The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm, the length of the open organ pipe is

Options

A. 12.5cm
B. 8 cm
C. 13.3cm
D. 16cm

Answer

To solve the problem, we need to establish the relationship between the fundamental frequency of an open organ pipe and the third harmonic of a closed organ pipe. Let's break it down step by step.

### Step 1: Understand the Frequencies
1. **Fundamental Frequency of an Open Organ Pipe**: The formula for the fundamental frequency (first harmonic) of an open organ pipe is given by:
\[ f_{open} = \frac{V}{2L} \]
where \( V \) is the speed of sound in air and \( L \) is the length of the open pipe.

2. **Third Harmonic of a Closed Organ Pipe**: The formula for the frequency of the nth harmonic of a closed organ pipe is given by:
\[ f_{closed} = \frac{nV}{4L_1} \]
For the third harmonic (\( n = 3 \)), this becomes:
\[ f_{closed} = \frac{3V}{4L_1} \]
where \( L_1 \) is the length of the closed pipe.

### Step 2: Set the Frequencies Equal
According to the problem, the fundamental frequency of the open pipe is equal to the third harmonic of the closed pipe:
\[ f_{open} = f_{closed} \]
Substituting the formulas from Step 1:
\[ \frac{V}{2L} = \frac{3V}{4L_1} \]

### Step 3: Simplify the Equation
We can cancel \( V \) from both sides (assuming \( V \neq 0 \)):
\[ \frac{1}{2L} = \frac{3}{4L_1} \]

### Step 4: Cross-Multiply
Cross-multiplying gives us:
\[ 4L_1 = 6L \]

### Step 5: Solve for \( L \)
Now, we can express \( L \) in terms of \( L_1 \):
\[ L = \frac{4L_1}{6} = \frac{2L_1}{3} \]

### Step 6: Substitute the Length of the Closed Pipe
We know from the problem that the length of the closed organ pipe \( L_1 = 20 \) cm. Substituting this value into the equation:
\[ L = \frac{2 \times 20 \text{ cm}}{3} = \frac{40 \text{ cm}}{3} \approx 13.33 \text{ cm} \]

### Final Answer
The length of the open organ pipe is approximately \( 13.33 \) cm.