Question

Fifth overtone of closed organ pipe is in unison with fifth overtone of open organ pipe The ratio of their lengths is

• A. 5/6
• B. 11/12
• C. 6/5
• D. 12/11

Answer 1

Answer: (B)

11/12

Answer 2

For a closed organ pipe, the frequency is given by

$$f = \frac{n v}{4L_{\text{closed}}} \quad \text{with } n = 1, 3, 5, \dots$$

Since the fundamental is the 0th overtone, the 5th overtone corresponds to the 11th harmonic (because $2 \times 5 + 1 = 11$). Thus,

$$f_{\text{closed}} = \frac{11v}{4L_{\text{closed}}}.$$

For an open organ pipe, the frequency is

$$f = \frac{n v}{2L_{\text{open}}} \quad \text{with } n = 1, 2, 3, \dots$$

Here, the 5th overtone means the 6th harmonic (fundamental is 0th overtone), so

$$f_{\text{open}} = \frac{6v}{2L_{\text{open}}} = \frac{3v}{L_{\text{open}}}.$$

Equate the two frequencies (since they are in unison):

$$\frac{11v}{4L_{\text{closed}}} = \frac{3v}{L_{\text{open}}}.$$

Cancel $v$ and solve:

$$\frac{11}{4L_{\text{closed}}} = \frac{3}{L_{\text{open}}} \quad \Longrightarrow \quad 11L_{\text{open}} = 12L_{\text{closed}}.$$

Thus, the ratio of their lengths is:

$$\frac{L_{\text{closed}}}{L_{\text{open}}} = \frac{11}{12}.$$