Question

The ratio between the wavelengths of the air column vibrating in the first two modes in an open organ pipe is

• A. \(2:1\)
• B. \(1:2\)
• C. \(1:1\)
• D. \(1:3\)
• E. \(6:1\)

Hint:

For an open organ pipe, the wavelength formula is \(\lambda_n=\frac{2L}{n}\). The first mode gives \(2L\), the second gives \(L\).

Answer 1

The Correct Option is A

Step 1: Recall the condition for an open organ pipe.
In an open organ pipe, both ends are open, so both ends are displacement antinodes. The pipe can vibrate in all harmonics.

Step 2: Write the wavelength formula for harmonics in an open pipe.
For an open organ pipe of length \(L\), the allowed wavelengths are:
\[ \lambda_n=\frac{2L}{n}, \quad n=1,2,3,\dots \]

Step 3: Find the wavelength in the first mode.
The first mode is the fundamental mode, corresponding to \(n=1\). Thus,
\[ \lambda_1=2L \]

Step 4: Find the wavelength in the second mode.
The second mode corresponds to the second harmonic, i.e. \(n=2\). Therefore,
\[ \lambda_2=\frac{2L}{2}=L \]

Step 5: Form the required ratio.
Now, \[ \lambda_1:\lambda_2=2L:L \]

Step 6: Simplify the ratio.
Dividing both terms by \(L\), we get:
\[ \lambda_1:\lambda_2=2:1 \]

Step 7: State the final answer.
Hence, the ratio between the wavelengths in the first two modes of an open organ pipe is:
\[ \boxed{2:1} \] which matches option \((1)\).