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Concepts Of Physics MCQ Edition [Volume 1]PhysicsSound Waves

In an open organ pipe, two successive resonance frequencies are 1620 Hz and 2268 Hz . Determine the length of the tube. The speed of sound in air is 324 m s ⁻¹ .

Options

  1. A20 cm
  2. B50 cm
  3. C12.5 cm
  4. D25 cm

Correct answer

D. 25 cm

Step-by-step solution

The difference between two successive resonance frequencies is: f = 2268 - 1620 = 648 Hz For an organ pipe, the difference between successive resonance frequencies is v 2L . v 2L = 648 324 2L = 648 2L = 324 648 = 1 2 L = 1 4 m = 25 cm

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