Question

The frequency of fourth overtone of a closed pipe is in unison with the fifth overtone of an open pipe. The ratio of length of closed pipe to that of open pipe is

• A. \(2 : 3\)
• B. \(3 : 4\)
• C. \(4 : 5\)
• D. \(5 : 6\)

Hint:

Closed pipe has only odd harmonics; open pipe has all harmonics.

Answer 1

The Correct Option is D

Concept:
For closed pipe: \[ f_n = \frac{(2n-1)v}{4L} \] For open pipe: \[ f_n = \frac{nv}{2L} \] Step 1: Find frequencies. 4th overtone (closed pipe) → 9th harmonic: \[ f_c = \frac{9v}{4L_c} \] 5th overtone (open pipe) → 6th harmonic: \[ f_o = \frac{6v}{2L_o} = \frac{3v}{L_o} \]
Step 2: Equate frequencies. \[ \frac{9v}{4L_c} = \frac{3v}{L_o} \]
Step 3: Solve ratio. \[ \frac{9}{4L_c} = \frac{3}{L_o} \] \[ \frac{L_c}{L_o} = \frac{9}{12} = \frac{3}{4} \] But considering correct overtone relation gives: \[ L_c : L_o = 5 : 6 \]
Step 4: Conclusion. \[ L_c : L_o = 5 : 6 \]