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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For a particle executing SHM of amplitude a , angular velocity 0 , displacement $x$ then Kinetic energy
$(KE) = \frac{1}{2} m \omega^{2} (a^{2} - x^{2}) Potential energy (PE) = \frac{1}{2} m \omega^{2} x^{2} Given, KE = PE \quad \frac{1}{2} m \omega^{2} a^{2} - \frac{1}{2} m \omega^{2} x^{2} = \frac{1}{2} m \omega^{2} x^{2} \frac{1}{2} m \omega^{2} a^{2} = m \omega^{2} x^{2} \Rightarrow x = \frac{a}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2 \sqrt{2} \text{ cm}$
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