The variation of PE of harmonic oscillator is as shown in figure. The spring constant is
A
1×102Nm−1
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B
1.5×102Nm−1
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C
2×102Nm−1
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D
3×102Nm−1
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Solution
The correct option is B1.5×102Nm−1 From the figure,
Amplitude of harmonic oscillator (A)=(20−0)mm=20mm
Potential energy of harmonivc oscillator 12KA2=(0.04−0.01)J ⇒12KA2=0.03J ⇒K=0.06A2=0.06(0.02)2=150N/m