Find the temperature at which the fundamental frequency of an organ pipe is independent of small variation in temperature in terms of the coefficient of linear expansion ( `alpha`) of the material of the tube.

Options

A. `1//3alpha`
B. `1//2alpha`
C. `1//4alpha`
D. `1//5alpha`

Explanation

To find the temperature at which the fundamental frequency of an organ pipe is independent of small variations in temperature in terms of the coefficient of linear expansion (α) of the material of the tube, we can follow these steps:

### Step-by-Step Solution:

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

2. **Length of the Organ Pipe**:
The length of the organ pipe at temperature \( T \) can be expressed as:
\[ L = L_0 (1 + \alpha (T - T_0)) \]
where \( L_0 \) is the original length at temperature \( T_0 \) and \( \alpha \) is the coefficient of linear expansion.

3. **Velocity of Sound**:
The velocity of sound in the medium can be expressed as:
\[ V = \sqrt{\frac{\gamma R T}{M}} \]
where \( \gamma \) is the adiabatic index, \( R \) is the gas constant, \( T \) is the absolute temperature, and \( M \) is the molar mass.

4. **Substituting into the Frequency Formula**:
Substituting the expressions for \( V \) and \( L \) into the frequency formula, we get:
\[ f(T) = \frac{\sqrt{\frac{\gamma R T}{M}}}{2L_0(1 + \alpha (T - T_0))} \]

5. **Condition for Independence from Temperature**:
We want to find the temperature \( T \) such that small variations in temperature do not affect the frequency. This means:
\[ f(T) \approx f(T_0) \]
for small \( T - T_0 \).

6. **Using the Binomial Approximation**:
For small \( T - T_0 \), we can use the binomial approximation:
\[ \sqrt{1 + x} \approx 1 + \frac{x}{2} \]
Applying this to our frequency equation, we can simplify it.

7. **Equating Terms**:
By equating the terms derived from the frequency expressions, we find:
\[ \frac{1}{2} \frac{T - T_0}{T_0} = \alpha (T - T_0) \]

8. **Solving for \( T_0 \)**:
Rearranging gives:
\[ \alpha = \frac{1}{2T_0} \]
Thus, we can express \( T_0 \) in terms of \( \alpha \):
\[ T_0 = \frac{1}{2\alpha} \]

### Final Answer:
The temperature at which the fundamental frequency of an organ pipe is independent of small variations in temperature is:
\[ T_0 = \frac{1}{2\alpha} \]