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

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$0.667 \times 10^{2} N / m$

$3 \times 10^{2} N / m$

$1 \times 10^{2} N / m$

$1.5 \times 10^{2} N / m$

answer is B.

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Detailed Solution

$P E_{\min} = 0.01 J$
$\begin{array}{l} P E_{\max} = 0.04 = T E \\ T E = 0.04 = P E_{\min} + \frac{1}{2} K A^{2} \\ 0.03 = \frac{1}{2} K {(2 c m)}^{2} \\ 0.06 = K \times 4 \times 10^{- 4} \\ k = 150 N / m \end{array}$

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