In a simple harmonic oscillation,what fractio | vedclass

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(C) The total mechanical energy $E$ of a particle in simple harmonic motion is given by $E = \frac{1}{2} m \omega^{2} A^{2}$.At any position $x$,the k

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$3/4$

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(C) The total mechanical energy $E$ of a particle in simple harmonic motion is given by $E = \frac{1}{2} m \omega^{2} A^{2}$.At any position $x$,the kinetic energy $K$ is given by $K = \frac{1}{2} m \omega^{2} (A^{2} - x^{2})$.The particle is midway between the mean position $(x = 0)$ and the extreme position $(x = A)$,so $x = \frac{A}{2}$.Substituting $x = \frac{A}{2}$ into the kinetic energy formula:$K = \frac{1}{2} m \omega^{2} (A^{2} - (\frac{A}{2})^{2})$$K = \frac{1}{2} m \omega^{2} (A^{2} - \frac{A^{2}}{4})$$K = \frac{1}{2} m \omega^{2} (\frac{3A^{2}}{4})$$K = \frac{3}{4} (\frac{1}{2} m \omega^{2} A^{2})$Since $E = \frac{1}{2} m \omega^{2} A^{2}$,we have $K = \frac{3}{4} E$.Thus,the fraction of total mechanical energy in the form of kinetic energy is $3/4$.

In a simple harmonic oscillation,what fraction of total mechanical energy is in the form of kinetic energy,when the particle is midway between mean and extreme position?

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