Question

The fundamental frequency of a closed pipe of length \(L\) is equal to the second overtone of a pipe open at both the ends of length (XL). The value of X is (Neglect end correction)

• A. \(4\)
• B. \(5\)
• C. \(6\)
• D. \(7\)

Hint:

Closed pipe → odd harmonics only.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Pipes produce standing waves. Closed pipes (one end closed) have only odd harmonics. Open pipes (both ends open) have all integer harmonics.
Step 2: Key Formula or Approach:
Fundamental frequency of closed pipe: \(f_c = \frac{v}{4L}\).
Harmonics of open pipe: \(f_n = \frac{nv}{2L'}\), where \(n = 1\) is fundamental, \(n = 2\) is 1st overtone, \(n = 3\) is 2nd overtone.
Step 3: Detailed Explanation:
Let length of closed pipe be \(L\). Its fundamental frequency is \(f_1 = \frac{v}{4L}\).
Let length of open pipe be \(L_{\text{open}} = XL\).
The second overtone of an open pipe is its 3rd harmonic (\(n=3\)): \[ f_2' = \frac{3v}{2L_{\text{open}}} = \frac{3v}{2XL} \] Given these frequencies are equal: \[ \frac{v}{4L} = \frac{3v}{2XL} \] Cancel \(v\) and \(L\) from both sides: \[ \frac{1}{4} = \frac{3}{2X} \] Cross-multiply to solve for \(X\): \[ 2X = 4 \times 3 = 12 \] \[ X = 6 \] Step 4: Final Answer:
The value of \(X\) is 6.