When an air column in a pipe open at both ends vibrates such that four antinodes and three nodes are formed, then the corresponding mode of vibration is
Show Hint
For an open pipe, a foolproof shortcut to find the overtone number is to simply look at the number of internal nodes. The number of nodes is always exactly equal to the harmonic number ($n$). Once you have $n$, just subtract 1 to get the overtone index ($3 - 1 = 2$).
Step 1: Understanding the Question:
An air column inside a cylindrical pipe open at both ends is vibrating in a specific steady standing wave pattern.
We are given that this pattern contains exactly four antinodes and three nodes. We need to identify the technical musical name for this specific mode of vibration (overtone level).
Step 2: Key Formula or Approach:
For an open-ended organ pipe, the boundary conditions demand that open ends always form displacement antinodes ($A$), while the stationary turning points form nodes ($N$).
1. The fundamental mode (first harmonic) contains $2\text{ antinodes}$ at the edges and $1\text{ node}$ in the center.
2. The $p$-th overtone corresponds to the $(p+1)$-th harmonic.
3. For an open pipe vibrating in its $n$-th harmonic:
$$\text{Number of Nodes} = n$$
$$\text{Number of Antinodes} = n + 1$$
Step 3: Detailed Explanation:
The problem states that the standing wave has 3 nodes and 4 antinodes.
Using our harmonic count rules:
$$\text{Number of Nodes} = n = 3$$
$$\text{Number of Antinodes} = n + 1 = 3 + 1 = 4$$
This confirms that the column is vibrating at its third harmonic ($n = 3$).
Now, convert this harmonic index into its corresponding overtone level. For an open pipe, the harmonics progress sequentially as integers ($n = 1, 2, 3, \dots$).
* $n = 1$ is the Fundamental frequency.
* $n = 2$ is the First Overtone.
* $n = 3$ is the Second Overtone.
Therefore, a pattern with 3 nodes and 4 antinodes represents the second overtone.
Step 4: Final Answer:
The corresponding mode of vibration is the second overtone, matching option (B).