Concepts Of Physics MCQ Edition [Volume 1]PhysicsSimple Harmonic Motion
A pendulum is formed by attaching a hollow sphere of radius 2 cm to a thread of length 18 cm . Determine the time period of oscillation of this pendulum. Also, how does this period compare to the value calculated using the simple pendulum formula?
Options
- A0.92 s ; it is about 1.5 % smaller than the simple pendulum value
- B0.89 s ; it is about 0.3 % larger than the simple pendulum value
- C0.89 s ; it is about 0.3 % smaller than the simple pendulum value
- D0.92 s ; it is about 1.5 % larger than the simple pendulum value
Correct answer
B. 0.89 s ; it is about 0.3 % larger than the simple pendulum value
Step-by-step solution
The distance from the pivot to the center of mass of the hollow sphere is d = L + R = 18 + 2 = 20 cm = 0.2 m . The moment of inertia of the hollow sphere about the pivot is given by the parallel axis theorem: I = I_ cm + md^2 = 2 3 mR^2 + md^2 The time period of the physical pendulum is: T = 2 I mgd = 2 2 3 mR^2 + md^2 mgd = 2 d g (1 + 2R^2 3d^2 ) The time period of a simple pendulum of length d is T₀ = 2 d g . Using the binomial expansion for R d , we get: T T₀ (1 + R^2 3d^2 ) Substituting R = 2 cm and d = 20 cm :