A close organ pipe of diameter 10 cm has length 42 cm. the air column in pipe vibrates in its second overtone with the maximum amplitude `DeltaP_(o)`. The pressure amplitude at middle of pipe is

Options

A. `(DeltaP_(@))/(2)`
B. `(DeltaP_(@))/(sqrt(2))`
C. `(sqrt(3)DeltaP_(@))/(2)`
D. `DeltaP_(@)`

Explanation

To find the pressure amplitude at the middle of a closed organ pipe vibrating in its second overtone, we can follow these steps:

### Step 1: Understand the Problem
We have a closed organ pipe with a given diameter and length. The pipe vibrates in its second overtone, which corresponds to the third harmonic. We need to find the pressure amplitude at the middle of the pipe.

### Step 2: Identify the Parameters
- Diameter of the pipe (d) = 10 cm = 0.1 m
- Length of the pipe (L) = 42 cm = 0.42 m
- End correction (e) = 0.3 times the diameter = 0.3 * 0.1 m = 0.03 m

### Step 3: Calculate the Effective Length
The effective length (L_eff) of the pipe considering the end correction is given by:
\[ L_{eff} = L + e = 0.42 m + 0.03 m = 0.45 m \]

### Step 4: Determine the Wavelength
For a closed organ pipe, the relationship between the length of the pipe and the wavelength (λ) for the nth harmonic is given by:
\[ L_{eff} = \frac{n \lambda}{4} \]
For the second overtone (third harmonic, n = 3):
\[ 0.45 = \frac{3 \lambda}{4} \]
Solving for λ:
\[ \lambda = \frac{4 \times 0.45}{3} = 0.6 m = 60 cm \]

### Step 5: Find the Pressure Amplitude at the Middle of the Pipe
The pressure amplitude (ΔP) at a point in the pipe can be expressed as:
\[ P(x) = \Delta P_0 \cdot \cos(kx) \]
where \( k = \frac{2\pi}{\lambda} \) and \( x \) is the position along the pipe.

### Step 6: Calculate the Wave Number
Calculating \( k \):
\[ k = \frac{2\pi}{\lambda} = \frac{2\pi}{0.6} = \frac{10\pi}{3} \text{ rad/m} \]

### Step 7: Determine the Position at the Middle of the Pipe
The middle of the pipe is at:
\[ x = \frac{L_{eff}}{2} = \frac{0.45}{2} = 0.225 m \]

### Step 8: Calculate the Pressure Amplitude at the Middle
Now substituting \( x \) into the pressure equation:
\[ P(0.225) = \Delta P_0 \cdot \cos\left(\frac{10\pi}{3} \cdot 0.225\right) \]
Calculating the angle:
\[ \frac{10\pi}{3} \cdot 0.225 = \frac{10\pi \cdot 0.225}{3} = \frac{2.25\pi}{3} = 0.75\pi \]
Thus,
\[ P(0.225) = \Delta P_0 \cdot \cos(0.75\pi) \]
Since \( \cos(0.75\pi) = -\frac{\sqrt{2}}{2} \):
\[ P(0.225) = \Delta P_0 \cdot \left(-\frac{\sqrt{2}}{2}\right) \]

### Step 9: Final Expression
The pressure amplitude at the middle of the pipe is:
\[ P(0.225) = \frac{\sqrt{2}}{2} \Delta P_0 \]