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(D) Given:Force constant $k = 6 	imes 10^5 \, N/m$Amplitude $A = 4 \, cm = 4 	imes 10^{-2} \, m$Total energy $E = 600 \, J$In a linear harmonic osci

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All of these

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(D) Given:Force constant $k = 6 	imes 10^5 \, N/m$Amplitude $A = 4 \, cm = 4 	imes 10^{-2} \, m$Total energy $E = 600 \, J$In a linear harmonic oscillator,the maximum potential energy is equal to the total energy of the system when the displacement is at its maximum (at the extreme positions).$U_{max} = \frac{1}{2} k A^2 = \frac{1}{2} 	imes (6 	imes 10^5) 	imes (4 	imes 10^{-2})^2 = 3 	imes 10^5 	imes 16 	imes 10^{-4} = 480 \, J$.Since the total energy $E = 600 \, J$ and the maximum potential energy (at amplitude) is $480 \, J$,this implies that the potential energy at the equilibrium position (minimum potential energy) is $E - U_{max} = 600 \, J - 480 \, J = 120 \, J$.Also,the maximum kinetic energy $K_{max}$ is equal to the total energy $E$ minus the minimum potential energy $U_{min}$.$K_{max} = E - U_{min} = 600 \, J - 120 \, J = 480 \, J$.Since all statements $A$,$B$,and $C$ are correct,the correct option is $D$.

Column-$I$	Column-$II$
$(a)$ $y = \frac{A}{\sqrt{2}}$	$(i)$ $K = \frac{3E}{4}$
$(b)$ $y = \frac{\sqrt{3}A}{2}$	$(ii)$ $K = \frac{E}{4}$
	$(iii)$ $K = \frac{E}{2}$

$A$ linear harmonic oscillator of force constant $6 \times 10^5 \, N/m$ and amplitude $4 \, cm$ has a total energy of $600 \, J$. Select the correct statement.

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