MHT CET · Physics · Waves and Sound
A sound wave is travelling with a frequency of \(50 \mathrm{~Hz}\). The phase difference between the two points in the path of a wave is \(\frac{\pi}{3}\). The distance between those two points is (Velocity of sound in air \(=330 \mathrm{~m} / \mathrm{s})\)
- A 1.1 m
- B 0.6 m
- C 2.2 m
- D 1.7 m
Answer & Solution
Correct Answer
(A) 1.1 m
Step-by-step Solution
Detailed explanation
Phase difference \(\phi=\frac{2 \pi}{\lambda} \mathrm{x}\)
\(
\begin{aligned}
& \lambda=\frac{\mathrm{v}}{\mathrm{f}}=\frac{330}{50}=6.6 \mathrm{~m} \\
& \therefore \mathrm{x}=\frac{\lambda \phi}{2 \pi}=\frac{6.6}{2 \pi} \times \frac{\pi}{3}=\frac{6.6}{6}=1.1 \mathrm{~m}
\end{aligned}
\)
\(
\begin{aligned}
& \lambda=\frac{\mathrm{v}}{\mathrm{f}}=\frac{330}{50}=6.6 \mathrm{~m} \\
& \therefore \mathrm{x}=\frac{\lambda \phi}{2 \pi}=\frac{6.6}{2 \pi} \times \frac{\pi}{3}=\frac{6.6}{6}=1.1 \mathrm{~m}
\end{aligned}
\)
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