them is $1.8 \pi$. If the speed of sound in air is $340 \ ms^{-1}$, the frequency of the sound wave is (in $Hz$)

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