On a day when the speed is `345 m//s`, the fundamental frequency of a closed organ pipe is `220 H_(Z)` . (a) How long is this closed pipe? (b) The second overtone of this pipe has the same wavelength as the third harmonic of an open pipe . How long is the open pipe ?

Explanation

To solve the problem step by step, we will first find the length of the closed organ pipe and then the length of the open organ pipe.

### Step 1: Calculate the Length of the Closed Organ Pipe

The fundamental frequency \( F \) of a closed organ pipe is given by the formula:

\[ F = \frac{V}{4L} \]

Where:
- \( F \) is the fundamental frequency (in Hz),
- \( V \) is the speed of sound (in m/s),
- \( L \) is the length of the closed pipe (in meters).

Given:
- \( F = 220 \, \text{Hz} \)
- \( V = 345 \, \text{m/s} \)

Rearranging the formula to solve for \( L \):

\[ L = \frac{V}{4F} \]

Substituting the known values:

\[ L = \frac{345}{4 \times 220} \]

Calculating:

\[ L = \frac{345}{880} \approx 0.392 \, \text{m} \]

### Step 2: Relate the Wavelengths of the Closed and Open Pipes

The second overtone of a closed pipe corresponds to the third harmonic, and its wavelength \( \lambda \) is given by:

\[ \lambda = \frac{4L}{3} \]

For an open pipe, the wavelength of the third harmonic is given by:

\[ \lambda' = \frac{2L'}{3} \]

Since the problem states that the second overtone of the closed pipe has the same wavelength as the third harmonic of the open pipe, we can set the two wavelengths equal:

\[ \frac{4L}{3} = \frac{2L'}{3} \]

Cancelling the \( \frac{1}{3} \) from both sides gives:

\[ 4L = 2L' \]

Rearranging gives:

\[ L' = 2L \]

### Step 3: Calculate the Length of the Open Organ Pipe

Now substituting the value of \( L \) we found earlier:

\[ L' = 2 \times 0.392 \approx 0.784 \, \text{m} \]

### Final Answers

(a) The length of the closed organ pipe is approximately \( 0.392 \, \text{m} \).

(b) The length of the open organ pipe is approximately \( 0.784 \, \text{m} \).

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