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KCET 2015 Physics Solved Paper
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© examsnet.com
Question : 13
Total: 60
The ratio of kinetic energy to the potential energy of a particle executing SHM at a distance equal to half its amplitude, the distance being measured from its equilibrium position is
3 : 1
4 : 1
2 : 1
8 : 1
Validate
Solution:
Kinetic energy of particle executing SHM is
K
E
=
1
2
m
ω
2
(
A
2
−
y
2
)
Hey
y
=
4
2
∴
K
E
=
1
2
m
ω
2
(
A
2
−
A
2
4
)
=
1
2
m
ω
2
3
A
2
4
Potential energy of particle executing SHM is
P
E
=
1
2
m
ω
2
y
2
=
1
2
m
ω
2
A
2
4
Then ratio of kinetic energy to potential energy is
K
E
P
E
=
(
1
2
)
m
ω
2
(
3
4
)
A
2
(
1
2
)
m
ω
2
A
2
4
=
3
1
Therefore, the ratio is 3 : 1.
© examsnet.com
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