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 solve the problem of finding 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 properties of the pipes
- An open organ pipe supports both ends open, while a closed organ pipe has one end closed.
- The fundamental frequency and overtones for each type of pipe are determined by their respective lengths and the speed of sound in air.

### Step 2: Write the formulas for the frequencies
1. **Open Organ Pipe**: The frequency of the nth overtone (f1) is given by:
\[ f_1 = (n + 1) \frac{v}{2l} \]
where \( v \) is the speed of sound, \( l \) is the length of the pipe, and \( n \) is the overtone number.

2. **Closed Organ Pipe**: The frequency of the nth overtone (f2) is given by:
\[ f_2 = (2n + 1) \frac{v}{4l} \]

### Step 3: Set up the ratio of the frequencies
We need to find the ratio \( \frac{f_1}{f_2} \):
\[ \frac{f_1}{f_2} = \frac{(n + 1) \frac{v}{2l}}{(2n + 1) \frac{v}{4l}} \]

### Step 4: Simplify the ratio
- Cancel out \( v \) and \( l \):
\[ \frac{f_1}{f_2} = \frac{(n + 1) \cdot 4}{(2n + 1) \cdot 2} \]
- This simplifies to:
\[ \frac{f_1}{f_2} = \frac{4(n + 1)}{2(2n + 1)} = \frac{2(n + 1)}{2n + 1} \]

### Step 5: Final result
Thus, the ratio of the frequencies of the nth overtone of the open organ pipe to that of the closed organ pipe is:
\[ \frac{f_1}{f_2} = \frac{2(n + 1)}{2n + 1} \]