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:
- $\pi/3$ s
- $2\pi/3$ s
- $\pi$ s
- $2\pi$ s
Show answer and 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.**