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A particle is vibrating in a simple harmonic motion with an amplitude of 4 cm. At what displacement from the equilibrium position, is its energy half potential and half kinetic

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$2 \sqrt{2} cm$

$\sqrt{2} cm$

1 cm

3 cm

answer is D.

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Detailed Solution

Let x be the point where K.E. = P.E.
Hence $\frac{1}{2} {mω}^{2} (a^{2} - x^{2}) = \frac{1}{2} {mω}^{2} x^{2}$

$\Rightarrow 2 x^{2} = a^{2}$ $\Rightarrow x = \frac{a}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2 \sqrt{2} cm$

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