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NEET PHYSICSMedium

Two points are located at a distance of 10 m10 \text{ m} and 15 m15 \text{ m} from the source of oscillation. The period of oscillation is 0.05 s0.05 \text{ s} and the velocity of the wave is 300 m/s300 \text{ m/s}. What is the phase difference between the oscillations of two points?

A

π3\frac{\pi}{3}

B

2π3\frac{2\pi}{3}

C

π\pi

D

π6\frac{\pi}{6}

Step-by-Step Solution

  1. Identify the Given Data: Path difference, Δx=x2−x1=15 m−10 m=5 m\Delta x = x_2 - x_1 = 15 \text{ m} - 10 \text{ m} = 5 \text{ m}. Time period, T=0.05 sT = 0.05 \text{ s}. Wave velocity, v=300 m/sv = 300 \text{ m/s}.
  2. Calculate Wavelength (λ\lambda): Wavelength is the distance travelled by the wave in one time period. λ=v×T=300×0.05=15 m\lambda = v \times T = 300 \times 0.05 = 15 \text{ m}.
  3. Calculate Phase Difference (Δϕ\Delta\phi): The relationship between phase difference and path difference is given by Δϕ=2πλΔx\Delta\phi = \frac{2\pi}{\lambda} \Delta x . Substitute the values: Δϕ=2π15×5=10π15=2π3 rad\Delta\phi = \frac{2\pi}{15} \times 5 = \frac{10\pi}{15} = \frac{2\pi}{3} \text{ rad}
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