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JEE MainPhysicsWaves and Sound

A uniform solid rod of length 2.0 m is clamped at its exact midpoint. It is stroked longitudinally to set up standing waves. If the Young's modulus of the rod's material is 2.0 10¹¹ N m ⁻² and its density is 8000 kg m ⁻³ , the fundamental frequency of longitudinal vibrations in the rod is

Options

  1. A625 Hz
  2. B2500 Hz
  3. C1250 Hz
  4. D6.25 10⁶ Hz

Correct answer

C. 1250 Hz

Step-by-step solution

First, we find the speed of the longitudinal wave in the material of the rod using the relation: v = Y Substituting the given values: v = 2.0 10¹¹ 8000 = 25 10⁶ = 5000 m s ⁻¹ When the rod is clamped at its midpoint, a node is formed at the center and antinodes are formed at the two free ends. For the fundamental mode of vibration, the length of the rod L corresponds to half a wavelength ( 2 ). Therefore, L = 2 = 2L Given L = 2.0 m , the wavelength is: = 2 2.0 = 4.0 m The fundamental frequency f is given by: f = v =

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