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(A) In a harmonically oscillating spring-block system,the total energy is the sum of potential energy $\left(\frac{1}{2} kx^{2}\right)$ and kinetic en

Vedclass · Accepted Answer

Electrostatic energy to potential energy and magnetic energy to kinetic energy.

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(A) In a harmonically oscillating spring-block system,the total energy is the sum of potential energy $\left(\frac{1}{2} kx^{2}\right)$ and kinetic energy $\left(\frac{1}{2} mv^{2}\right)$.In an $LC$ circuit,the total energy is the sum of electrostatic energy stored in the capacitor $\left(\frac{1}{2} \frac{q^{2}}{C}\right)$ and magnetic energy stored in the inductor $\left(\frac{1}{2} LI^{2}\right)$.Comparing these,the electrostatic energy $\left(\frac{1}{2} \frac{q^{2}}{C}\right)$ is analogous to the potential energy of the spring $\left(\frac{1}{2} kx^{2}\right)$,as both depend on the displacement/charge.The magnetic energy $\left(\frac{1}{2} LI^{2}\right)$ is analogous to the kinetic energy of the block $\left(\frac{1}{2} mv^{2}\right)$,as both depend on the velocity/current.

If a $LC$ circuit is considered analogous to a harmonically oscillating spring-block system,which energy of the $LC$ circuit would be analogous to potential energy and which one analogous to kinetic energy?

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