The third overtone of a closed pipe of length '$L_c$' has the same frequency as the third overtone of the open pipe of length '$L_0$'. Both the pipes have same diameters. The ratio $L_c:L_0$ is equal to

Question

A pipe open at both ends has a fundamental frequency $f$ in air. The pipe is now dipped vertically in water drum to half of its length. The fundamental frequency of the air column is now equal to:

Answer 1

Step 1: Fundamentals:
Closed pipe: $f_1=\dfrac v{4L_c}$, overtones at $3f_1,5f_1,7f_1$. Open pipe: $f_1'=\dfrac v{2L_0}$, overtones at $2f_1',3f_1',4f_1'$.

Step 2: Third Overtones:
Closed: $7f_1=\dfrac{7v}{4L_c}$. Open: $4f_1'=\dfrac{4v}{2L_0}$.

Step 3: Ratio:
Equal frequencies give $\dfrac7{4L_c}=\dfrac2{L_0}$, so $L_c:L_0=7:8$. Option (B).

Final Answer:
Option (B). \[ \boxed{\text{(B) } 7:8} \]

The third overtone of a closed pipe of length '\(L_c\)' has the same frequency as the third overtone of the open pipe of length '\(L_0\)'. Both the pipes have same diameters. The ratio \(L_c:L_0\) is equal to

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The Correct Option is B

Solution and Explanation

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