A pipe of length 20 cm is closed at one end.Which harmonic mode ofthe pipe is resonantly excited by a 425 Hz source ? The speed of sound = `340m//s^(-1)`

Options

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

Explanation

To solve the problem, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Pipe Configuration**:
- The pipe is closed at one end, which means that at the closed end, there is a node (point of no displacement) and at the open end, there is an antinode (point of maximum displacement).

2. **Determine the Fundamental Frequency**:
- For a pipe closed at one end, the fundamental frequency corresponds to the first harmonic, where the length of the pipe (L) is equal to one-fourth of the wavelength (λ).
- Thus, we can express this relationship as:
\[ L = \frac{\lambda}{4} \]

3. **Calculate Wavelength**:
- Rearranging the equation gives us:
\[ \lambda = 4L \]
- Given the length of the pipe, \( L = 20 \, \text{cm} = 0.2 \, \text{m} \), we can find the wavelength:
\[ \lambda = 4 \times 0.2 \, \text{m} = 0.8 \, \text{m} \]

4. **Use the Wave Speed Formula**:
- The speed of sound (v) is related to frequency (f) and wavelength (λ) by the equation:
\[ v = f \cdot \lambda \]
- Rearranging this gives us:
\[ f = \frac{v}{\lambda} \]

5. **Substituting Known Values**:
- We know the speed of sound \( v = 340 \, \text{m/s} \) and the wavelength \( \lambda = 0.8 \, \text{m} \). Substituting these values into the equation gives:
\[ f = \frac{340 \, \text{m/s}}{0.8 \, \text{m}} = 425 \, \text{Hz} \]

6. **Identify the Harmonic Mode**:
- The frequency we calculated matches the given frequency of the source, which is 425 Hz. Since this frequency corresponds to the fundamental frequency (first harmonic) of the pipe, we conclude that the harmonic mode resonantly excited by the 425 Hz source is the first harmonic.

### Final Answer:
The harmonic mode of the pipe that is resonantly excited by a 425 Hz source is the **first harmonic**.

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