Question

When an open pipe is suddenly closed at one end, the frequency of the third harmonic of the closed pipe is found to be $50\text{ Hz}$ more than the fundamental frequency of the open pipe. The fundamental frequency of the open pipe is

• A. 100 Hz
• B. 50 Hz
• C. 200 Hz
• D. 300 Hz
• E. 350 Hz

Hint:

Always remember that for the same length, an open pipe has a fundamental frequency twice that of a closed pipe. Also, closed pipes only produce odd harmonics.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We are dealing with standing waves in pipes. An open pipe is open at both ends, and a closed pipe is closed at one end. The boundary conditions change the harmonic series.
Step 2: Key Formula or Approach:
Fundamental frequency of an open pipe: \(f_{\text{open}} = \frac{v}{2L}\).
Harmonic frequencies of a closed pipe: \(f_n = n \left(\frac{v}{4L}\right)\), where \(n\) must be an odd integer (1, 3, 5...).
Third harmonic of closed pipe: \(f_{c3} = \frac{3v}{4L}\).
Given condition: \(f_{c3} = f_{\text{open}} + 50\).
Step 3: Detailed Explanation:
Let's express \(f_{c3}\) in terms of \(f_{\text{open}}\).
We know \(f_{\text{open}} = \frac{v}{2L}\).
We can rewrite \(f_{c3}\) as:
\[ f_{c3} = \frac{3v}{4L} = \frac{3}{2} \left(\frac{v}{2L}\right) = 1.5 f_{\text{open}} \] Substitute this into the given condition equation:
\[ 1.5 f_{\text{open}} = f_{\text{open}} + 50 \] Subtract \(f_{\text{open}}\) from both sides:
\[ 0.5 f_{\text{open}} = 50 \] Multiply by 2 to solve for the fundamental frequency:
\[ f_{\text{open}} = 100 \text{ Hz} \] Step 4: Final Answer:
The fundamental frequency of the open pipe is 100 Hz.