Question

Consider two points lying at a distance of $10m$ and $15m$ from an oscillating source. If the periodic time of oscillation is $0.05s$ and the velocity of the wave produced is $300m{s}^{-1}$, then what will be the phase difference between the two points?

• A. $\mathrm{\pi }$
• B. $\frac{\mathrm{\pi }}{6}$
• C. $\frac{\mathrm{\pi }}{3}$
• D. $\frac{2\mathrm{\pi }}{3}$

Answer 1

The correct option is D

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

Step 1. Given data

The distance of points from the source is $10m$ and $15m$

The velocity of the wave, $v=300m{s}^{-1}$

The time period of the oscillation, $T=0.05s$

Step 2. Finding the wavelength, $\lambda$

By using the formula,

$Velocity=Frequency\times Wavelength$

$v=f\times \lambda$

Also, $f=\frac{1}{T}$

$\lambda =v\times T$

$\lambda =300\times 0.05$

$\lambda =15m$

Step 3. Finding the Phase difference

Path difference$=15-10$

Path difference$=5m$

By using the formula of the Phase difference

$Phasediffernce=\frac{2\mathrm{\pi }}{\lambda }\times pathdiffrence$

$Phasedifference=\frac{2\mathrm{\pi }}{15}\times 5$

$Phasediffrence=\frac{2\mathrm{\pi }}{3}$

Hence the correct option is $\left(D\right)$.