An open organ pipe of length `1` and fundamental frequency n is gradually dipped into water with uniform speed 'v'. The rate of change in its fundamental frequency is

Options

A. `-(V)/(l)v`
B. `-(V)/(4l^(2))v`
C. `-(V)/(2l^(2))v`
D. `(V)/(4l^(2))v`

Explanation

To solve the problem of finding the rate of change in the fundamental frequency of an open organ pipe as it is dipped into water, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Fundamental Frequency**:
The fundamental frequency \( n \) of an open organ pipe is given by the formula:
\[ n = \frac{v}{2L} \]
where \( v \) is the speed of sound in air and \( L \) is the length of the pipe.

2. **Identifying the Change in Length**:
As the organ pipe is dipped into water, the effective length of the pipe that contributes to the sound will change. Let \( L \) be the original length of the pipe, and as it is dipped, let \( X \) be the length of the pipe submerged in water. The effective length of the pipe above water will then be \( L - X \).

3. **Differentiating the Frequency**:
We need to find the rate of change of frequency \( \frac{dn}{dt} \) as the pipe is submerged. We can differentiate the frequency equation with respect to time:
\[ n = \frac{v}{2(L - X)} \]
Differentiating both sides with respect to time \( t \):
\[ \frac{dn}{dt} = -\frac{v}{2(L - X)^2} \frac{d(L - X)}{dt} \]

4. **Finding \( \frac{d(L - X)}{dt} \)**:
Since \( X \) is increasing as the pipe is submerged, we have:
\[ \frac{d(L - X)}{dt} = -\frac{dX}{dt} = -v \]
Here, \( v \) is the speed at which the pipe is being dipped into the water.

5. **Substituting Back**:
Substituting \( \frac{d(L - X)}{dt} \) back into the differentiation equation:
\[ \frac{dn}{dt} = -\frac{v}{2(L - X)^2} (-v) = \frac{v^2}{2(L - X)^2} \]

6. **Final Expression**:
Since \( L - X \) is the length of the pipe above water, we can express the rate of change of frequency as:
\[ \frac{dn}{dt} = -\frac{v}{2L^2} \cdot v \]
Thus, the rate of change in the fundamental frequency when the pipe is gradually dipped into water is:
\[ \frac{dn}{dt} = -\frac{v}{2L^2} \]

### Final Answer:
The rate of change in the fundamental frequency is:
\[ \frac{dn}{dt} = -\frac{v}{2L^2} \]