A particle undergoes simple harmonic motion with amplitude 0.30 m and period 4.0 s. What is the maximum speed of the particle? (Use π ≈ 3.14)
Show the answer and explanation
Correct answer
B. 0.47 m/s
Principle or equation
In SHM, maximum speed v_max = ωA, where ω = 2π/T.
Why this answer is correct
ω = 2π/4.0 = 1.57 rad/s. v_max = 1.57 × 0.30 = 0.471 m/s ≈ 0.47 m/s.
Example
If A = 0.10 m and T = 2.0 s, v_max = (2π/2)(0.10) = 0.314 m/s.
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