Question:
A pipe of length 80 cm is open at both ends. A second pipe, closed at one end, has the same fundamental frequency as the open pipe. The length of the second pipe is:
A pipe of length 80 cm is open at both ends. A second pipe, closed at one end, has the same fundamental frequency as the open pipe. The length of the second pipe is:
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Open pipe fundamental: \(v/2L\), closed pipe fundamental: \(v/4L\).
Updated On: Jun 19, 2026
- 80 cm
- 160 cm
- 40 cm
- 120 cm
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The Correct Option is C
Solution and Explanation
Step 1: Fundamental frequency of open pipe.
For open pipe: \[ f_1 = \frac{v}{2L_1} \] Given \(L_1 = 80\,cm\): \[ f_1 = \frac{v}{160} \]
Step 2: Fundamental frequency of closed pipe.
For closed pipe: \[ f_2 = \frac{v}{4L_2} \]
Step 3: Equate frequencies.
\[ \frac{v}{160} = \frac{v}{4L_2} \]
Step 4: Solve for \(L_2\).
\[ 4L_2 = 160 \Rightarrow L_2 = 40\,cm \]
Step 5: Physical interpretation.
Closed pipe has only odd harmonics, so its fundamental is half that of open pipe for same length scale.
Step 6: Final conclusion.
Thus, required length is \(40\,cm\).
Final Answer: \[ \boxed{40\,cm} \]
For open pipe: \[ f_1 = \frac{v}{2L_1} \] Given \(L_1 = 80\,cm\): \[ f_1 = \frac{v}{160} \]
Step 2: Fundamental frequency of closed pipe.
For closed pipe: \[ f_2 = \frac{v}{4L_2} \]
Step 3: Equate frequencies.
\[ \frac{v}{160} = \frac{v}{4L_2} \]
Step 4: Solve for \(L_2\).
\[ 4L_2 = 160 \Rightarrow L_2 = 40\,cm \]
Step 5: Physical interpretation.
Closed pipe has only odd harmonics, so its fundamental is half that of open pipe for same length scale.
Step 6: Final conclusion.
Thus, required length is \(40\,cm\).
Final Answer: \[ \boxed{40\,cm} \]
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