If the fifth overtone of a closed pipe is in unison with the fifth overtone of an open pipe, then the ratio of the length of closed pipe to open pipe will be

Options

A. `11 : 12`
B. ` 12 : 11`
C. `13 : 11`
D. `11 : 13`

Explanation

To solve the problem, we need to find the ratio of the lengths of a closed pipe to an open pipe when their fifth overtones are in unison. Let's break down the steps:

### Step 1: Understand the overtone frequencies for closed and open pipes
- The frequency of the nth overtone in a closed pipe is given by the formula:
\[ F_{n,c} = \frac{(2n + 1)v}{4L_c} \]
where \( L_c \) is the length of the closed pipe, \( v \) is the speed of sound, and \( n \) is the overtone number.

- The frequency of the nth overtone in an open pipe is given by the formula:
\[ F_{n,o} = \frac{(n + 1)v}{2L_o} \]
where \( L_o \) is the length of the open pipe.

### Step 2: Set the overtone numbers to 5
Since we are looking for the fifth overtone, we set \( n = 5 \):
- For the closed pipe:
\[ F_{5,c} = \frac{(2 \cdot 5 + 1)v}{4L_c} = \frac{11v}{4L_c} \]

- For the open pipe:
\[ F_{5,o} = \frac{(5 + 1)v}{2L_o} = \frac{6v}{2L_o} = \frac{3v}{L_o} \]

### Step 3: Set the frequencies equal
Since the fifth overtone of the closed pipe is in unison with the fifth overtone of the open pipe, we equate the two frequencies:
\[ \frac{11v}{4L_c} = \frac{3v}{L_o} \]

### Step 4: Cancel out the common term
We can cancel \( v \) from both sides:
\[ \frac{11}{4L_c} = \frac{3}{L_o} \]

### Step 5: Rearrange to find the ratio of lengths
Cross-multiplying gives:
\[ 11L_o = 12L_c \]
Rearranging this gives:
\[ \frac{L_c}{L_o} = \frac{11}{12} \]

### Conclusion
Thus, the ratio of the length of the closed pipe to the length of the open pipe is:
\[ \frac{L_c}{L_o} = \frac{11}{12} \]