MHT CET · Physics · Waves and Sound
A sound of frequency \(480 \mathrm{~Hz}\) is emitted from the stringed instrument. The velocity of sound in air is \(320 \mathrm{~m} / \mathrm{s}\). After completing 180 vibrations, the distance covered by a wave is
- A \(60 \mathrm{~m}\)
- B \(90 \mathrm{~m}\)
- C \(120 \mathrm{~m}\)
- D \(180 \mathrm{~m}\)
Answer & Solution
Correct Answer
(C) \(120 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
Given: \(\mathrm{v}=320 \mathrm{~m} / \mathrm{s}, \mathrm{f}=480 \mathrm{~Hz}, \mathrm{~N}=180\)
\(\mathrm{v}=\mathrm{f} \lambda\)
\(\therefore \quad \lambda=\frac{\mathrm{v}}{\mathrm{f}}\)
Substituting the values, we get
\(\lambda=\frac{320}{480}=\frac{2}{3}\)
\(\therefore \quad\) The total distance covered after 180 vibrations is
\(\begin{aligned}
& \mathrm{D}=\mathrm{N} \times \lambda \\
& \mathrm{D}=180 \times \frac{2}{3} \\
& \mathrm{D}=120 \mathrm{~m}
\end{aligned}\)
\(\mathrm{v}=\mathrm{f} \lambda\)
\(\therefore \quad \lambda=\frac{\mathrm{v}}{\mathrm{f}}\)
Substituting the values, we get
\(\lambda=\frac{320}{480}=\frac{2}{3}\)
\(\therefore \quad\) The total distance covered after 180 vibrations is
\(\begin{aligned}
& \mathrm{D}=\mathrm{N} \times \lambda \\
& \mathrm{D}=180 \times \frac{2}{3} \\
& \mathrm{D}=120 \mathrm{~m}
\end{aligned}\)
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