MHT CET · Physics · Waves and Sound
Stationary wave is produced along the stretched string of length 80 cm . The resonant frequencies of string are \(90 \mathrm{~Hz}, 150 \mathrm{~Hz}\) and 210 Hz . The speed of transverse wave in the string is
- A \(45 \mathrm{~m} / \mathrm{s}\)
- B \(75 \mathrm{~m} / \mathrm{s}\)
- C \(48 \mathrm{~m} / \mathrm{s}\)
- D \(80 \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(C) \(48 \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{f}_1=90 \mathrm{~Hz}, \mathrm{f}_2=150 \mathrm{~Hz}, \mathrm{f}_3=210 \mathrm{~Hz}\)
Fundamental frequency, \(\mathrm{f}=30 \mathrm{~Hz}\)
\(\mathrm{f}_1=3 \mathrm{f}\)
Length of string, \(L=\frac{n \lambda}{2}\)
\(\lambda=\frac{2 \mathrm{~L}}{\mathrm{n}}=\frac{2 \times 80}{3 \times 100}=\frac{160}{300} \mathrm{~m}\)
\(\begin{aligned} & \text { Speed, } v=f \lambda \\ & v=90 \times \frac{160}{300}=48 \mathrm{~m} / \mathrm{s}\end{aligned}\)
Fundamental frequency, \(\mathrm{f}=30 \mathrm{~Hz}\)
\(\mathrm{f}_1=3 \mathrm{f}\)
Length of string, \(L=\frac{n \lambda}{2}\)
\(\lambda=\frac{2 \mathrm{~L}}{\mathrm{n}}=\frac{2 \times 80}{3 \times 100}=\frac{160}{300} \mathrm{~m}\)
\(\begin{aligned} & \text { Speed, } v=f \lambda \\ & v=90 \times \frac{160}{300}=48 \mathrm{~m} / \mathrm{s}\end{aligned}\)
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