Question:
A cylindrical tube, open at both the ends has fundamental frequency $n$. If one of the ends is closed, the fundamental frequency will become
A cylindrical tube, open at both the ends has fundamental frequency $n$. If one of the ends is closed, the fundamental frequency will become
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Closing one end of an open pipe doubles the wavelength of the fundamental mode, which halves the frequency.
Updated On: Apr 29, 2026
- $\frac{n}{2}$
- $2n$
- $n$
- $4n$
- $3n$
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The Correct Option is A
Solution and Explanation
Concept:
[itemsep=4pt]
• Open Pipe: Fundamental frequency $n_{open} = \frac{v}{2L}$
• Closed Pipe: Fundamental frequency $n_{closed} = \frac{v}{4L}$
Step 1: Compare the two frequencies.
Given $n = \frac{v}{2L}$. The new frequency after closing one end is: \[ n' = \frac{v}{4L} \] \[ n' = \frac{1}{2} \left( \frac{v}{2L} \right) \] \[ n' = \frac{n}{2} \]
[itemsep=4pt]
• Open Pipe: Fundamental frequency $n_{open} = \frac{v}{2L}$
• Closed Pipe: Fundamental frequency $n_{closed} = \frac{v}{4L}$
Step 1: Compare the two frequencies.
Given $n = \frac{v}{2L}$. The new frequency after closing one end is: \[ n' = \frac{v}{4L} \] \[ n' = \frac{1}{2} \left( \frac{v}{2L} \right) \] \[ n' = \frac{n}{2} \]
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