Q.
The potential energy of a particle oscillating on x-axis is given as U=20+(x–2)2 where ‘U’ is in joules and ‘x’ is in metres. Its mean position is
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a
0
b
1
c
2
d
4
answer is C.
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Detailed Solution
at mean position, PE is minimum,ie 2 (x – 2) = 0 x – 2 = 0 x = 2
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