Question

An open organ pipe is closed such that the third overtone of the closed pipe is found to be higher in frequency by 200 Hz than the second overtone of the original pipe. The fundamental frequency of the open pipe is (Neglect end correction)

• A. 150 Hz
• B. 200 Hz
• C. 400 Hz
• D. 500 Hz

Hint:

In open pipes, overtones are $2n, 3n, 4n...$ In closed pipes, only odd harmonics exist ($3n, 5n, 7n...$).

Answer 1

The Correct Option is C

Step 1: Open Pipe Frequencies
Fundamental $n_o = \frac{v}{2L}$. Second overtone $n_2 = 3 n_o = 3(\frac{v}{2L})$.
Step 2: Closed Pipe Frequencies
Fundamental $n_c = \frac{v}{4L} = \frac{1}{2} n_o$.
Third overtone $n'_3 = 7 n_c = 7(\frac{1}{2} n_o) = 3.5 n_o$.
Step 3: Set up Equation
$n'_3 - n_2 = 200 \text{ Hz}$
$3.5 n_o - 3 n_o = 200 \Rightarrow 0.5 n_o = 200$.
Step 4: Final Calculation
$n_o = \frac{200}{0.5} = 400 \text{ Hz}$.
Final Answer:(C)