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MHT CET Medical202621 April 2026Evening ShiftPhysicsWaves and SoundActual

A string fixed at both the ends, forms a standing wave with node separation of 5 cm . If the velocity of the wave on the string is 2 m/s , then the frequency of vibration of the string is

Options

  1. A40 Hz
  2. B20 Hz
  3. C10 Hz
  4. D0.2 Hz

Correct answer

B. 20 Hz

Step-by-step solution

The distance between two consecutive nodes in a standing wave is 2 . Given that the node separation is 5 cm , we have: 2 = 5 cm = 0.05 m = 0.1 m The velocity of the wave is given as v = 2 m/s . The frequency of vibration f is related to the wave velocity and wavelength by the formula: f = v Substituting the values: f = 2 0.1 = 20 Hz Answer: 20 Hz

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