The third overtone of an open organ pipe of length `l_(0)` has the same frequency as the third overtone of a closed pipe of length `l_(c)`. The ratio `l_(0)//l_(c)` is equal to

Options

A. 2
B. `3//2`
C. `5//3`
D. `8//7`

Answer

To solve the problem, we need to find the ratio \( \frac{l_0}{l_c} \) where \( l_0 \) is the length of the open organ pipe and \( l_c \) is the length of the closed organ pipe. We know that the third overtone of both pipes has the same frequency.

### Step-by-Step Solution:

1. **Identify the Frequency Formula for Open Organ Pipe**:
The frequency of the \( n \)-th overtone in an open organ pipe is given by:
\[ f_{open} = \frac{n \cdot v}{2L_0} \]
For the third overtone, \( n = 4 \):
\[ f_{open} = \frac{4v}{2L_0} = \frac{2v}{L_0} \]

2. **Identify the Frequency Formula for Closed Organ Pipe**:
The frequency of the \( n \)-th overtone in a closed organ pipe is given by:
\[ f_{closed} = \frac{n \cdot v}{4L_c} \]
For the third overtone, \( n = 7 \):
\[ f_{closed} = \frac{7v}{4L_c} \]

3. **Set the Frequencies Equal**:
Since the frequencies of the third overtone of both pipes are equal, we can set the two equations equal to each other:
\[ \frac{2v}{L_0} = \frac{7v}{4L_c} \]

4. **Cancel \( v \) from Both Sides**:
We can cancel \( v \) from both sides of the equation:
\[ \frac{2}{L_0} = \frac{7}{4L_c} \]

5. **Cross-Multiply**:
Cross-multiplying gives us:
\[ 2 \cdot 4L_c = 7 \cdot L_0 \]
Simplifying this, we have:
\[ 8L_c = 7L_0 \]

6. **Rearranging for the Ratio**:
To find the ratio \( \frac{L_0}{L_c} \):
\[ \frac{L_0}{L_c} = \frac{8}{7} \]

### Final Answer:
The ratio \( \frac{l_0}{l_c} \) is:
\[ \frac{l_0}{l_c} = \frac{8}{7} \]