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

A wave travelling on a string has the equation y = (0.10 mm ) [(31.4 m ⁻¹)x + (314 s ⁻¹)t] . What is the maximum displacement and the maximum speed of a portion of the string?

Options

  1. A0.20 mm and 3.14 cm s ⁻¹
  2. B0.10 mm and 1.57 cm s ⁻¹
  3. C0.10 mm and 3.14 cm s ⁻¹
  4. D0.10 mm and 31.4 cm s ⁻¹

Correct answer

C. 0.10 mm and 3.14 cm s ⁻¹

Step-by-step solution

The standard equation of a travelling wave is given by y = A (kx + t) . Comparing the given equation y = (0.10 mm ) [(31.4 m ⁻¹)x + (314 s ⁻¹)t] with the standard equation, we get: Amplitude A = 0.10 mm Angular frequency = 314 s ⁻¹ The maximum displacement of a portion of the string is its amplitude, which is A = 0.10 mm . The maximum speed of a portion of the string (maximum particle velocity) is given by v_ max = A . Substituting the values: v_ max = (0.10 mm ) (314 s ⁻¹) = 31.4 mm s ⁻¹ Converting to cm s ⁻¹ : v_

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