A tube of diameter d and of length l is open a both ends. Its fundamental frequnecy of resonance is found to be `f_(1)`. One end of the tube is now closed. The lowrst frequency of resonance of this closed tube is now `f_(2)`. Taking into consideration the end correction ,`(f2)/(f_(1)` is

Options

A. `((l +0.6d))/((l+0.3 d))`
B. `((l +0.3d))/(2(l+0.6 d))`
C. `((l +0.6d))/(2(l+0.3 d))`
D. `(1(d +0.3l))/(2(d+0.6l))`

Answer

To solve the problem, we need to find the ratio of the lowest frequency of resonance \( f_2 \) in a closed tube to the fundamental frequency of resonance \( f_1 \) in an open tube. We will take into account the end correction for both cases.

### Step-by-Step Solution:

1. **Understanding the Open Tube Case**:
- For a tube open at both ends, the fundamental frequency of resonance occurs when there is one full wavelength fitting in the length of the tube.
- The effective length of the tube considering end corrections is given by:
\[ L_{effective} = L + 0.3d + 0.3d = L + 0.6d \]
- The wavelength \( \lambda_1 \) for the fundamental frequency \( f_1 \) can be expressed as:
\[ \frac{\lambda_1}{2} = L + 0.6d \implies \lambda_1 = 2(L + 0.6d) \]

2. **Understanding the Closed Tube Case**:
- For a tube closed at one end, the fundamental frequency of resonance occurs with a quarter of a wavelength fitting in the length of the tube.
- The effective length of the tube considering the end correction at the closed end is:
\[ L_{effective} = L + 0.3d \]
- The wavelength \( \lambda_2 \) for the fundamental frequency \( f_2 \) can be expressed as:
\[ \frac{\lambda_2}{4} = L + 0.3d \implies \lambda_2 = 4(L + 0.3d) \]

3. **Finding the Ratio of Frequencies**:
- The relationship between frequency and wavelength is given by:
\[ f \propto \frac{1}{\lambda} \]
- Therefore, the ratio of the frequencies \( \frac{f_2}{f_1} \) can be expressed as:
\[ \frac{f_2}{f_1} = \frac{\lambda_1}{\lambda_2} \]
- Substituting the expressions for \( \lambda_1 \) and \( \lambda_2 \):
\[ \frac{f_2}{f_1} = \frac{2(L + 0.6d)}{4(L + 0.3d)} = \frac{L + 0.6d}{2(L + 0.3d)} \]

4. **Final Expression**:
- Thus, the final expression for the ratio of the frequencies is:
\[ \frac{f_2}{f_1} = \frac{L + 0.6d}{2(L + 0.3d)} \]

### Conclusion:
The ratio \( \frac{f_2}{f_1} \) is given by \( \frac{L + 0.6d}{2(L + 0.3d)} \).