Question

If the lengths of the open and closed pipes are in the ratio of $2:3$, then the ratio of the frequencies of the third harmonic of the open pipe and the fifth harmonic of the closed pipe is

• A. $3:5$
• B. $9:5$
• C. $2:3$
• D. $4:9$

Hint:

Open pipe harmonics: $v/2L, 2v/2L, 3v/2L \dots$ (All integers). Closed pipe harmonics: $v/4L, 3v/4L, 5v/4L \dots$ (Odd integers only).

Answer 1

The Correct Option is B

Step 1: Formulas for Harmonics:
For an Open Pipe of length $L_o$: Fundamental frequency $f_o = \frac{v}{2L_o}$. $n^{th}$ harmonic frequency $= n f_o$. Third harmonic ($n=3$): $f_{open} = 3 \left( \frac{v}{2L_o} \right)$. For a Closed Pipe of length $L_c$: Fundamental frequency $f_c = \frac{v}{4L_c}$. Only odd harmonics exist: $(2n-1)f_c$. Fifth harmonic corresponds to frequency $5 f_c$: $f_{closed} = 5 \left( \frac{v}{4L_c} \right)$.
Step 2: Calculate Ratio:
Given ratio of lengths $\frac{L_o}{L_c} = \frac{2}{3}$. Ratio of frequencies: \[ \frac{f_{open}}{f_{closed}} = \frac{3v / 2L_o}{5v / 4L_c} \] \[ = \frac{3}{2} \times \frac{4}{5} \times \frac{L_c}{L_o} \] \[ = \frac{6}{5} \times \left( \frac{3}{2} \right) \quad \text{(Substituting } \frac{L_c}{L_o} = \frac{3}{2} \text{)} \] \[ = \frac{18}{10} = \frac{9}{5} \] Final Answer:
The ratio is $9:5$.