KEAM 2019 Physics and Chemistry Paper

In the given simple harmonic motion, maximum acceleration amax=8πm/s2 and maximum speed vmax=1.6m/s .
As, we know that for simple harmonic motion
amax=ω2A
and ‌‌vmax=ωA
where, A= amplitude of SHM,
⇒‌‌amax=vmaxω .....(i)
So, from Eq.(i), we get
∴ Time period,
T=‌

2π

ω

=‌

2π

5

=1.25s