An open organ pipe is closed suddenly with the result that the second overtone of the closed pipe is found to be higher in frequency by 100 than the first overtone of the original pipe. Then, the fundamental frequency of the open pipe is

Options

A. `200 s^-1`
B. `100 s^(-1)`
C. `250 s^(-1)`
D. `150 s^(-1)`

Answer

To solve the problem, we need to analyze the relationship between the frequencies of the open and closed organ pipes. Let's break it down step by step.

### Step 1: Understand the Frequencies of Open and Closed Pipes
- The fundamental frequency \( f_0 \) of an open pipe is given by:
\[ f_0 = \frac{V}{2L} \]
where \( V \) is the speed of sound in air and \( L \) is the length of the pipe.

- The first overtone (which is the second harmonic) of an open pipe is:
\[ f_1 = 2f_0 = \frac{V}{L} \]

- For a closed pipe, the fundamental frequency \( f_c \) is:
\[ f_c = \frac{V}{4L} \]

- The first overtone (which is the third harmonic) of a closed pipe is:
\[ f_{c1} = 3f_c = \frac{3V}{4L} \]

- The second overtone (which is the fifth harmonic) of a closed pipe is:
\[ f_{c2} = 5f_c = \frac{5V}{4L} \]

### Step 2: Set Up the Equation Based on the Given Information
According to the problem, the second overtone of the closed pipe is higher in frequency by 100 Hz than the first overtone of the open pipe. This can be expressed as:
\[ f_{c2} = f_1 + 100 \]

Substituting the expressions for \( f_{c2} \) and \( f_1 \):
\[ \frac{5V}{4L} = \frac{V}{L} + 100 \]

### Step 3: Solve the Equation
To solve for \( V \) in terms of \( L \):
1. Multiply through by \( 4L \) to eliminate the denominators:
\[ 5V = 4V + 400L \]

2. Rearranging gives:
\[ 5V - 4V = 400L \]
\[ V = 400L \]

### Step 4: Find the Fundamental Frequency of the Open Pipe
Now that we have \( V \) in terms of \( L \), we can find the fundamental frequency of the open pipe:
\[ f_0 = \frac{V}{2L} = \frac{400L}{2L} = 200 \text{ Hz} \]

### Conclusion
Thus, the fundamental frequency of the open pipe is:
\[ \boxed{200 \text{ Hz}} \]