Question

In fundamental mode, the time required for the sound wave to reach up to the closed end of a pipe filled with air is \( t \) second. The frequency of vibration of air column is

• A. \( \frac{1}{3t} \)
• B. \( \frac{1}{t} \)
• C. \( \frac{1}{4t} \)
• D. \( \frac{1}{2t} \)

Hint:

In closed-end pipes, the frequency of the fundamental mode is related to the time taken for the sound wave to travel to the closed end. The wavelength is four times the length of the pipe in this mode.

Answer 1

The Correct Option is C

Step 1: Understanding the fundamental mode of a pipe.
In the fundamental mode of vibration in a pipe closed at one end, the pipe behaves as a quarter-wavelength resonator. The length \( L \) of the pipe is related to the wavelength \( \lambda \) of the sound wave by: \[ L = \frac{\lambda}{4} \] The speed of sound \( v \) is related to the frequency \( f \) and the wavelength by: \[ v = f \lambda \] Since \( \lambda = 4L \), we substitute this into the equation: \[ v = f \cdot 4L \] Step 2: Solving for the frequency.
Given that the time \( t \) is the time it takes for the sound wave to travel from the open end to the closed end, the frequency of the vibration is the inverse of the time for one complete oscillation, which gives \( f = \frac{1}{4t} \).
Step 3: Conclusion.
Thus, the correct answer is (C) \( \frac{1}{4t} \).