SATHEE: Wave Motion 2 Question 42

Wave Motion 2 Question 42

42. 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. $(2002, 5 M)$

(a) If the frequency to the second harmonic of pipe $A$ is equal to the frequency of the third harmonic of the fundamental mode in pipe $B$ , determine the value of $M_{A} / M_{B}$ .

(b) Now the open end of the pipe $B$ is closed (so that the pipe is closed at both ends). Find the ratio of the fundamental frequency in pipe $A$ to that in pipe $B$ .

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Answer:

Correct Answer: 42. (a) $\frac{400}{189}$

(b) $\frac{3}{4}$

Solution:

(a) Frequency of second harmonic in pipe $A$

= frequency of third harmonic in pipe $B$

$∴ 2 (\frac{v_{A}}{2 l_{A}}) = 3 (\frac{v_{B}}{4 l_{B}})$

$\frac{v_{A}}{v_{B}} = \frac{3}{4} or \frac{\sqrt{\frac{γ_{A} R T_{A}}{M_{A}}}}{\sqrt{\frac{γ_{B} R T_{B}}{M_{B}}}} = \frac{3}{4} (as l_{A} = l_{B})$

or $\sqrt{\frac{γ_{A}}{γ_{B}}} \sqrt{\frac{M_{B}}{M_{A}}} = \frac{3}{4} ($ as $T_{A} = T_{B}) \Rightarrow \frac{M_{A}}{M_{B}} = \frac{γ_{A}}{γ_{B}} (\frac{16}{9})$

$= (\frac{5 / 3}{7 / 5}) (\frac{16}{9}) (γ_{A} = \frac{5}{3} and γ_{B} = \frac{7}{5})$

$\frac{M_{A}}{M_{B}} = (\frac{25}{21}) (\frac{16}{9}) = \frac{400}{189}$

(b) Ratio of fundamental frequency in pipe $A$ and in pipe $B$ is

$\begin{aligned} \frac{f_{A}}{f_{B}} & = \frac{v_{A} / 2 l_{A}}{v_{B} / 2 l_{B}} = \frac{v_{A}}{v_{B}} (as l_{A} = l_{B}) \\ = \frac{\sqrt{\frac{γ_{A} R T_{A}}{M_{A}}}}{\sqrt{\frac{γ_{B} R T_{B}}{M_{B}}}} = \sqrt{\frac{γ_{A}}{γ_{B}} \cdot \frac{M_{B}}{M_{A}}} (as T_{A} = T_{B}) \end{aligned}$