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(B) Given frequency $n = 500 \, Hz$ and velocity $v = 360 \, m/s$.First,calculate the wavelength $\lambda$ using the formula $v = n\lambda$:$\lambda =

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$12$

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(B) Given frequency $n = 500 \, Hz$ and velocity $v = 360 \, m/s$.First,calculate the wavelength $\lambda$ using the formula $v = n\lambda$:$\lambda = \frac{v}{n} = \frac{360}{500} = 0.72 \, m$.The phase difference $\phi$ is given as $60^o$. Converting this to radians:$\phi = 60^o 	imes \frac{\pi}{180^o} = \frac{\pi}{3} \, 	ext{radians}$.The path difference $\Delta x$ is related to the phase difference $\phi$ by the formula:$\Delta x = \frac{\lambda}{2\pi} 	imes \phi$.Substituting the values:$\Delta x = \frac{0.72}{2\pi} 	imes \frac{\pi}{3} = \frac{0.72}{6} = 0.12 \, m$.Converting the distance to centimeters:$\Delta x = 0.12 	imes 100 = 12 \, cm$.

$A$ wave of frequency $500 \, Hz$ has a velocity of $360 \, m/s$. The distance between two nearest points $60^o$ out of phase is .... $cm$.

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