How much is KE for displacement equal to half the amplitude?

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Solution

We know that
$v = 2 \sqrt{A^{2} - x^{2}} w h e n x = \frac{A}{2} v = w \sqrt{A^{2} - \frac{A^{2}}{4}} = \frac{\sqrt{3}}{2} w A ∴ k E = \frac{1}{2} m v^{2} = \frac{1}{2} m \frac{3}{4} w^{2} A^{2} = \frac{3}{8} m w^{2} A^{2}$

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How much is KE for displacement equal to half the amplitude?

We know that v=2√A2−x2whenx=A2v=w√A2−A24=√32wA∴kE=12mv2=12m34w2A2=38mw2A2

We know that
$v = 2 \sqrt{A^{2} - x^{2}} w h e n x = \frac{A}{2} v = w \sqrt{A^{2} - \frac{A^{2}}{4}} = \frac{\sqrt{3}}{2} w A ∴ k E = \frac{1}{2} m v^{2} = \frac{1}{2} m \frac{3}{4} w^{2} A^{2} = \frac{3}{8} m w^{2} A^{2}$