Question

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 $\dfrac{{{M_A}}}{{{M_B}}}$ ?

Answer 1

First obtain the equations for second and third harmonic frequency and equate them. The ratio obtained can be substituted by applying Laplace formula. Use, for monoatomic gas ${\gamma _{\text{A}}} = 1.67$ and for diatomic gas B, ${\gamma _{\text{B}}} = 1.4$.

Complete step by step answer:
Given, the frequency of a second fundamental model harmonic in pipe $A$ is equal to the corresponding frequency of the third fundamental mode harmonic in pipe $B$.
Here, $- {I_{\text{A}}} = {I_{\text{B}}} = I$

Now, the equation of second harmonic frequency of open pipe is given by,
${n_2} = \dfrac{{{V_{\text{A}}}}}{I}$, ${V_{\text{A}}}$ is speed of sound in pipe A.

Again the equation of third harmonic frequency of closed pipe is given by,
$n_2^/ = \dfrac{{3{V_{\text{B}}}}}{{4I}}$, ${V_{\text{B}}}$ is speed of sound in pipe B.

From the question,
${n_2} = n_2^/$
Substitute the respective values of ${n_2}$ and $n_2^/$ in the above equation
\dfrac{{{V_{\text{A}}}}}{I} = \dfrac{{3{V_{\text{B}}}}}{{4I}} \\\ \implies \dfrac{{{V_{\text{B}}}}}{{{V_{\text{A}}}}} = \dfrac{4}{3} \\\ …… (1)

Apply the Laplace formula.
${V_{\text{A}}} = \sqrt {\dfrac{{{\gamma _{\text{A}}}RT}}{{{M_{\text{A}}}}}} and {V_{\text{B}}} = \sqrt {\dfrac{{{\gamma _{\text{B}}}RT}}{{{M_{\text{B}}}}}}$
Substitute these values in equation (1) and we get,
$\dfrac{{\sqrt {\dfrac{{{\gamma _{\text{B}}}RT}}{{{M_{\text{B}}}}}} }}{{\sqrt {\dfrac{{{\gamma _{\text{A}}}RT}}{{{M_{\text{A}}}}}} }} = \dfrac{4}{3}$
Further solving,
$\dfrac{{\sqrt {\dfrac{{{\gamma _{\text{B}}}RT}}{{{M_{\text{B}}}}}} }}{{\sqrt {\dfrac{{{\gamma _{\text{A}}}RT}}{{{M_{\text{A}}}}}} }} = \dfrac{4}{3} \\\ \implies\dfrac{{\dfrac{{{\gamma_{\text{B}}}RT}}{{{M_{\text{B}}}}}}}{{\dfrac{{{\gamma _{\text{A}}}RT}}{{{M_{\text{A}}}}}}} = \dfrac{{16}}{9} \\\ \implies\dfrac{{\dfrac{{{\gamma_{\text{B}}}}}{{{M_{\text{B}}}}}}}{{\dfrac{{{\gamma _{\text{A}}}}}{{{M_{\text{A}}}}}}} = \dfrac{{16}}{9} \\\ \implies\dfrac{{{\gamma _{\text{B}}}{M_{\text{A}}}}}{{{\gamma _{\text{A}}}{M_{\text{B}}}}} = \dfrac{{16}}{9} \\\\\dfrac{{{M_{\text{A}}}}}{{{M_{\text{B}}}}} = \dfrac{{16{\gamma _{\text{A}}}}}{{9{\gamma _{\text{B}}}}}$ …… (2)

Now, in case of a monatomic gas A, ${\gamma _{\text{A}}} = 1.67$
And in case of a diatomic gas B, ${\gamma _{\text{B}}} = 1.4$

Substitute these values in equation (2) and solve.
$\dfrac{{{M_{\text{A}}}}}{{{M_{\text{B}}}}} = \dfrac{{16 \times 1.67}}{{9 \times 1.4}} \\\ \dfrac{{{M_{\text{A}}}}}{{{M_{\text{B}}}}} = 2.12 \\\$

Hence, the required answer is $2.12$.

Additional Information:
Harmonic function: Any part of the harmonic series is a harmonic one. The word is used in several areas, including music, mechanics, acoustics, transmission of electrical power, radio technology, and other fields. It is usually applied, like sinusoidal waves, to repeated signals. A vibrating object's lowest resonant frequency is considered the fundamental frequency. A harmonic is specified as a multi of the fundamental frequency of an integer (whole number).

Note:
To obtain the value of $\dfrac{{{M_{\text{A}}}}}{{{M_{\text{B}}}}}$ we first deduce the relationship between the equations for second and third harmonic frequency. Which is ${n_2} = n_2^/$. From there we obtain the ratio of the corresponding speeds. To ease the problem we use the Laplace formula. Remember that, for monoatomic gas ${\gamma _{\text{A}}} = 1.67$ and for diatomic gas B, ${\gamma _{\text{B}}} = 1.4$.