Question

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 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

Answer 1

(a) Frequency of second harmonic in pipe A
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, $= frequency of third harmonic in pipe B
$\therefore 2 \bigg (\frac {v_A}{2l_A}\bigg )=3 \bigg (\frac {v_B}{4l_B}\bigg ) $
or $ \frac {v_A}{v_B}= \frac {3}{4} \, \, \, \, \, \, \, \, \, \, \, (as \, l_A=l_B) $
or $ \frac {\sqrt {\frac {\gamma _ART_A}{M_A}}}{\sqrt {\frac {\gamma_BRT_B}{M_B}}}= \frac {3}{4} $
or $\sqrt {\frac {\gamma _A}{\gamma _B}}\sqrt {\frac {M_B}{M_A}}= \frac {3}{4} \, (as \, T_A=T_B) $
$\therefore \, \, \, \, \, \, \, \frac {M_A}{M_B}= \frac {\gamma_A}{\gamma _B}\bigg (\frac {16}{9}\bigg ) $
$ =\bigg (\frac {5/3}{7/5}\bigg )\bigg (\frac {16}{9}\bigg ) \, \, \, \, \, \, \bigg (\gamma_A=\frac {5}{3} \, and \, \gamma_B= \frac {7}{5} \bigg ) $
or $\, \, \, \, \, \, \, \, \frac {M_A}{M_B}= \bigg (\frac {25}{21}\bigg )\bigg (\frac {16}{9}\bigg )= \frac {400}{189} $
(b) Ratio of fundamental frequency in pipe A and in pipe B is
$\, \, \, \, \, \frac {f_A}{f_B}= \frac {v_A/2l_A}{v_B/2l_B}= \frac {v_A}{v_B} \, \, \, \, \, \, (as \, l_A=l_B) $
$\, \, \, \, \, \, =\frac {\sqrt {\frac {\gamma_ART_A}{M_A}}}{\sqrt {\frac {\gamma_BRT_B}{M_B}}}= \sqrt {\frac {\gamma_A}{\gamma_B}. \frac {M_B}{M_A}}\, \, \, \, (as T_A=T_B) $
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} $