A point particle of mass 200 text{ g} is exec | vedclass

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(D) Given: Mass $m = 200 	ext{ g} = 0.2 	ext{ kg}$,Amplitude $A = 0.2 	ext{ m}$,Kinetic Energy at mean position $K_{max} = 16 	imes 10^{-3} 	ext{

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$Y = 0.2 \sin(2t)$

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(D) Given: Mass $m = 200 	ext{ g} = 0.2 	ext{ kg}$,Amplitude $A = 0.2 	ext{ m}$,Kinetic Energy at mean position $K_{max} = 16 	imes 10^{-3} 	ext{ J}$.At the mean position,the velocity is maximum,$v_{max} = A\omega$.The maximum kinetic energy is given by $K_{max} = \frac{1}{2} m v_{max}^2 = \frac{1}{2} m (A\omega)^2$.Substituting the values: $16 	imes 10^{-3} = \frac{1}{2} 	imes 0.2 	imes (0.2 	imes \omega)^2$.$16 	imes 10^{-3} = 0.1 	imes 0.04 	imes \omega^2$.$16 	imes 10^{-3} = 0.004 	imes \omega^2$.$\omega^2 = \frac{16 	imes 10^{-3}}{4 	imes 10^{-3}} = 4$.$\omega = 2 	ext{ rad/s}$.The general equation for $S.H.M.$ is $Y = A \sin(\omega t + \phi)$.Given initial phase $\phi = 0^{\circ}$,so $Y = 0.2 \sin(2t)$.

$A$ point particle of mass $200 \text{ g}$ is executing $S.H.M.$ of amplitude $0.2 \text{ m}$. When the particle passes through the mean position,its kinetic energy is $16 \times 10^{-3} \text{ J}$. The equation of motion of this particle is (Initial phase of oscillation $= 0^{\circ}$)

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