MHT CET · Physics · Waves and Sound
A closed pipe and an open pipe have their first overtone equal in frequency. Then the lengths of these pipes are in the ratio
- A \(1: 2\)
- B \(2: 3\)
- C \(3: 4\)
- D \(4: 5\)
Answer & Solution
Correct Answer
(C) \(3: 4\)
Step-by-step Solution
Detailed explanation
The first overtone for open pipe is
\(\mathrm{f}_{\mathrm{o}}=\frac{2 \mathrm{v}}{2 l_{\mathrm{o}}}\)
The first overtone for closed pipe is
\(\mathrm{f}_{\mathrm{c}}=\frac{3 \mathrm{v}}{4 l_{\mathrm{c}}}\)
Equating the frequencies,
\(\frac{2 \mathrm{v}}{2 l_{\mathrm{o}}}=\frac{3 \mathrm{v}}{4 l_{\mathrm{c}}}\)
\(\frac{l_{\mathrm{c}}}{l_{\mathrm{o}}}=\frac{3}{4}\)
\(\mathrm{f}_{\mathrm{o}}=\frac{2 \mathrm{v}}{2 l_{\mathrm{o}}}\)
The first overtone for closed pipe is
\(\mathrm{f}_{\mathrm{c}}=\frac{3 \mathrm{v}}{4 l_{\mathrm{c}}}\)
Equating the frequencies,
\(\frac{2 \mathrm{v}}{2 l_{\mathrm{o}}}=\frac{3 \mathrm{v}}{4 l_{\mathrm{c}}}\)
\(\frac{l_{\mathrm{c}}}{l_{\mathrm{o}}}=\frac{3}{4}\)
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