AP EAMCET Engineering 22 Sep 2020 Shift 1 Paper

Kinetic energy of a body performing simple harmonic motion is given as
K=‌

1

2

mv2 ....(i)
where, v=‌

dy

dt

=‌

d

dt

⋅a‌sin‌ω‌t [∵y=a‌sin‌ω‌t]
v=aω‌cos‌ω‌t
∴ From Eq. (i), we get
K=‌

1

2

m(aω‌cos‌ω‌t)2=‌

1

2

ma2ω2cos2ωt
K=‌

1

2

ma2ω2(‌

1+cos‌2‌ω‌t

2

)
The graph represented in option (a) is correct for expression of kinetic energy represented by Eq. (i).