When open pipe is closed from one end, then third overtone of closed pipe is higher in frequency by 150 Hz than second overtone of open pipe. The fundamental frequency of open end pipe will be

Options

A. 75 Hz
B. 150 Hz
C. 225 Hz
D. 300 Hz

Answer

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the Problem
We have a closed pipe and an open pipe. The problem states that the third overtone of the closed pipe is 150 Hz higher than the second overtone of the open pipe. We need to find the fundamental frequency of the open pipe.

### Step 2: Identify the Frequencies
1. **Closed Pipe Frequencies**: The frequency of the nth harmonic for a closed pipe is given by:
\[ f_n = \frac{(2n-1)V}{4L} \]
where \( n \) is the harmonic number, \( V \) is the speed of sound, and \( L \) is the length of the pipe.

2. **Open Pipe Frequencies**: The frequency of the nth harmonic for an open pipe is given by:
\[ f_n = \frac{nV}{2L} \]

### Step 3: Determine the Harmonics
- The **third overtone** of the closed pipe corresponds to the **fourth harmonic** (\( n=4 \)).
- The **second overtone** of the open pipe corresponds to the **third harmonic** (\( n=3 \)).

### Step 4: Write the Frequency Equations
1. For the closed pipe (fourth harmonic):
\[ f_4 = \frac{(2 \cdot 4 - 1)V}{4L} = \frac{7V}{4L} \]

2. For the open pipe (third harmonic):
\[ f_3 = \frac{3V}{2L} \]

### Step 5: Set Up the Equation
According to the problem, the frequency of the fourth harmonic of the closed pipe is 150 Hz higher than the frequency of the third harmonic of the open pipe:
\[ f_4 = f_3 + 150 \]
Substituting the expressions we derived:
\[ \frac{7V}{4L} = \frac{3V}{2L} + 150 \]

### Step 6: Solve for \( \frac{V}{L} \)
Rearranging the equation:
\[ \frac{7V}{4L} - \frac{3V}{2L} = 150 \]
Finding a common denominator (which is \( 4L \)):
\[ \frac{7V}{4L} - \frac{6V}{4L} = 150 \]
This simplifies to:
\[ \frac{V}{4L} = 150 \]

### Step 7: Find \( \frac{V}{2L} \)
To find the fundamental frequency of the open pipe, we need \( \frac{V}{2L} \):
\[ \frac{V}{2L} = 2 \cdot \frac{V}{4L} = 2 \cdot 150 = 300 \text{ Hz} \]

### Step 8: Conclusion
The fundamental frequency of the open pipe is:
\[ \boxed{300 \text{ Hz}} \]