Question

A closed organ pipe of length \( L_c \) and an open organ pipe of length \( L_o \) contain different gases of densities \( \rho_1 \) and \( \rho_2 \) respectively. The compressibility of the gases is the same in both the pipes. The gases are vibrating in their first overtone with the same frequency. What is the length of open organ pipe?

• A. \(\frac{4L_c}{3} \sqrt{\frac{\rho}{\rho_2}}\)
• B. \(\frac{3L_c}{4} \sqrt{\frac{\rho_2}{\rho_1}}\)
• C. \(\frac{4L_c}{3} \sqrt{\frac{\rho_2}{\rho_1}}\)
• D. \(\frac{2L_c}{3} \sqrt{\frac{\rho_2}{\rho}}\)

Hint:

Remember: For a closed pipe, only odd harmonics exist. First overtone = third harmonic (3f₁). For an open pipe, first overtone = second harmonic (2f₁). Speed of sound in a gas is inversely proportional to the square root of density when bulk modulus is constant.

Answer 1

The Correct Option is A

Step 1: Lay out the situation.
A closed pipe of length $L_c$ and an open pipe of length $L_o$ hold gases of densities $\rho_1$ and $\rho_2$. The gases have the same compressibility, so the same bulk modulus $B$. Both pipes sound their first overtone at the same frequency. We want $L_o$.

Step 2: Write each first overtone.
For a closed pipe the first overtone is the third harmonic: $f_c = \dfrac{3v_1}{4L_c}$. For an open pipe the first overtone is the second harmonic: $f_o = \dfrac{v_2}{L_o}$.

Step 3: Speed of sound in each gas.
Sound speed is $v = \sqrt{\dfrac{B}{\rho}}$. With the same $B$, $v_1 = \sqrt{\dfrac{B}{\rho_1}}$ and $v_2 = \sqrt{\dfrac{B}{\rho_2}}$.

Step 4: Match the frequencies.
Set $f_c = f_o$: $\dfrac{3v_1}{4L_c} = \dfrac{v_2}{L_o}$.

Step 5: Solve for $L_o$.
Rearranging, $L_o = \dfrac{4L_c}{3}\cdot\dfrac{v_2}{v_1}$. The speed ratio is $\dfrac{v_2}{v_1} = \sqrt{\dfrac{\rho_1}{\rho_2}}$.

Step 6: Write the result.
So $L_o = \dfrac{4L_c}{3}\sqrt{\dfrac{\rho_1}{\rho_2}}$, which is option (A).
\[ \boxed{L_o = \frac{4L_c}{3}\sqrt{\frac{\rho_1}{\rho_2}}} \]