Question

Two open organ pipes A and B of length 22 cm and 22.5 cm respectively produce 2 beats per sec when sounded together. The frequency of the shorter pipe is:

• A. 92 Hz
• B. 90 Hz
• C. 88 Hz
• D. 86 Hz

Hint:

The beat frequency can be found by subtracting the frequencies of two pipes and finding the difference. Use the inverse proportionality between the frequency and length of an open pipe.

Answer 1

The Correct Option is C

The beat frequency \( f_b \) is given by the absolute difference in the frequencies of the two pipes: \[ f_b = |f_A - f_B| \] where \( f_A \) is the frequency of the longer pipe, and \( f_B \) is the frequency of the shorter pipe. We are given: - The beat frequency \( f_b = 2 \, \text{beats/sec} \), - The length of the first pipe, \( L_A = 22 \, \text{cm} \), - The length of the second pipe, \( L_B = 22.5 \, \text{cm} \). For open pipes, the frequency \( f \) is inversely proportional to the length \( L \), i.e., \[ f \propto \frac{1}{L} \] Thus, we can write: \[ \frac{f_A}{f_B} = \frac{L_B}{L_A} \] Now, substituting the lengths of the pipes: \[ \frac{f_A}{f_B} = \frac{22.5}{22} \] \[ f_A = f_B \times \frac{22.5}{22} = f_B \times 1.0227 \] Now, we know the difference in their frequencies is equal to the beat frequency: \[ f_A - f_B = 2 \, \text{beats/sec} \] Substitute \( f_A = 1.0227 f_B \) into the equation: \[ 1.0227 f_B - f_B = 2 \] \[ 0.0227 f_B = 2 \] Solving for \( f_B \): \[ f_B = \frac{2}{0.0227} = 88 \, \text{Hz} \] Thus, the frequency of the shorter pipe is \( 88 \, \text{Hz} \).