A pipe of length 20 cm is open at both ends. Which harmonic mode ofthe pipe is resonantly excited by a 1700 Hz source? The speed of sound= `340ms^(-1)`

Options

A. First harmonic
B. Second harmonic
C. Third harmonic
D. Fourth harmonic

Explanation

To solve the problem, we need to determine which harmonic mode of a pipe of length 20 cm (0.2 m) open at both ends is excited by a sound source of frequency 1700 Hz. The speed of sound is given as 340 m/s.

### Step-by-Step Solution:

1. **Understand the Harmonic Modes of a Pipe Open at Both Ends:**
- A pipe open at both ends supports standing waves with displacement nodes at the ends and antinodes in the middle. The fundamental frequency (first harmonic) has one antinode in the middle and two nodes at the ends.

2. **Determine the Wavelength (λ):**
- For a pipe open at both ends, the fundamental frequency corresponds to a wavelength given by:
\[ L = \frac{\lambda}{2} \]
- Rearranging gives:
\[ \lambda = 2L \]
- Substituting the length of the pipe (L = 0.2 m):
\[ \lambda = 2 \times 0.2 = 0.4 \, \text{m} \]

3. **Calculate the Fundamental Frequency (f₁):**
- The speed of sound (v) is related to frequency (f) and wavelength (λ) by:
\[ v = f \cdot \lambda \]
- For the fundamental frequency:
\[ f_1 = \frac{v}{\lambda} = \frac{340 \, \text{m/s}}{0.4 \, \text{m}} = 850 \, \text{Hz} \]

4. **Determine the Harmonic Frequencies:**
- The frequency of the nth harmonic mode is given by:
\[ f_n = n \cdot f_1 \]
- Therefore, substituting for \( f_1 \):
\[ f_n = n \cdot 850 \, \text{Hz} \]

5. **Set Up the Equation for the Given Frequency:**
- We need to find n such that:
\[ f_n = 1700 \, \text{Hz} \]
- Thus:
\[ n \cdot 850 = 1700 \]
- Solving for n:
\[ n = \frac{1700}{850} = 2 \]

6. **Conclusion:**
- The harmonic mode that is resonantly excited by the 1700 Hz source is the second harmonic (n = 2).

### Final Answer:
The second harmonic mode of the pipe is resonantly excited by the 1700 Hz source.