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 $B$
$\therefore \quad 2\left(\frac{v _A}{2 l _A}\right)=3\left(\frac{v _B}{4 l _B}\right)$
or
$$ \frac{v _A}{v _B}=\frac{3}{4} \text { or } \frac{\sqrt{\frac{\gamma _A R T _A}{M _A}}}{\sqrt{\frac{\gamma _B R T _B}{M _B}}}=\frac{3}{4} \quad\left(\text { as } l _A=l _B\right) $$
or $\sqrt{\frac{\gamma _A}{\gamma _B}} \sqrt{\frac{M _B}{M _A}}=\frac{3}{4}\left(\right.$ as $\left.T _A=T _B\right) \Rightarrow \frac{M _A}{M _B}=\frac{\gamma _A}{\gamma _B}\left(\frac{16}{9}\right)$
$$ =\left(\frac{5 / 3}{7 / 5}\right)\left(\frac{16}{9}\right) \quad\left(\gamma _A=\frac{5}{3} \text { and } \gamma _B=\frac{7}{5}\right) $$
or
$$ \frac{M _A}{M _B}=\left(\frac{25}{21}\right)\left(\frac{16}{9}\right)=\frac{400}{189} $$
(b) Ratio of fundamental frequency in pipe $A$ and in pipe $B$ is
$$ \begin{aligned} \frac{f _A}{f _B} & =\frac{v _A / 2 l _A}{v _B / 2 l _B}=\frac{v _A}{v _B} \quad\left(\text { as } l _A=l _B\right) \\ & =\frac{\sqrt{\frac{\gamma _A R T _A}{M _A}}}{\sqrt{\frac{\gamma _B R T _B}{M _B}}}=\sqrt{\frac{\gamma _A}{\gamma _B} \cdot \frac{M _B}{M _A}} \quad\left(\text { as } T _A=T _B\right) \end{aligned} $$
Substituting $\frac{M _B}{M _A}=\frac{189}{400}$ from part (a), we get
$$ \frac{f _A}{f _B}=\sqrt{\frac{25}{21} \times \frac{189}{400}}=\frac{3}{4} $$
