Consider the propagating sound (with velocity `330 ms^(-1)`) in a pipe of length 1.5 m with one end closed and the other open. The frequency associated with the fundamental mode is

Options

A. 11 Hz
B. 55 Hz
C. 110 Hz
D. 165 Hz

Answer

To solve the problem, we need to find the frequency associated with the fundamental mode of a pipe that is closed at one end and open at the other.

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a pipe of length \( L = 1.5 \, \text{m} \).
- The speed of sound in the pipe is \( v = 330 \, \text{m/s} \).
- The pipe is closed at one end and open at the other.

2. **Determine the Wavelength for the Fundamental Mode**:
- For a pipe closed at one end and open at the other, the fundamental mode (first harmonic) has a node at the closed end and an antinode at the open end.
- The length of the pipe corresponds to a quarter of the wavelength (\( \lambda \)) of the sound wave:
\[ L = \frac{\lambda}{4} \]
- Rearranging this gives us:
\[ \lambda = 4L \]

3. **Calculate the Wavelength**:
- Substitute \( L = 1.5 \, \text{m} \) into the equation:
\[ \lambda = 4 \times 1.5 \, \text{m} = 6 \, \text{m} \]

4. **Use the Wave Equation to Find Frequency**:
- The frequency \( f \) can be calculated using the formula:
\[ f = \frac{v}{\lambda} \]
- Substitute \( v = 330 \, \text{m/s} \) and \( \lambda = 6 \, \text{m} \):
\[ f = \frac{330 \, \text{m/s}}{6 \, \text{m}} = 55 \, \text{Hz} \]

5. **Final Answer**:
- The frequency associated with the fundamental mode is \( 55 \, \text{Hz} \).