An organ pipe of length `L` is open at one end and closed at other end. The wavelengths of the three lowest resonating frequencies that can be produced by this pipe are

Options

A. (a)4L,2L,L
B. (b)2L,L,L/2
C. (c)2L,L,2L/3
D. (d)4L,4L/3,4L/5

Answer

To find the wavelengths of the three lowest resonating frequencies that can be produced by an organ pipe that is open at one end and closed at the other, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Pipe Configuration**:
- An organ pipe that is open at one end and closed at the other end supports standing waves. The closed end will have a displacement node (minimum displacement), while the open end will have a displacement antinode (maximum displacement).

2. **Relating Length to Wavelength**:
- For a pipe closed at one end, the length \( L \) of the pipe is related to the wavelength \( \lambda \) of the sound waves by the formula:
\[ L = \frac{(2n - 1) \lambda}{4} \]
where \( n \) is the harmonic number (1 for the fundamental frequency, 2 for the first overtone, etc.).

3. **Finding the Wavelengths**:
- Rearranging the formula gives:
\[ \lambda = \frac{4L}{2n - 1} \]
- Now, we can calculate the wavelengths for the first three harmonics (n = 1, 2, 3).

4. **Calculating the First Wavelength** (n = 1):
- For \( n = 1 \):
\[ \lambda_1 = \frac{4L}{2(1) - 1} = \frac{4L}{1} = 4L \]

5. **Calculating the Second Wavelength** (n = 2):
- For \( n = 2 \):
\[ \lambda_2 = \frac{4L}{2(2) - 1} = \frac{4L}{3} \]

6. **Calculating the Third Wavelength** (n = 3):
- For \( n = 3 \):
\[ \lambda_3 = \frac{4L}{2(3) - 1} = \frac{4L}{5} \]

7. **Final Result**:
- The wavelengths of the three lowest resonating frequencies are:
\[ \lambda_1 = 4L, \quad \lambda_2 = \frac{4L}{3}, \quad \lambda_3 = \frac{4L}{5} \]

### Summary of Wavelengths:
- \( \lambda_1 = 4L \)
- \( \lambda_2 = \frac{4L}{3} \)
- \( \lambda_3 = \frac{4L}{5} \)