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Concepts Of Physics MCQ Edition [Volume 1]PhysicsWave Motion and Waves on a String

As shown in the figure, a uniform horizontal rod of length 40 cm and mass 1.2 kg is supported by two identical wires. Where must a mass of 4.8 kg be positioned on the rod such that the same tuning fork can excite the left wire into its fundamental vibrations and the right wire into its first overtone? Take g = 10 m s ⁻² .

Options

  1. A10 cm from the left end
  2. B20 cm from the left end
  3. C5 cm from the left end
  4. D5 cm from the right end

Correct answer

C. 5 cm from the left end

Step-by-step solution

Let T_L and T_R be the tensions in the left and right wires respectively. The frequency of the fundamental mode of the left wire is f_L = 1 2l T_L . The frequency of the first overtone of the right wire is f_R = 2 2l T_R . Since both wires are excited by the same tuning fork, their frequencies are equal: 1 2l T_L = 2 2l T_R T_L = 2 T_R T_L = 4T_R For translational equilibrium of the rod, the total upward force equals the total downward force: T_L + T_R = mg + Mg 4T_R + T_R = 1.2g + 4.8g 5T_R = 6.0g T_R = 1.2g Then,

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