An open organ pipe and closed organ pipe of same length produce \(2\) beats per second in fundamental mode. The length of open pipe is made half and that of closed pipe is doubled. The number of beats produced per second will be:
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Open pipe frequency is twice that of a closed pipe of the same length in fundamental mode.
Step 1: Understanding the Question:
Initially, two pipes of equal length \(L\) produce \(2\) beats per second.
We need to find the new beat frequency after the lengths are changed. Step 2: Key Formula or Approach:
Fundamental frequency of an open pipe: \(f_o = \frac{v}{2L}\)
Fundamental frequency of a closed pipe: \(f_c = \frac{v}{4L}\)
Beat frequency: \(\Delta f = |f_1 - f_2|\) Step 3: Detailed Explanation:
Initially:
\(f_o = \frac{v}{2L}\) and \(f_c = \frac{v}{4L}\)
The beat frequency is \(f_o - f_c = \frac{v}{2L} - \frac{v}{4L} = \frac{v}{4L}\)
Given \(\frac{v}{4L} = 2 \text{ Hz}\). This means \(\frac{v}{2L} = 4 \text{ Hz}\).
Now, the lengths are changed:
New length of open pipe: \(L_o' = \frac{L}{2} \implies f_o' = \frac{v}{2(L/2)} = \frac{v}{L}\)
New length of closed pipe: \(L_c' = 2L \implies f_c' = \frac{v}{4(2L)} = \frac{v}{8L}\)
Calculate new frequencies in terms of the initial beat frequency (\(B = \frac{v}{4L} = 2\)):
\(f_o' = \frac{v}{L} = 4 \times \left( \frac{v}{4L} \right) = 4 \times 2 = 8 \text{ Hz}\)
\(f_c' = \frac{v}{8L} = \frac{1}{2} \times \left( \frac{v}{4L} \right) = \frac{1}{2} \times 2 = 1 \text{ Hz}\)
The new beat frequency is:
\[ \Delta f' = |f_o' - f_c'| = |8 - 1| = 7 \text{ Hz} \] Step 4: Final Answer:
The number of beats produced per second will be \(7\).