Question

An open pipe is suddenly closed at one end. The frequency of the third overtone of the closed pipe differs by $f_D$ from the frequency of the third overtone of the open pipe. Find $n$ if the fundamental frequency of the open pipe is $(n f_D / 2)$.

Answer 1

4

Answer 2

Explanation:
For an open pipe, the harmonics are given by

$$f_{\text{open}, k} = k\frac{v}{2L}$$

where the third overtone is the 4th harmonic:

$$f_{\text{open}, 4} = 4\frac{v}{2L} = 2\frac{v}{L}.$$

For a pipe closed at one end, only odd harmonics occur. The third overtone is the 7th harmonic:

$$f_{\text{closed}, 7} = 7\frac{v}{4L}.$$

The difference between the frequencies is

$$f_D = f_{\text{open}, 4} - f_{\text{closed}, 7} = 2\frac{v}{L} - 7\frac{v}{4L} = \frac{v}{4L}.$$

The fundamental frequency of the open pipe is given as

$$f_1 = \frac{v}{2L}.$$

According to the problem,

$$\frac{v}{2L} = \frac{n f_D}{2} = \frac{n}{2} \times \frac{v}{4L} = \frac{n v}{8L}.$$

Equate and solve for $n$:

$$\frac{v}{2L} = \frac{n v}{8L} \quad \Rightarrow \quad n = 4.$$

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Answer:
$n = 4$ (Option 4)

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