A closed organ pipe of length L is vibrating in its first overtone there is a point Q inside the pipe at a distance `7L//9` form the open end the ratio of pressure amplitude at Q to the maximum pressure amplitude in the pipe is

Options

A. `1:2`
B. `2:1`
C. `1:1`
D. `2:3`

Explanation

To solve the problem, we need to find the ratio of the pressure amplitude at point Q (located at a distance of \( \frac{7L}{9} \) from the open end) to the maximum pressure amplitude in a closed organ pipe vibrating in its first overtone.

### Step-by-Step Solution:

1. **Understand the Closed Organ Pipe:**
- A closed organ pipe has one end closed and one end open. In the first overtone (which is the second harmonic for a closed pipe), the length of the pipe \( L \) corresponds to \( \frac{3\lambda}{4} \), where \( \lambda \) is the wavelength.

2. **Determine the Wavelength:**
- From the relationship for the first overtone in a closed pipe:
\[ L = \frac{3\lambda}{4} \]
- Rearranging gives:
\[ \lambda = \frac{4L}{3} \]

3. **Identify the Pressure Amplitude:**
- The pressure amplitude \( P(x) \) in a closed organ pipe varies as a cosine function due to the phase difference of \( \frac{\pi}{2} \) between displacement and pressure:
\[ P(x) = P_{\text{max}} \cos(kx) \]
- Here, \( k \) is the wave number given by:
\[ k = \frac{2\pi}{\lambda} = \frac{2\pi}{\frac{4L}{3}} = \frac{3\pi}{2L} \]

4. **Calculate the Pressure Amplitude at Point Q:**
- The position of point Q is at \( x = \frac{7L}{9} \).
- Substitute \( x \) into the pressure amplitude equation:
\[ P\left(\frac{7L}{9}\right) = P_{\text{max}} \cos\left(\frac{3\pi}{2L} \cdot \frac{7L}{9}\right) \]
- Simplifying the argument of the cosine:
\[ P\left(\frac{7L}{9}\right) = P_{\text{max}} \cos\left(\frac{3\pi \cdot 7}{18}\right) = P_{\text{max}} \cos\left(\frac{7\pi}{6}\right) \]

5. **Evaluate the Cosine:**
- The cosine of \( \frac{7\pi}{6} \) is:
\[ \cos\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2} \]
- Therefore, the pressure amplitude at point Q is:
\[ P\left(\frac{7L}{9}\right) = P_{\text{max}} \left(-\frac{\sqrt{3}}{2}\right) \]

6. **Find the Ratio:**
- The ratio of the pressure amplitude at Q to the maximum pressure amplitude is:
\[ \text{Ratio} = \frac{P\left(\frac{7L}{9}\right)}{P_{\text{max}}} = -\frac{\sqrt{3}}{2} \]
- Since we are interested in the magnitude, we take the absolute value:
\[ \text{Ratio} = \frac{\sqrt{3}}{2} \]

### Final Answer:
The ratio of the pressure amplitude at point Q to the maximum pressure amplitude in the pipe is \( \frac{\sqrt{3}}{2} \).