AP EAPCET 25-Aug-2021 Shift 2 Paper

Given, maximum speed of SHM is 40ms−1.
vmax=aω=40ms−1 ...(i)
Maximum acceleration of SHM is 60ms−2.
As, amax=aω2=60ms−2
Dividing Eq. (ii) by Eq. (i), we get

ω2

ω

=

60

40

⇒ ω=

3

2

s−1
We know, ω=2πv
⇒

3

2

=2π

1

T

⇒

1

T

=

3

4π

⇒ T=

4π

3

s
Hence, time period of the oscillation is

4π

3

s.