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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

Options

  1. A88 Hz
  2. B90 Hz
  3. C92 Hz
  4. D86 Hz

Correct answer

B. 90 Hz

Step-by-step solution

The fundamental frequency of an open organ pipe of length L is given by f = v 2L , where v is the speed of sound in air. Let v = 330 m/s be the standard speed of sound. For pipe A with length L_A = 22 cm = 0.22 m , the frequency is f_A = 330 2 0.22 = 330 0.44 = 750 Hz . For pipe B with length L_B = 22.5 cm = 0.225 m , the frequency is f_B = 330 2 0.225 = 330 0.45 = 733.33 Hz . The beat frequency is |f_A - f_B| = |750 - 733.33| = 16.67 Hz , which does not match the given 2 beats per second. This implies the speed of

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