`n th` harmonic of a closed organ is equal to `m th` harmonic of an pipe . First overtone frequency of the closed organ pipe is also equal to first overtone frequaency of an organ pipe . Find the value of `n`, if `m = 6`.

Explanation

To solve the problem, we need to analyze the relationship between the harmonics of a closed organ pipe and an open organ pipe. Let's go through the solution step by step.

### Step 1: Understand the Frequencies of the Pipes
- The fundamental frequency of a closed organ pipe (closed at one end) is given by:
\[ f_c = \frac{V}{4L_c} \]
where \( V \) is the speed of sound and \( L_c \) is the length of the closed organ pipe.

- The fundamental frequency of an open organ pipe (open at both ends) is given by:
\[ f_o = \frac{V}{2L_o} \]
where \( L_o \) is the length of the open organ pipe.

### Step 2: Harmonics of the Pipes
- For a closed organ pipe, the harmonics are given by:
\[ f_n = n \cdot f_c = n \cdot \frac{V}{4L_c} \]
where \( n \) can take odd values (1, 3, 5, ...).

- For an open organ pipe, the harmonics are given by:
\[ f_m = m \cdot f_o = m \cdot \frac{V}{2L_o} \]
where \( m \) can take any positive integer value (1, 2, 3, ...).

### Step 3: Setting Up the Equation
According to the problem, the \( n \)-th harmonic of the closed organ pipe is equal to the \( m \)-th harmonic of the open organ pipe:
\[ f_n = f_m \]
Substituting the expressions for \( f_n \) and \( f_m \):
\[ n \cdot \frac{V}{4L_c} = m \cdot \frac{V}{2L_o} \]
We can cancel \( V \) from both sides (assuming \( V \neq 0 \)):
\[ \frac{n}{4L_c} = \frac{m}{2L_o} \]

### Step 4: Rearranging the Equation
Rearranging the equation gives:
\[ \frac{n}{m} = \frac{4L_c}{2L_o} = \frac{2L_c}{L_o} \]

### Step 5: Using the First Overtone Condition
The first overtone frequency of a closed organ pipe is:
\[ f_{1c} = 3 \cdot f_c = 3 \cdot \frac{V}{4L_c} \]
The first overtone frequency of an open organ pipe is:
\[ f_{1o} = 2 \cdot f_o = 2 \cdot \frac{V}{2L_o} = \frac{V}{L_o} \]
Setting these equal gives:
\[ 3 \cdot \frac{V}{4L_c} = \frac{V}{L_o} \]
Cancelling \( V \):
\[ \frac{3}{4L_c} = \frac{1}{L_o} \]
Rearranging gives:
\[ L_o = \frac{4}{3}L_c \]

### Step 6: Substituting Back
Now substituting \( L_o \) into the equation from Step 4:
\[ \frac{n}{m} = \frac{2L_c}{\frac{4}{3}L_c} = \frac{2 \cdot 3}{4} = \frac{6}{4} = \frac{3}{2} \]
Thus, we have:
\[ n = \frac{3}{2}m \]

### Step 7: Finding \( n \) with \( m = 6 \)
Substituting \( m = 6 \):
\[ n = \frac{3}{2} \cdot 6 = 9 \]

### Final Answer
The value of \( n \) is:
\[ \boxed{9} \]