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Question

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


A
1×102 Nm1
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B
1.5×102 Nm1
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C
2×102 Nm1
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D
3×102 Nm1
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Solution

The correct option is B 1.5×102 Nm1
From the figure,
Amplitude of harmonic oscillator (A)=(200) mm=20 mm
Potential energy of harmonivc oscillator
12KA2=(0.040.01) J
12KA2=0.03 J
K=0.06A2=0.06(0.02)2=150 N/m

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