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)
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In open pipes, overtones are $2n, 3n, 4n...$ In closed pipes, only odd harmonics exist ($3n, 5n, 7n...$).
Step 1: Understanding the Question:
We need to compare the harmonics of an open pipe and a closed pipe of the same length $L$. Step 2: Key Formula or Approach:
1. Open pipe: $n^{th}$ overtone is $(n+1)^{th}$ harmonic. Frequency $f_p = (n+1) \left(\frac{v}{2L}\right)$.
2. Closed pipe: $n^{th}$ overtone is $(2n+1)^{th}$ harmonic. Frequency $f'_c = (2n+1) \left(\frac{v}{4L}\right)$. Step 3: Detailed Explanation:
Let the fundamental frequency of the open pipe be $f = \frac{v}{2L}$.
Second overtone of open pipe $(n=2)$:
\[ f_{o2} = (2+1) \cdot f = 3f \]
Now, consider the closed pipe of same length. Its fundamental frequency is $f_c = \frac{v}{4L} = \frac{f}{2}$.
Third overtone of closed pipe $(n=3)$:
\[ f'_{c3} = (2 \times 3 + 1) \cdot f_c = 7 \cdot \left(\frac{f}{2}\right) = 3.5f \]
Given: $f'_{c3} - f_{o2} = 200$ Hz.
\[ 3.5f - 3f = 200 \]
\[ 0.5f = 200 \implies f = 400 \text{ Hz} \] Step 4: Final Answer:
The fundamental frequency of the open pipe is 400 Hz.