The fundamental frequency of a closed pipe is `220 H_(Z)`.
(a) Find the length of this pipe.
(b) The second overtone of this pipe has the same frequency as the third harmonic of an open pipe. Find the length of this open pipe. Take speed of sound in air `345 m//s`.

Explanation

To solve the problem step by step, we will break it down into two parts as given in the question.

### Part (a): Find the length of the closed pipe

1. **Identify the formula for the fundamental frequency of a closed pipe**:
The fundamental frequency (f) of a closed pipe is given by the formula:
\[ f = \frac{V}{4L} \]
where \( V \) is the speed of sound in air, and \( L \) is the length of the pipe.

2. **Rearrange the formula to find the length (L)**:
Rearranging the formula gives:
\[ L = \frac{V}{4f} \]

3. **Substitute the values into the formula**:
Given:
- Speed of sound \( V = 345 \, \text{m/s} \)
- Fundamental frequency \( f = 220 \, \text{Hz} \)

Substituting these values:
\[ L = \frac{345}{4 \times 220} \]

4. **Calculate the length**:
Performing the calculation:
\[ L = \frac{345}{880} \approx 0.392 \, \text{m} \]

Thus, the length of the closed pipe is approximately **0.392 meters**.

### Part (b): Find the length of the open pipe

1. **Determine the frequency of the second overtone of the closed pipe**:
The second overtone of a closed pipe corresponds to the frequency given by:
\[ f_{2nd \, overtone} = 5 \frac{V}{4L} \]
Since we already know \( L \) from part (a), we can find this frequency:
\[ f_{2nd \, overtone} = 5 \times 220 \, \text{Hz} = 1100 \, \text{Hz} \]

2. **Identify the formula for the frequency of the third harmonic of an open pipe**:
The frequency of the third harmonic (f) of an open pipe is given by:
\[ f = 3 \frac{V}{2L_0} \]
where \( L_0 \) is the length of the open pipe.

3. **Set the two frequencies equal**:
Since the second overtone of the closed pipe has the same frequency as the third harmonic of the open pipe, we can set them equal:
\[ 1100 = 3 \frac{345}{2L_0} \]

4. **Rearrange to find \( L_0 \)**:
Rearranging gives:
\[ L_0 = \frac{3 \times 345}{2 \times 1100} \]

5. **Calculate the length of the open pipe**:
Performing the calculation:
\[ L_0 = \frac{1035}{2200} \approx 0.470 \, \text{m} \]

Thus, the length of the open pipe is approximately **0.470 meters**.

### Summary of Answers:
- Length of the closed pipe: **0.392 m**
- Length of the open pipe: **0.470 m**