A uniform rod of length 1.5 m is pivoted at one end and swings as a physical pendulum (I = mL 2 /3, centre of mass at L/2). Taking g = 10 m/s 2 , its time period is:

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A uniform rod of length 1.5 m is pivoted at one end and swings as a physical pendulum (I = mL2/3, centre of mass at L/2). Taking g = 10 m/s2, its time period is:

Answer: 1.99 s.

A 1.99 s
B 2.43 s
C 1.57 s
D 2.81 s

Correct answer: A. 1.99 s

Explanation: T = 2π√(I/mgd) = 2π√((mL2/3)/(mg·L/2)) = 2π√(2L/3g) = 2π√(2×1.5/30) = 2π√0.1 = 1.99 s.

In SHM, displacement and velocity are 90° out of phase (v is maximum when x=0, and zero when x is extreme), while acceleration is always exactly opposite in sign to displacement, consistent with a = −ω²x.

Concept context

Simple harmonic motion, spring-mass, pendulum, energy in SHM, and resonance.

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A uniform rod of length 1.5 m is pivoted at one end and swings as a physical pendulum (I = mL<sup>2</sup>/3, centre of mass at L/2). Taking g = 10 m/s<sup>2</sup>, its time period is:

Concept context

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