Question

A rod of mass M and length L is pivoted about its end O as shown in the figure. Find the period of SHM.

• A. (A) 2π3L2g
• B. (B) 2π2L3g
• C. (C) 2πE9
• D. (D) 2πL3g

Answer 1

Answer: (B)

(B) 2π2L3g

Answer 2

Explanation:
τ=MgL2sin⁡θFor the small angle, sin⁡θ≈θAnd we have τ=Iα=Id2θdt2So, (ML23)d2θdt2=−MgL2θd2θdt2=−3g2LθSo, the angular frequency is ω=3g2L∴T=2π2L3gHence, the correct option is (B).