A standing wave in second overtone is maintained in a open pipe of length `L`. The distance between consecutive displacement node and pressure node is `L//x`. Find `x`

Options

A. 4
B. 5
C. 6
D. 7

Explanation

To solve the problem, we need to understand the relationship between the standing wave in an open pipe and the nodes and antinodes of the wave. Here’s a step-by-step solution:

### Step 1: Understand the standing wave in an open pipe
In an open pipe, standing waves are formed due to the interference of waves traveling in opposite directions. The harmonics in an open pipe are given by the formula:
\[ L = n \frac{\lambda}{2} \]
where \( n \) is the harmonic number (1 for the fundamental frequency, 2 for the first overtone, 3 for the second overtone, etc.), \( \lambda \) is the wavelength, and \( L \) is the length of the pipe.

### Step 2: Identify the second overtone
The second overtone corresponds to the third harmonic (n = 3). Thus, we can write:
\[ L = 3 \frac{\lambda}{2} \]

### Step 3: Calculate the wavelength
From the equation above, we can rearrange to find the wavelength \( \lambda \):
\[ \lambda = \frac{2L}{3} \]

### Step 4: Find the distance between nodes and antinodes
In a standing wave, the distance between a node and the nearest antinode is given by:
\[ \text{Distance} = \frac{\lambda}{4} \]
Substituting the value of \( \lambda \) we found:
\[ \text{Distance} = \frac{1}{4} \left( \frac{2L}{3} \right) = \frac{L}{6} \]

### Step 5: Relate the distance to the given expression
According to the problem, the distance between consecutive displacement node and pressure node is given as \( \frac{L}{x} \). We have calculated that this distance is \( \frac{L}{6} \).

### Step 6: Equate and solve for \( x \)
Setting the two expressions for distance equal gives:
\[ \frac{L}{x} = \frac{L}{6} \]
By cross-multiplying, we find:
\[ x = 6 \]

### Conclusion
Thus, the value of \( x \) is:
\[ \boxed{6} \]