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(B) For a particle executing simple harmonic motion,the force is given by $F = -kx$,where $k$ is the force constant.From the graph,at $x = 20\, cm = 0

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$\left( \frac{5}{2\pi} \right)\, s^{-1}$

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(B) For a particle executing simple harmonic motion,the force is given by $F = -kx$,where $k$ is the force constant.From the graph,at $x = 20\, cm = 0.2\, m$,the force $F = -2.0\, N$.Thus,the magnitude of the force constant is $k = \frac{|F|}{|x|} = \frac{2.0\, N}{0.2\, m} = 10\, N/m$.The mass of the particle is $m = 400\, g = 0.4\, kg$.The frequency of oscillation $f$ is given by the formula $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$.Substituting the values,we get $f = \frac{1}{2\pi} \sqrt{\frac{10}{0.4}} = \frac{1}{2\pi} \sqrt{25} = \frac{5}{2\pi}\, s^{-1}$.

The figure shows the variation of force acting on a particle of mass $400\, g$ executing simple harmonic motion. The frequency of oscillation of the particle is

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