A particle moves in simple harmonic motion with an amplitude of 0.20 m and a frequency of 2.0 Hz. What is the maximum speed of the particle? (Use π ≈ 3.14)
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Correct answer
C. 2.51 m/s
Principle or equation
For simple harmonic motion, the maximum speed v_max is given by v_max = Aω, where A is the amplitude and ω = 2πf is the angular frequency.
Why this answer is correct
Given A = 0.20 m and f = 2.0 Hz, ω = 2πf = 2 × 3.14 × 2.0 = 12.56 rad/s. Therefore v_max = Aω = 0.20 × 12.56 = 2.512 m/s, which rounds to 2.51 m/s.
Example
For a particle with amplitude 0.10 m and frequency 5.0 Hz, ω = 2π × 5.0 = 31.4 rad/s, so v_max = 0.10 × 31.4 = 3.14 m/s.
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