MHT CET · Physics · Waves and Sound
Velocity of sound waves in air is \(330 \mathrm{~m} / \mathrm{s}\). For a particular sound wave in air, path difference of 40 cm is equivalent to phase difference of (1.6) \(\pi\). The frequency of this wave is
- A 165 Hz
- B 150 Hz
- C 660 Hz
- D 330 Hz
Answer & Solution
Correct Answer
(C) 660 Hz
Step-by-step Solution
Detailed explanation
Given: \(\phi=1.6 \pi\) and \(\lambda=\frac{\mathrm{V}}{\mathrm{f}}\)
We know,
Phase difference \(\phi=\frac{2 \pi x}{\lambda}\)
\(\begin{aligned} \therefore \quad & 1.6 \pi=\frac{2 \pi x f}{330} \Rightarrow f=\frac{1.6 \times 330}{2 \times 40 \times 0.01} \\ & f=660 \mathrm{~Hz}\end{aligned}\)
We know,
Phase difference \(\phi=\frac{2 \pi x}{\lambda}\)
\(\begin{aligned} \therefore \quad & 1.6 \pi=\frac{2 \pi x f}{330} \Rightarrow f=\frac{1.6 \times 330}{2 \times 40 \times 0.01} \\ & f=660 \mathrm{~Hz}\end{aligned}\)
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