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KEAM 2013 Physics and Chemistry Paper
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© examsnet.com
Question : 29
Total: 120
A particle is executing simple harmonic motion with amplitude A. When the ratio of its kinetic energy to the potential energy is
1
4
,
its displacement from its mean position is
2
√
5
A
√
3
2
A
3
4
A
1
4
A
2
5
A
Validate
Solution:
Given,
K
E
P
E
=
1
4
But we know that in SHM
K
E
=
1
2
K
(
A
2
−
x
2
)
and
P
E
=
1
2
K
x
2
Hence,
1
2
K
(
A
2
−
x
2
)
1
2
K
x
2
=
1
4
A
2
−
x
2
x
2
=
1
4
4
A
2
−
4
x
2
=
x
2
4
A
2
−
5
x
2
=
0
4
A
2
=
5
x
2
x
2
=
4
5
A
2
x
=
2
√
5
A
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