Fifth overtone of an open pipe of length $L_0$ is in unison with the fifth overtone of the pipe closed at one end of length $L_c$.
The ratio $L_0$ to $L_c$ is

Question

A steel wire with mass per unit length $ 7.0 \times 10^{-3} \, \text{kg/m} $ is under a tension of $ 70 \, \text{N} $. The speed of transverse waves in the wire will be:

Answer 1

Step 1: Count overtones:
The first overtone of a closed pipe is the 3rd harmonic. The fifth overtone is the harmonic $2 \times 5 + 1 = 11$. For an open pipe the first overtone is the 2nd harmonic, so the fifth is the 6th.

Step 2: Equate wavelengths:
$\lambda_o = \frac{2L_o}{6} = \frac{L_o}{3}$ and $\lambda_c = \frac{4L_c}{11}$. Equal frequency means equal wavelength, so $\frac{L_o}{3} = \frac{4L_c}{11}$, giving $\frac{L_o}{L_c} = \frac{12}{11}$.

Final Answer:
The ratio is $12:11$, option (D). \[ \boxed{12:11} \]

Fifth overtone of an open pipe of length \(L_0\) is in unison with the fifth overtone of the pipe closed at one end of length \(L_c\).
The ratio \(L_0\) to \(L_c\) is

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The Correct Option is D

Solution and Explanation

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Fifth overtone of an open pipe of length \(L_0\) is in unison with the fifth overtone of the pipe closed at one end of length \(L_c\).The ratio \(L_0\) to \(L_c\) is