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

A simple pendulum measuring 40 cm in length is placed inside a deep mine. Assuming the mine's depth to be 1600 km , determine the time period of the pendulum at that location. The radius of the earth is 6400 km .

Options

  1. A1.13 s
  2. B1.27 s
  3. C1.47 s
  4. D1.79 s

Correct answer

C. 1.47 s

Step-by-step solution

The acceleration due to gravity at a depth d is given by: g' = g (1 - d R ) Substituting d = 1600 km and R = 6400 km : g' = g (1 - 1600 6400 ) = g (1 - 1 4 ) = 3g 4 The time period of a simple pendulum is: T = 2 L g' Given L = 40 cm = 0.4 m and taking g = 9.8 m/s ^2 : T = 2 0.4 3 9.8 4 = 2 1.6 29.4 T = 2 0.0544 2 3.14 0.233 1.466 s Rounding off to two decimal places, T 1.47 s .

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