A particle executes simple harmonic motion with an amplitude of 0.25 m and a period of 2.0 s. What is the maximum speed of the particle? (Use π ≈ 3.14)
Show the answer and explanation
Correct answer
C. 0.79 m/s
Principle or equation
In SHM, the maximum speed is v_max = Aω = A(2π/T).
Why this answer is correct
Given A = 0.25 m and T = 2.0 s, ω = 2π/T = 2 × 3.14 / 2.0 = 3.14 rad/s. v_max = 0.25 × 3.14 = 0.785 m/s ≈ 0.79 m/s.
Example
For amplitude 0.10 m and period 1.0 s, v_max = 0.10 × 2π ≈ 0.63 m/s.
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