Question

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

• A. \(\frac{1}{t}\)
• B. \(\frac{0.25}{t}\)
• C. \(\frac{λ}{4t}\)
• D. \(t\)

Hint:

In the fundamental mode a closed pipe has length lambda over 4, so the travel time to the closed end is a quarter of the period.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A closed pipe has a node at the closed end and an antinode at the open end. In the fundamental mode, the length of the pipe is \(L = \dfrac{\lambda}{4}\).

Step 2: Travel time.
Sound travels the length \(L\) at speed \(v\), so \(t = \dfrac{L}{v} = \dfrac{\lambda}{4v}\).

Step 3: Frequency.
Since \(f = \dfrac{v}{\lambda}\), we get \(t = \dfrac{1}{4f}\), so
\[ f = \frac{1}{4t} = \frac{0.25}{t} \]

Step 4: Check the options.
Option (C) \(\dfrac{\lambda}{4t}\) is the speed, not the frequency. Options (A) and (D) do not match the fundamental mode relation.

Final Answer:
The frequency is \(\dfrac{0.25}{t}\), option (B). \[ \boxed{\frac{0.25}{t}} \]