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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 = \frac{1}{2} {KA}^{2} = \frac{1}{2} \times 200 \times 1 = 100 J

$Minimum PE = 150 - 100 = 50 J$

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