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(C) Given: Frequency $f = 500 \,Hz$, Velocity $v = 360 \,ms^{-1}$, Phase difference $\Delta\phi = 60^{\circ}$.First, calculate the wavelength $\lambda

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$0.12$

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(C) Given: Frequency $f = 500 \,Hz$, Velocity $v = 360 \,ms^{-1}$, Phase difference $\Delta\phi = 60^{\circ}$.First, calculate the wavelength $\lambda$ using the formula $\lambda = \frac{v}{f}$.$\lambda = \frac{360}{500} = 0.72 \,m$.The relationship between path difference $\Delta x$ and phase difference $\Delta\phi$ is given by $\Delta\phi = \frac{2\pi}{\lambda} \Delta x$.Converting $60^{\circ}$ to radians: $60^{\circ} = \frac{\pi}{3} \,rad$.Substituting the values: $\frac{\pi}{3} = \frac{2\pi}{\lambda} \Delta x$.$\Delta x = \frac{\lambda}{6} = \frac{0.72}{6} = 0.12 \,m$.Therefore, the distance between the two points is $0.12 \,m$.

$A$ progressive wave of frequency $500 \,Hz$ is travelling with a velocity of $360 \,ms^{-1}$. The distance between two points, having a phase difference of $60^{\circ}$, is ............. (in $\,m$)

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