AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper

The maximum kinetic energy at an extreme position is given as,
(K.E)max=

1

2

kx2
(T.E−P.E)=

1

2

kx2
Here, the total energy is T.E and the Potential energy is P.E.
Substitute the value in above expression.
(9J−5J)=

1

2

k(10−2)2
4J=

1

2

k10−4
k=8×104J∕m2
The time period of the oscillator is given as,
T=2π√

m

k

T=2π√

2‌kg

8×104J∕m2

T=

π

100

s