Two consecutive harmonics of an air column in a pipe closed at one end are of frequencies 150 Hz and 250 Hz.
What is the fundamental frequency ?

Options

A. 50 Hz
B. 100 Hz
C. 75 Hz
D. 25 Hz

Explanation

To find the fundamental frequency of a pipe closed at one end, we can use the properties of harmonics in such a pipe. In a closed pipe, the harmonics are odd multiples of the fundamental frequency.

### Step-by-Step Solution:

1. **Identify the Harmonics**:
- Given frequencies are 150 Hz and 250 Hz. Let's denote these frequencies as \( f_1 = 150 \, \text{Hz} \) and \( f_2 = 250 \, \text{Hz} \).
- Since these are consecutive harmonics, we can denote them as \( f_n \) and \( f_{n+1} \).

2. **Determine the Relationship**:
- For a closed pipe, the harmonics are given by the formula:
\[ f_n = n \cdot f_1 \]
where \( n \) is an odd integer (1, 3, 5, ...).
- Thus, we can express the two frequencies as:
\[ f_n = 150 \, \text{Hz} \quad (n = 3) \quad \text{and} \quad f_{n+1} = 250 \, \text{Hz} \quad (n + 1 = 4) \]

3. **Calculate the Fundamental Frequency**:
- The fundamental frequency \( f_1 \) corresponds to \( n = 1 \) in the harmonic series.
- The relationship between the harmonics can be expressed as:
\[ f_n = 3 \cdot f_1 \quad \text{and} \quad f_{n+1} = 4 \cdot f_1 \]
- We can set up the equations:
\[ 3 \cdot f_1 = 150 \quad \Rightarrow \quad f_1 = \frac{150}{3} = 50 \, \text{Hz} \]
\[ 4 \cdot f_1 = 250 \quad \Rightarrow \quad f_1 = \frac{250}{4} = 62.5 \, \text{Hz} \]

4. **Verification**:
- Since \( f_1 \) must be consistent, we check both calculations. The fundamental frequency derived from the first harmonic (150 Hz) gives \( f_1 = 50 \, \text{Hz} \), while the second harmonic (250 Hz) gives \( f_1 = 62.5 \, \text{Hz} \).
- The correct fundamental frequency must be the one that fits both harmonics, which is \( 50 \, \text{Hz} \).

5. **Conclusion**:
- Therefore, the fundamental frequency of the air column in the pipe closed at one end is:
\[ \boxed{50 \, \text{Hz}} \]