Step 1: Understanding the Concept:
An organ pipe that is closed at one end and open at the other behaves as a "quarter-wave resonator" in its fundamental mode.
When air inside the pipe vibrates, a stationary (standing) wave is formed.
At the closed end, the air molecules are trapped and cannot move, creating a "displacement node" (zero amplitude).
At the open end, the air molecules move with maximum freedom, creating a "displacement antinode" (maximum amplitude).
In the fundamental mode (simplest pattern), the length of the pipe spans exactly from one node to the very next antinode.
Key Formula or Approach:
1. Node-to-Antinode distance = \(\lambda/4\).
2. Relationship: \(L = \frac{\lambda}{4}\) or \(\lambda = 4L\).
3. Wave equation: \(v = f\lambda\).
4. Frequency: \(f = \frac{v}{4L}\).
Step 2: Detailed Explanation:
The problem provides the fundamental frequency \(f = 100\) Hz.
We need to find the length \(L\) in terms of the speed of sound \(v\).
Using the fundamental frequency formula for a closed pipe:
\[ f = \frac{v}{4L} \]
Substitute 100 for \(f\):
\[ 100 = \frac{v}{4L} \]
To isolate \(L\), we first multiply both sides by \(4L\):
\[ 100 \times 4L = v \]
\[ 400L = v \]
Now, divide both sides by 400 to find the expression for length:
\[ L = \frac{v}{400} \]
This algebraic result relates the physical length of the resonating air column to the speed of the sound wave and the given frequency.
Step 3: Final Answer:
The length of the pipe is \(v/400\), which is option (B).