Question

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

• A. \(8:7\)
• B. \(7:8\)
• C. \(5:3\)
• D. \(3:2\)

Hint:

Third overtone is the 7th harmonic for a closed pipe and the 4th harmonic for an open pipe.

Answer 1

The Correct Option is B

Step 1: Closed Pipe:
A closed pipe has only odd harmonics: \(f=\dfrac{(2k+1)v}{4L_c}\). The first overtone is the 3rd harmonic, the second is the 5th and the third overtone is the 7th harmonic: \(f_c=\dfrac{7v}{4L_c}\).

Step 2: Open Pipe:
An open pipe has all harmonics: \(f=\dfrac{nv}{2L_0}\). The third overtone is the 4th harmonic: \(f_0=\dfrac{4v}{2L_0}=\dfrac{2v}{L_0}\).

Step 3: Equate:
\[ \frac{7v}{4L_c}=\frac{2v}{L_0}\Rightarrow\frac{L_c}{L_0}=\frac{7}{8} \]

Step 4: Check the Other Options:
\(8:7\) is the inverse. \(5:3\) and \(3:2\) would come from using the wrong harmonic numbers, such as 5th for the closed pipe and 3rd for the open pipe. So (B) is correct.

Final Answer:
The ratio is \(7:8\), option (B). \[ \boxed{\text{(B) } 7:8} \]