A particle undergoes simple harmonic motion with amplitude A = 0.2 m and period T = 2.0 s. What is its maximum acceleration?
Answer and explanation
Correct answer: B. 1.97 m/s^2
- Step 1: Angular frequency: omega = 2pi / T = 2pi / 2.0 = pi rad/s = 3.1416 rad/s.
- Step 2: Maximum acceleration in SHM is a_{max} = omega^2 * A.
- Step 3: Calculate: a_{max} = pi^2 0.2 = 9.8696 0.2 = 1.974 m/s^2.
Answer
1.97 m/s^2.
Why not the others: 0.63 m/s^2 is the maximum velocity v_{max} = omega A = pi 0.2; 3.95 m/s^2 uses period T = 1 s; 0.99 m/s^2 computes omega * A / 2.
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