Two points are located at a distance of $10 ext{ m}$ and $15 ext{ m}$ from the source of oscillation. The

Question

Two points are located at a distance of $10 	ext{ m}$ and $15 	ext{ m}$ from the source of oscillation. The period of oscillation is $0.05 	ext{ s}$ and the velocity of the wave is $300 	ext{ m/s}$. What is the phase difference between the oscillations of two points?

Accepted Answer

1. **Identify the Given Data:** Path difference, $\Delta x = x_2 - x_1 = 15 	ext{ m} - 10 	ext{ m} = 5 	ext{ m}$. Time period, $T = 0.05 	ext{ s}$. Wave velocity, $v = 300 	ext{ m/s}$.
2. **Calculate Wavelength ($\lambda$):** Wavelength is the distance travelled by the wave in one time period. $\lambda = v 	imes T = 300 	imes 0.05 = 15 	ext{ m}$.
3. **Calculate Phase Difference ($\Delta\phi$):** The relationship between phase difference and path difference is given by $\Delta\phi = \frac{2\pi}{\lambda} \Delta x$ .
   Substitute the values:
   $$\Delta\phi = \frac{2\pi}{15} 	imes 5 = \frac{10\pi}{15} = \frac{2\pi}{3} 	ext{ rad}$$

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