A closed organ pipe and an open organ pipe are tuned to the same fundamental frequency. The ratio of their lengths is

Options

A. `1:2`
B. `2:1`
C. `2:3`
D. `4:3`

Answer

To solve the problem of finding the ratio of the lengths of a closed organ pipe and an open organ pipe that are tuned to the same fundamental frequency, we can follow these steps:

### Step 1: Write the formula for the fundamental frequency of an open organ pipe.
The fundamental frequency \( f_0 \) of an open organ pipe is given by:
\[ f_0 = \frac{V}{2L_O} \]
where \( V \) is the speed of sound in air and \( L_O \) is the length of the open organ pipe.

### Step 2: Write the formula for the fundamental frequency of a closed organ pipe.
The fundamental frequency \( f_0 \) of a closed organ pipe is given by:
\[ f_0 = \frac{V}{4L_C} \]
where \( L_C \) is the length of the closed organ pipe.

### Step 3: Set the frequencies equal to each other.
Since both pipes are tuned to the same fundamental frequency, we can set the two equations equal to each other:
\[ \frac{V}{2L_O} = \frac{V}{4L_C} \]

### Step 4: Simplify the equation.
We can cancel \( V \) from both sides (assuming \( V \neq 0 \)):
\[ \frac{1}{2L_O} = \frac{1}{4L_C} \]

### Step 5: Cross-multiply to find the relationship between \( L_O \) and \( L_C \).
Cross-multiplying gives:
\[ 4L_C = 2L_O \]

### Step 6: Rearrange to find the ratio of the lengths.
Rearranging the equation gives:
\[ \frac{L_C}{L_O} = \frac{2}{4} = \frac{1}{2} \]

### Conclusion
The ratio of the length of the closed organ pipe to the length of the open organ pipe is:
\[ \frac{L_C}{L_O} = \frac{1}{2} \]

Thus, the ratio of their lengths is \( 1:2 \) (closed to open).