The speed of sound in air is 333 m/s. The fundamental frequency of the open pipe is 333 Hz. The second oevertone of the open organ pipe can be produced with a pipe of length

Options

A. 0.5 m
B. 1.0 m
C. 1.5 m
D. 2.0 m

Explanation

To find the length of the open pipe that produces the second overtone, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between frequency, speed of sound, and length of the pipe**:
- The fundamental frequency (first harmonic) of an open pipe is given by the formula:
\[ f_1 = \frac{v}{2L} \]
- Where:
- \( f_1 \) is the fundamental frequency,
- \( v \) is the speed of sound,
- \( L \) is the length of the pipe.

2. **Identify the frequency of the second overtone**:
- The second overtone corresponds to the third harmonic (n=3) for an open pipe. The frequency for the nth harmonic is given by:
\[ f_n = \frac{nv}{2L} \]
- For the second overtone (third harmonic), we have \( n = 3 \):
\[ f_3 = \frac{3v}{2L} \]

3. **Substitute the known values**:
- Given:
- Speed of sound, \( v = 333 \, \text{m/s} \)
- Fundamental frequency, \( f_1 = 333 \, \text{Hz} \)
- From the fundamental frequency formula, we can find the length \( L \):
\[ 333 = \frac{333}{2L} \]

4. **Rearranging to find \( L \)**:
- Rearranging the equation gives:
\[ 2L \cdot 333 = 333 \]
\[ 2L = 1 \quad \Rightarrow \quad L = \frac{1}{2} = 0.5 \, \text{m} \]

5. **Now, find the length for the second overtone**:
- Using the formula for the second overtone:
\[ f_3 = \frac{3v}{2L} \]
- We know \( f_3 = 333 \, \text{Hz} \):
\[ 333 = \frac{3 \cdot 333}{2L} \]
- Simplifying gives:
\[ 2L = 3 \quad \Rightarrow \quad L = \frac{3}{2} = 1.5 \, \text{m} \]

### Final Answer:
The length of the open pipe that produces the second overtone is **1.5 meters**.

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