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: Understanding the Concept:
An open organ pipe is a pipe that is open at both ends. When an air column in such a pipe vibrates, it forms standing waves. For a pipe open at both ends, antinodes (points of maximum displacement) are formed at the open ends. We need to find the wavelengths of the first two modes of vibration (harmonics).
Step 2: Key Formula or Approach:
For an organ pipe of length L open at both ends, the condition for standing waves is that the length of the pipe must be an integer multiple of half-wavelengths.
\[ L = n \frac{\lambda_n}{2} \] where n = 1, 2, 3, ... is the mode number (or harmonic number).
From this, the wavelength of the n-th mode is:
\[ \lambda_n = \frac{2L}{n} \] Step 3: Detailed Explanation:
We need to find the wavelengths for the first two modes, n=1 and n=2.
First Mode (Fundamental, n=1):
This is the fundamental frequency or the first harmonic. The wavelength, \( \lambda_1 \), is:
\[ \lambda_1 = \frac{2L}{1} = 2L \] This corresponds to a standing wave with an antinode at each end and one node in the middle.
Second Mode (Second Harmonic, n=2):
This is the first overtone or the second harmonic. The wavelength, \( \lambda_2 \), is:
\[ \lambda_2 = \frac{2L}{2} = L \] This corresponds to a standing wave with antinodes at the ends and two nodes in between.
Ratio of Wavelengths:
The question asks for the ratio \( \lambda_1 : \lambda_2 \).
\[ \frac{\lambda_1}{\lambda_2} = \frac{2L}{L} = \frac{2}{1} \] So, the ratio is 2:1.
Step 4: Final Answer:
The ratio between the wavelengths of the first two modes is 2:1. This corresponds to option (A).