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Byju's Answer
Standard XII
Physics
Potential Energy Formulae
Potential ene...
Question
Potential energy of a particle at mean position is
4
J
and at extreme position is
20
J
. Given that amplitude of oscillation is
A
. Match the following two columns.
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Solution
U
0
=
1
2
K
A
2
=
U
m
a
x
4
+
1
2
k
A
2
=
20
or
1
2
k
A
2
=
16
J
(1)
U
=
U
0
+
1
2
k
X
2
=
U
0
+
1
2
k
(
A
2
)
2
=
4
+
16
4
=
8
J
(2)
K
E
=
1
2
k
(
A
2
−
A
2
16
)
=
15
16
(
1
2
k
A
2
)
=
15
16
×
16
=
15
J
(3)
K
E
=
1
2
k
(
A
2
−
0
)
=
1
2
k
A
2
=
16
J
(4)
K
E
=
1
2
k
(
A
2
−
A
2
4
)
=
3
4
(
1
2
k
A
2
)
=
3
4
(
16
J
)
=
12
J
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Similar questions
Q.
Force acting on a particle is F = –8x in S.H.M. The amplitude of oscillations is 2 (in m) and mass of the particle is 0.5 kg. The total mechanical energy of the particle is 20 J. The potential energy of the particle at mean position (in J) is
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Q.
When a body is in SHM, match the statements in
Column A with that in Column B
Column A Column B
a) Velocity is maximum e) At half of the amplitude
b) Kinetic energy is f) At the mean position
3/4th of total energy
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√
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/
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Potential energy at mean position of a particle performing oscillation in mean position is
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A particle is describing SHM with amplitude
′
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′
.
When the potential energy of particle is one fourth of the maximum energy during oscillation, then its displacement from mean position will be:
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The ratio of the kinetic energy of a particle executing SHM at its mean position to its potential energy at extreme position is
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