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

Question

Aicharya · Accepted Answer

Key Idea : Phase difference
$=\frac{2 \pi}{\lambda} \times \text { path difference }$
Path difference between two points,
$\Delta x=15-10=5 \mathrm{~m}$
Time period, $T=0.05 \mathrm{~s}$
$\Rightarrow$ frequency $v=\frac{1}{T}=\frac{1}{0.05}=20 \mathrm{~Hz}$
Velocity, $y=300 \mathrm{~m} / \mathrm{s}$
$\therefore$ Wavelength, $\lambda=\frac{v}{v}=\frac{300}{20}=15 \mathrm{~m}$
Hence, phase difference
$\begin{aligned}
\Delta \phi & =\frac{2 \pi}{\lambda} \times \Delta x \\
& =\frac{2 \pi}{15} \times 5=\frac{2 \pi}{3}
\end{aligned}$ :

Two points are located at a distance of $10 \mathrm{~m}$ and $15 \mathrm{~m}$ from the source of oscillation…

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