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Concepts Of Physics MCQ Edition [Volume 1]PhysicsSimple Harmonic Motion

A particle having a mass of 10 g oscillates according to the equation x = (2.0 cm ) [(100 s ⁻¹) t + /6] . Determine the position, the velocity, and the acceleration of the particle at t = 0 .

Options

  1. A2.0 cm , 0 m/s , -100 m/s ^2
  2. B1.0 cm , -1.73 m/s , 100 m/s ^2
  3. C1.0 cm , 1.73 m/s , -100 m/s ^2
  4. D1.73 cm , 1.0 m/s , -200 m/s ^2

Correct answer

C. 1.0 cm , 1.73 m/s , -100 m/s ^2

Step-by-step solution

The given equation of motion is x = 2.0 (100t + /6) cm . At t = 0 , the position is: x = 2.0 ( /6) = 2.0 1 2 = 1.0 cm The velocity is the rate of change of position: v = dx dt = d dt [2.0 (100t + /6)] = 200 (100t + /6) cm/s At t = 0 , the velocity is: v = 200 ( /6) = 200 3 2 = 100 3 cm/s = 3 m/s 1.73 m/s The acceleration is the rate of change of velocity: a = dv dt = d dt [200 (100t + /6)] = -20000 (100t + /6) cm/s ^2 At t = 0 , the acceleration is: a = -20000 ( /6) = -20000 1 2 = -10000 cm/s ^2 = -100 m/s ^2

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