MHT CET · Physics · Waves and Sound
A string fixed at both the ends forms standing wave with node separation of \(5 \mathrm{~cm}\). If the velocity of the wave on the string is \(2 \mathrm{~m} / \mathrm{s}\), then the frequency of vibration of the string is
- A \(0.2 \mathrm{~Hz}\)
- B \(10 \mathrm{~Hz}\)
- C \(20 \mathrm{~Hz}\)
- D \(40 \mathrm{~Hz}\)
Answer & Solution
Correct Answer
(C) \(20 \mathrm{~Hz}\)
Step-by-step Solution
Detailed explanation
Separation between consecutive nodes, \(\frac{\lambda}{2}=5 \mathrm{~cm}\)
\(
\therefore \lambda=10 \mathrm{~cm}=0.1 \mathrm{~m}
\)
The frequency of vibration is given as:
\(
\mathrm{n}=\frac{\mathrm{v}}{\lambda}=\frac{2}{0.1}=20 \mathrm{~Hz}
\)
\(
\therefore \lambda=10 \mathrm{~cm}=0.1 \mathrm{~m}
\)
The frequency of vibration is given as:
\(
\mathrm{n}=\frac{\mathrm{v}}{\lambda}=\frac{2}{0.1}=20 \mathrm{~Hz}
\)
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