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Question

In a SHM, potential energy of a particle at mean position is E1 and kinetic energy is E2, then

A
E1=E2
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B
total potential energy at x=3A2 is E1+34E2
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C
total potential energy at x=3A2 is 34E2
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D
None of these
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Solution

The correct option is B total potential energy at x=3A2 is E1+34E2
Potential energy of the particle executing SHM is given by,
UUmean=12kx2 .........(1)
From the data given in the question,
Umean=E1
Kinetic of the particle at the mean position is given by
E2=12kA2 {maximum kinetic energy}
So, at x=32A
from (1) we get,
UE1=12k×(3A2)2
U=E1+12kA2×34U=E1+34E2
Thus, option (b) is the correct answer.

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