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 \(1.6 \pi\) : 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 \(\lambda=\frac{\mathrm{V}}{\mathrm{f}}=\frac{330}{\mathrm{f}}\)...(i)
We know,
Phase difference \(\phi=\frac{2 \pi \times x}{\lambda}\)
where x is path difference.
\(\therefore \quad 1.6 \pi=\frac{2 \pi \times\left(40 \times 10^{-2}\right) \times \mathrm{f}}{330}\)
...[From(i)]
\(\mathrm{f}=660 \mathrm{~Hz}\)
We know,
Phase difference \(\phi=\frac{2 \pi \times x}{\lambda}\)
where x is path difference.
\(\therefore \quad 1.6 \pi=\frac{2 \pi \times\left(40 \times 10^{-2}\right) \times \mathrm{f}}{330}\)
...[From(i)]
\(\mathrm{f}=660 \mathrm{~Hz}\)
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