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Q.

The variation of potential energy of harmonic oscillator is as shown in figure. The spring constant is 

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a

1×102N/m

b

150 N/m

c

0.667×102N/m

d

3×102N/m

answer is B.

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Detailed Solution

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Total potential energy = 0.04 J
Resting potential energy =0.01 J
Maximum kinetic energy =(0.04–0.01)
=0.03J=12mω2a2=12ka2
0.03=12×k×2010002
k=0.06×2500N/m=150N/m

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