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Question

(A) The kinetic energy of a particle in simple harmonic motion is given by $K = K_0 \cos^2(\omega t)$.Since the maximum value of $\cos^2(\omega t)$ is

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$K_0, K_0$

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(A) The kinetic energy of a particle in simple harmonic motion is given by $K = K_0 \cos^2(\omega t)$.Since the maximum value of $\cos^2(\omega t)$ is $1$,the maximum kinetic energy is $K_{max} = K_0$.In simple harmonic motion,the total energy $E$ remains constant and is equal to the maximum kinetic energy or the maximum potential energy.Thus,$E = K_{max} = K_0$.Since the total energy $E = K + U$,where $U$ is the potential energy,we have $U = E - K = K_0 - K_0 \cos^2(\omega t) = K_0 \sin^2(\omega t)$.The maximum value of potential energy is $U_{max} = K_0$.Therefore,the maximum potential energy is $K_0$ and the total energy is $K_0$.

$A$ particle executing simple harmonic motion has a kinetic energy $K = K_0 \cos^2(\omega t)$. The maximum values of the potential energy and the total energy are respectively:

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