A particle in SHM has a maximum speed of 30 cm/s and a maximum acceleration of 90 cm/s². The period of oscillation is:
A$\pi/3$ s
B$2\pi/3$ s
C$\pi$ s
D$2\pi$ s
Answer & Solution
Correct answer: B. $2\pi/3$ s
$v_{max} = A\omega = 30$; $a_{max} = A\omega^2 = 90$. Dividing: $\omega = a_{max}/v_{max} = 90/30 = 3$ rad/s. So $T = 2\pi/\omega = 2\pi/3$ s.
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