Consider two pipes $A$ and $B$ of identical length. $A$ has one end closed and one end open. $B$ has both ends open. Each tube is immersed in a closed chamber of ideal gas having volume $V$. The chamber containing tube $A$ is at temperature $T_A$ and the chamber containing tube $B$ is at temperature $T_B$. The sound frequencies corresponding to the $n_A$-th harmonic in tube $A$ and the $n_B$-th harmonic in tube $B$ are the same. What is the relation between the temperatures $T_A$ and $T_B$?
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Remember that for identical lengths, the fundamental frequency of an open pipe is twice that of a closed pipe: $f_{\text{open}} = 2 f_{\text{closed}}$ at the same temperature.
Including temperature dependence ($v \propto \sqrt{T}$), this leads to the relation $\sqrt{T_A} / \sqrt{T_B} \propto 2 n_B / n_A$.
Squaring this immediately gives the factor of $4$.
Step 1: Understanding the Question:
This problem links the harmonics of open and closed organ pipes with the speed of sound in ideal gases at different temperatures. Step 2: Key Formulas and Approach:
1. Frequency of the $n_A$-th harmonic for a pipe closed at one end (length $L$, sound speed $v_A$):
\[ f_A = n_A \frac{v_A}{4L} \]
2. Frequency of the $n_B$-th harmonic for a pipe open at both ends (length $L$, sound speed $v_B$):
\[ f_B = n_B \frac{v_B}{2L} \]
3. Speed of sound in an ideal gas:
\[ v = \sqrt{\frac{\gamma R T}{M}} \implies v \propto \sqrt{T} \] Step 3: Detailed Explanation:
• Let $L$ be the identical length of both pipes $A$ and $B$.
• For pipe $A$ (one end closed, one end open), the allowed frequencies are given by the odd harmonics:
\[ f_A = n_A \frac{v_A}{4L} \]
• For pipe $B$ (both ends open), the allowed frequencies are given by:
\[ f_B = n_B \frac{v_B}{2L} \]
• Since we are given that the frequencies are equal ($f_A = f_B$):
\[ n_A \frac{v_A}{4L} = n_B \frac{v_B}{2L} \]
• Simplify this relation by canceling out $L$ and reducing the constants:
\[ \frac{n_A v_A}{2} = n_B v_B \implies \frac{v_A}{v_B} = \frac{2 n_B}{n_A} \]
• The speed of sound in an ideal gas is related to the absolute temperature $T$ of the gas by:
\[ v = \sqrt{\frac{\gamma R T}{M}} \]
Assuming both chambers contain the same ideal gas (same $\gamma$ and molecular mass $M$), we have:
\[ \frac{v_A}{v_B} = \sqrt{\frac{T_A}{T_B}} \]
• Substituting this back into our frequency relation:
\[ \sqrt{\frac{T_A}{T_B}} = \frac{2 n_B}{n_A} \]
• Squaring both sides to solve for the temperature ratio:
\[ \frac{T_A}{T_B} = \frac{4 n_B^2}{n_A^2} \implies T_A = \left( \frac{4 n_B^2}{n_A^2} \right) T_B \] Step 4: Final Answer:
The relation between the temperatures is $T_A = \left( \frac{4 n_B^2}{n_A^2} \right) T_B$, which corresponds to Option (A).
