Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

$\frac{\pi}{3}$

$\frac{2\pi}{3}$

$\pi$

$\frac{\pi}{6}$

Step-by-Step Solution

Identify Given Values: The distances of the two points from the source are $x_1 = 10\text{ m}$ and $x_2 = 15\text{ m}$ . The path difference is $\Delta x = 15 - 10 = 5\text{ m}$ . The time period is $T = 0.05\text{ s}$ and wave velocity is $v = 300\text{ m/s}$ .
Calculate Wavelength ( $\lambda$ ): The wavelength is the distance covered by the wave in one time period. $\lambda = v \times T = 300 \times 0.05 = 15\text{ m}$ .
Calculate Phase Difference ( $\Delta \phi$ ): The relation between phase difference and path difference is given by $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$ . Substituting the values: $\Delta \phi = \frac{2\pi}{15} \times 5 = \frac{2\pi}{3} \text{ radians}$

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