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 inside a closed organ pipe to the maximum pressure amplitude in the pipe when vibrating in its first overtone. Let's break down the solution step by step:

### Step 1: Understand the closed organ pipe and its harmonics
A closed organ pipe has one end closed and the other end open. The first overtone corresponds to the second harmonic, which has a displacement node at the closed end and a displacement antinode at the open end. This means that the pressure amplitude will be maximum at the closed end and minimum at the open end.

### Step 2: Determine the position of point Q
Point Q is located at a distance of \( \frac{7L}{9} \) from the open end. The total length of the pipe is \( L \).

### Step 3: Find the wavelength of the sound wave
For a closed organ pipe, the relationship between the length of the pipe and the wavelength \( \lambda \) is given by:
\[ L = \frac{n \lambda}{4} \]
where \( n \) is the harmonic number. For the first overtone (second harmonic), \( n = 2 \):
\[ L = \frac{2\lambda}{4} \implies \lambda = \frac{L}{\frac{1}{2}} = \frac{L}{2} \]

### Step 4: Calculate the phase difference at point Q
The distance from the closed end to point Q is:
\[ L - \frac{7L}{9} = \frac{2L}{9} \]
The phase difference \( \Delta \phi \) at point Q can be calculated using:
\[ \Delta \phi = \frac{2\pi}{\lambda} \times \text{path difference} \]
Substituting \( \lambda = \frac{L}{2} \) and the path difference:
\[ \Delta \phi = \frac{2\pi}{\frac{L}{2}} \times \frac{2L}{9} = \frac{4\pi}{L} \times \frac{2L}{9} = \frac{8\pi}{9} \]

### Step 5: Calculate the pressure amplitude at point Q
The pressure amplitude \( P_Q \) at point Q can be expressed in terms of the maximum pressure amplitude \( P_{max} \) (which occurs at the closed end) as:
\[ P_Q = P_{max} \sin(\Delta \phi) \]
Substituting \( \Delta \phi = \frac{8\pi}{9} \):
\[ P_Q = P_{max} \sin\left(\frac{8\pi}{9}\right) \]

### Step 6: Evaluate the sine function
Using the sine function properties:
\[ \sin\left(\frac{8\pi}{9}\right) = \sin\left(\pi - \frac{\pi}{9}\right) = \sin\left(\frac{\pi}{9}\right) \]
However, we can also use the fact that \( \sin\left(\frac{8\pi}{9}\right) \) is positive and can be approximated or calculated directly.

### Step 7: Find the ratio of pressure amplitudes
The maximum pressure amplitude \( P_{max} \) occurs at the closed end, and we can express the ratio:
\[ \frac{P_Q}{P_{max}} = \sin\left(\frac{8\pi}{9}\right) \]

### Step 8: Final ratio calculation
Using the sine value:
\[ \sin\left(\frac{8\pi}{9}\right) = \sin\left(\frac{\pi}{9}\right) \approx \frac{1}{2} \]
Thus, the ratio of pressure amplitude at point Q to the maximum pressure amplitude is:
\[ \frac{P_Q}{P_{max}} = \frac{1}{2} \]

### Conclusion
The ratio of the pressure amplitude at point Q to the maximum pressure amplitude in the pipe is \( \frac{1}{2} \).