A travelling acoustic wave of frequency 500 H | vedclass

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(C) The frequency of the wave is $f = 500 Hz$ and the velocity is $v = 300 ms^{-1}$.First,we calculate the wavelength $\lambda$ using the relation $v

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$10 cm$

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(C) The frequency of the wave is $f = 500 Hz$ and the velocity is $v = 300 ms^{-1}$.First,we calculate the wavelength $\lambda$ using the relation $v = f \lambda$:$\lambda = \frac{v}{f} = \frac{300}{500} = 0.6 m = 60 cm$.The phase difference $\Delta \phi$ is given as $60^{\circ}$,which is $\frac{\pi}{3}$ radians.The relationship between phase difference $\Delta \phi$ and path difference $\Delta x$ is given by $\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$.Substituting the values: $\frac{\pi}{3} = \frac{2 \pi}{60 cm} \Delta x$.Solving for $\Delta x$: $\Delta x = \frac{\pi}{3} 	imes \frac{60 cm}{2 \pi} = \frac{60}{6} cm = 10 cm$.

$A$ travelling acoustic wave of frequency $500 Hz$ is moving along the positive $x$-direction with a velocity of $300 ms^{-1}$. The phase difference between two points $x_{1}$ and $x_{2}$ is $60^{\circ}$. The minimum separation between the two points is:

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