A resonance pipe is open at both ends and 30 cm of its length is in resonance with an external frequency 1.1 kHz. If the speed of sound is 330 m/s which harmonic is in resonance

Options

A. first
B. second
C. third
D. fouth

Explanation

To determine which harmonic is in resonance in a resonance pipe that is open at both ends, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Length of the pipe (L) = 30 cm = 0.3 m
- Frequency (f) = 1.1 kHz = 1100 Hz
- Speed of sound (v) = 330 m/s

2. **Understand the Formula for Frequency of an Open Organ Pipe**:
The frequency of an open organ pipe is given by the formula:
\[ f = \frac{n \cdot v}{2L} \]
where:
- \( f \) is the frequency,
- \( n \) is the harmonic number,
- \( v \) is the speed of sound,
- \( L \) is the length of the pipe.

3. **Rearrange the Formula to Solve for n**:
We need to find \( n \). Rearranging the formula gives:
\[ n = \frac{2Lf}{v} \]

4. **Substitute the Known Values**:
Substitute \( L = 0.3 \, \text{m} \), \( f = 1100 \, \text{Hz} \), and \( v = 330 \, \text{m/s} \) into the equation:
\[ n = \frac{2 \cdot 0.3 \cdot 1100}{330} \]

5. **Calculate the Value**:
- Calculate the numerator:
\[ 2 \cdot 0.3 \cdot 1100 = 660 \]
- Now divide by the speed of sound:
\[ n = \frac{660}{330} = 2 \]

6. **Conclusion**:
The harmonic that is in resonance is the **2nd harmonic**.