A particle executing SHM has a maximum speed of \(30\) cm/s and a maximum acceleration of \(60\) cm/s2.  The period of oscillation is:
1. \(\pi \) s
2. \(\dfrac{\pi }{2}\) s
3. \(2\pi\) s
4. \(\dfrac{\pi }{4}\) s

(1) Hint: The particle will have the maximum speed at the mean position and the maximum acceleration at the extreme position.
Step 1: Find the maximum velocity.
Let equation of an SHM is represented by, y = a sin ωt
                                                v=dydt=aωcosωt
                    (v)max=aω=aω=30                               ...(i)
Step 2: Find the maximum acceleration.
Acceleration, (A)=dx2dt2=-aω2 sinωt
                               Amax=ω2a=60                                    ...(ii)
Step 3: Find the period of the oscillation.
From Eqs. (i) and (ii), we get, ω(ωa)=60ω(30)=60
                          ω=2rad/s
                           2πT=2 rad/s T=πsec