An open organ pipe and closed pipe have same length. The ratio of frequencies of their `n^(th)` over tone is

Options

A. `(n+1)/(2n + 1)`
B. `(2(n+1))/(2n + 1)`
C. `(n)/(2n+ 1)`
D. `(n+1)/(2n)`

Answer

To find the ratio of frequencies of the nth overtone of an open organ pipe and a closed organ pipe of the same length, we can follow these steps:

### Step 1: Understand the Frequencies of the Pipes
1. **Open Organ Pipe Frequency**: The frequency of the nth overtone for an open organ pipe is given by the formula:
\[ f_1 = \frac{(n + 1) V}{2L} \]
where:
- \( f_1 \) is the frequency of the open pipe,
- \( n \) is the overtone number,
- \( V \) is the speed of sound,
- \( L \) is the length of the pipe.

2. **Closed Organ Pipe Frequency**: The frequency of the nth overtone for a closed organ pipe is given by the formula:
\[ f_2 = \frac{(2n + 1) V}{4L} \]
where:
- \( f_2 \) is the frequency of the closed pipe.

### Step 2: Set Up the Ratio of Frequencies
To find the ratio of the frequencies \( f_1 \) and \( f_2 \):
\[ \text{Ratio} = \frac{f_1}{f_2} = \frac{\frac{(n + 1) V}{2L}}{\frac{(2n + 1) V}{4L}} \]

### Step 3: Simplify the Ratio
1. Cancel out common terms in the ratio:
\[ \text{Ratio} = \frac{(n + 1) V}{2L} \cdot \frac{4L}{(2n + 1) V} \]
- The \( V \) and \( L \) terms cancel out:
\[ \text{Ratio} = \frac{(n + 1) \cdot 4}{2n + 1} \cdot \frac{1}{2} \]

2. Simplify further:
\[ \text{Ratio} = \frac{4(n + 1)}{2(2n + 1)} = \frac{2(n + 1)}{2n + 1} \]

### Step 4: Final Result
Thus, the ratio of the frequencies of the nth overtone of the open organ pipe to the closed organ pipe is:
\[ \text{Ratio} = \frac{2(n + 1)}{2n + 1} \]