NECO 2009 Physics · Question 30

A body executing simple harmonic motion has an angular speed of 2π radians. Its period of oscillation is (π 3.14).

Options

A.0.05s
B.1.00s
C.2.00s
D.3.14s
E.6.28s

Answer: B. 1.00s

Explanation written with AI and checked against the answer key

Analyze the question: We are given the angular speed (angular frequency) ω = 2π rad/s (the unit is presumably per second) and asked for the period T. Concept: In simple harmonic motion, angular frequency and period are related by ω = 2π/T, so T = 2π/ω Also, ω = 2πf, where f is the frequency. Solution: T = 2π / ω T = 2π / 2π T = 1.00 s Check using frequency: f = ω/2π = 1 Hz, so T = 1/f = 1 s ✔ Answer: The period is 1.00 s, which is option B. Tip: When ω is given as a multiple of π, such as 2π, the π cancels in T = 2π/ω. You don't need to use 3.14 at all. Whenever ω = 2π, T = 1 s. Always check whether the π cancels before substituting the number.

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