Two narrow cylindrical pipes `A and B` have the same length. Pipe `A` is open at both ends and is filled with a monoatomic gas of molar mass `M_(A)`. Pipe `B` is open at one end and closed at the other end, and is filled with a diatomic gas of molar mass `M_(B)`. Both gases are at the same temperature.
(a) If the frequency of the second harmonic of the fundamental mode in pipe `A` is equal to the frequency of the third harmonic of the fundamental mode in pipe `B`, determine the value of `M_(B)//M_(B)`.
(b) Now the open end of pipe `B` is also closed (so that the pipe is closed at both ends). Find the ratio of the fundamental frequency in pipe `A` to that in pipe `B`.

Explanation

To solve the problem, we need to analyze the frequencies of the sound waves in the two pipes filled with different gases.

### Part (a): Finding the ratio of molar masses \( \frac{M_A}{M_B} \)

1. **Identify the frequencies of the harmonics:**
- For pipe A (open at both ends), the frequency of the second harmonic is given by:
\[ f_A = \frac{2V_A}{L} \]
- For pipe B (open at one end and closed at the other), the frequency of the third harmonic is given by:
\[ f_B = \frac{3V_B}{4L} \]

2. **Express the speed of sound in each pipe:**
- The speed of sound in a gas is given by:
\[ V = \sqrt{\frac{\gamma RT}{M}} \]
- For pipe A:
\[ V_A = \sqrt{\frac{\gamma_A RT}{M_A}} \]
- For pipe B:
\[ V_B = \sqrt{\frac{\gamma_B RT}{M_B}} \]

3. **Substitute the expressions for \( V_A \) and \( V_B \) into the frequency equations:**
- For pipe A:
\[ f_A = \frac{2}{L} \sqrt{\frac{\gamma_A RT}{M_A}} \]
- For pipe B:
\[ f_B = \frac{3}{4L} \sqrt{\frac{\gamma_B RT}{M_B}} \]

4. **Set the frequencies equal to each other:**
\[ \frac{2}{L} \sqrt{\frac{\gamma_A RT}{M_A}} = \frac{3}{4L} \sqrt{\frac{\gamma_B RT}{M_B}} \]

5. **Cancel common terms and rearrange:**
- Cancel \( \frac{1}{L} \) and \( RT \):
\[ 2 \sqrt{\frac{\gamma_A}{M_A}} = \frac{3}{4} \sqrt{\frac{\gamma_B}{M_B}} \]

6. **Square both sides to eliminate the square roots:**
\[ 4 \frac{\gamma_A}{M_A} = \frac{9}{16} \frac{\gamma_B}{M_B} \]

7. **Rearrange to find the ratio of molar masses:**
\[ \frac{M_A}{M_B} = \frac{4 \cdot 16 \cdot \gamma_A}{9 \cdot \gamma_B} \]

8. **Substitute the values of \( \gamma_A \) and \( \gamma_B \):**
- For monatomic gas \( \gamma_A = \frac{5}{3} \)
- For diatomic gas \( \gamma_B = \frac{7}{5} \)
\[ \frac{M_A}{M_B} = \frac{64 \cdot \frac{5}{3}}{9 \cdot \frac{7}{5}} = \frac{64 \cdot 25}{27 \cdot 7} = \frac{1600}{189} \]

### Part (b): Finding the ratio of fundamental frequencies \( \frac{f_A}{f_B} \)

1. **Calculate the fundamental frequency for pipe A:**
\[ f_A = \frac{V_A}{2L} = \frac{1}{2L} \sqrt{\frac{\gamma_A RT}{M_A}} \]

2. **Calculate the fundamental frequency for pipe B (now closed at both ends):**
- For a pipe closed at both ends, the fundamental frequency is:
\[ f_B = \frac{V_B}{2L} = \frac{1}{2L} \sqrt{\frac{\gamma_B RT}{M_B}} \]

3. **Take the ratio of the two frequencies:**
\[ \frac{f_A}{f_B} = \frac{\sqrt{\frac{\gamma_A RT}{M_A}}}{\sqrt{\frac{\gamma_B RT}{M_B}}} \]

4. **Cancel common terms:**
\[ \frac{f_A}{f_B} = \sqrt{\frac{\gamma_A M_B}{\gamma_B M_A}} \]

5. **Substitute the known values:**
\[ \frac{f_A}{f_B} = \sqrt{\frac{\frac{5}{3} \cdot M_B}{\frac{7}{5} \cdot M_A}} = \sqrt{\frac{25 M_B}{21 M_A}} \]

6. **Substituting \( \frac{M_A}{M_B} = \frac{1600}{189} \):**
\[ \frac{f_A}{f_B} = \sqrt{\frac{25 \cdot 189}{21 \cdot 1600}} = \sqrt{\frac{4725}{33600}} = \sqrt{\frac{315}{2240}} = \frac{\sqrt{315}}{\sqrt{2240}} \]