Question

The frequency of fourth overtone of a closed pipe is in unison with the fifth overtone of an open pipe. The ratio of length of closed pipe to that of open pipe is

• A. \(2 : 3\)
• B. \(3 : 4\)
• C. \(4 : 5\)
• D. \(5 : 6\)

Hint:

Write the frequency of the required overtone for each pipe and equate them.

Answer 1

The Correct Option is B

Step 1: Harmonic numbers
Closed pipe: the $n$-th overtone is the $(2n+1)$-th harmonic, so the fourth overtone is the 9th. Open pipe: the fifth overtone is the 6th harmonic.

Step 2: Equal frequencies
$9\cdot\frac{v}{4L_c}=6\cdot\frac{v}{2L_o}$, so $\frac{9}{4L_c}=\frac{3}{L_o}$.

Step 3: Result
$\frac{L_c}{L_o}=\frac{3}{4}$, the second option.

Final Answer:
The length ratio is $3:4$. \[ \boxed{\text{(B) }3:4} \]