A particle moves in simple harmonic motion with amplitude 0.40 m and angular frequency 4.0 rad/s. What is the magnitude of the maximum acceleration of the particle?
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Correct answer
C. 6.4 m/s²
Principle or equation
In simple harmonic motion, acceleration a = -ω²x, so maximum acceleration magnitude is a_max = ω²A, where A is amplitude and ω is angular frequency.
Why this answer is correct
Given A = 0.40 m, ω = 4.0 rad/s. a_max = ω²A = (4.0)² × 0.40 = 16 × 0.40 = 6.4 m/s².
Example
If amplitude is 0.10 m and angular frequency is 5.0 rad/s, a_max = (5.0)² × 0.10 = 2.5 m/s².
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