Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

$0$ and $2K_0$

$K_0/2$ and $K_0$

$K_0$ and $2K_0$

$K_0$ and $K_0$

Step-by-Step Solution

Analyze Kinetic Energy: The given kinetic energy is $K = K_0 \cos^2(\omega t)$ . The maximum value of $\cos^2(\omega t)$ is 1. Therefore, the maximum kinetic energy is $K_{max} = K_0$ .
Total Energy ( $E$ ): In simple harmonic motion, the total mechanical energy is conserved and is equal to the maximum kinetic energy (at the mean position) or the maximum potential energy (at the extreme position) . Thus, Total Energy $E = K_{max} = K_0$ .
Maximum Potential Energy ( $U_{max}$ ): Since the total energy is the sum of kinetic and potential energies ( $E = K + U$ ), the potential energy is maximum when the kinetic energy is minimum (zero). At this point, the entire energy is potential. Therefore, $U_{max} = E = K_0$ .
Conclusion: Both the maximum potential energy and the total energy are equal to $K_0$ .

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