Question

The fundamental frequency of a closed organ pipe of length L is equal to the frequency of the first overtone of an open organ pipe of length L'. The relation between their lengths is:

• A. \( L' = 4L \)
• B. \( L' = 2L \)
• C. \( L = L' \)
• D. \( L' = \frac{L}{2} \)

Hint:

Closed pipe: odd harmonics only Open pipe: all harmonics First overtone of open pipe = second harmonic

Answer 1

The Correct Option is A

Step 1: Understand the two pipes.
A closed organ pipe is closed at one end. An open pipe is open at both ends. They make sound at different sets of frequencies. We compare two of these.

Step 2: Write the closed pipe fundamental.
For a closed pipe of length $L$, the lowest frequency is \[ f_{\text{closed}} = \frac{v}{4L}, \] where $v$ is the speed of sound.

Step 3: Write the open pipe first overtone.
For an open pipe of length $L'$, the fundamental is $\dfrac{v}{2L'}$. The first overtone is the next allowed one, which is twice that: \[ f_{\text{open}} = \frac{2v}{2L'} = \frac{v}{L'}. \]

Step 4: Set them equal.
The problem says these two frequencies match: \[ \frac{v}{4L} = \frac{v}{L'}. \]

Step 5: Solve for the length.
The speed $v$ cancels. Cross multiply to get \[ L' = 4L. \]

Step 6: State the answer.
The open pipe must be four times as long as the closed pipe. \[ \boxed{L' = 4L} \]