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Q.

The total mechanical energy of a harmonic oscillator of amplitude 1m and force constant 200 N/m is 150J. Then

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a

The minimum P E is Zero

b

The minimum P E is 50 J

c

The maximum P E is 50 J

d

The maximum P E is 100 J

answer is C.

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Detailed Solution

TE=12KA2=12×200×1=100J

MinimumPE=150100=50J

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