An open organ pipe filled with air has a fundamental frequency 500 Hz. The first harmonic of another organ pipe closed at one end and filled with carbon dioxide has the same frequency as that of the first harmonic of the open organ pipe. Calculate the length of each pipe. Assume that the velocity of sound in air and in carbondioxide to be 330 and 264 m/s respectively.

Explanation

To solve the problem, we need to find the lengths of two different organ pipes: one that is open at both ends and another that is closed at one end. We will use the given fundamental frequency and the velocities of sound in air and carbon dioxide to derive the lengths.

### Step 1: Calculate the length of the open organ pipe

1. **Identify the fundamental frequency (f)**: The fundamental frequency of the open organ pipe is given as \( f = 500 \, \text{Hz} \).
2. **Velocity of sound in air (v)**: The velocity of sound in air is given as \( v = 330 \, \text{m/s} \).
3. **Use the formula for the fundamental frequency of an open organ pipe**:
\[ f = \frac{v}{\lambda} \]
where \( \lambda \) is the wavelength. For an open pipe, the relationship between the length (L) and the wavelength is:
\[ L = \frac{\lambda}{2} \]
Thus, we can express \( \lambda \) in terms of \( L \):
\[ \lambda = 2L \]
4. **Substitute \( \lambda \) into the frequency formula**:
\[ f = \frac{v}{2L} \]
5. **Rearranging to find L**:
\[ L = \frac{v}{2f} \]
6. **Substituting the known values**:
\[ L = \frac{330 \, \text{m/s}}{2 \times 500 \, \text{Hz}} = \frac{330}{1000} = 0.33 \, \text{m} = 330 \times 10^{-3} \, \text{m} \]

### Step 2: Calculate the length of the closed organ pipe

1. **Identify the velocity of sound in carbon dioxide (v')**: The velocity of sound in carbon dioxide is given as \( v' = 264 \, \text{m/s} \).
2. **For a closed organ pipe, the relationship between the length (L') and the wavelength is**:
\[ L' = \frac{\lambda'}{4} \]
where \( \lambda' \) is the wavelength for the closed pipe.
3. **Using the frequency formula for the closed pipe**:
\[ f = \frac{v'}{\lambda'} \]
and substituting \( \lambda' = 4L' \):
\[ f = \frac{v'}{4L'} \]
4. **Rearranging to find L'**:
\[ L' = \frac{v'}{4f} \]
5. **Substituting the known values**:
\[ L' = \frac{264 \, \text{m/s}}{4 \times 500 \, \text{Hz}} = \frac{264}{2000} = 0.132 \, \text{m} = 132 \times 10^{-3} \, \text{m} \]

### Final Results
- Length of the open organ pipe: \( L = 330 \times 10^{-3} \, \text{m} \)
- Length of the closed organ pipe: \( L' = 132 \times 10^{-3} \, \text{m} \)