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(B) The total energy $(E)$ of a particle in simple harmonic motion is given by $E = \frac{1}{2} m \omega^2 a^2$,where $a$ is the amplitude.The kinetic

Vedclass · Accepted Answer

$3/4$

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(B) The total energy $(E)$ of a particle in simple harmonic motion is given by $E = \frac{1}{2} m \omega^2 a^2$,where $a$ is the amplitude.The kinetic energy $(K)$ at any displacement $(x)$ is given by $K = \frac{1}{2} m \omega^2 (a^2 - x^2)$.Given that the displacement $x = \frac{a}{2}$,we substitute this 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 $K = \frac{3}{4} E$.Therefore,the fraction of the total energy that is kinetic is $3/4$.

In a simple harmonic motion,when the displacement is one-half the amplitude,what fraction of the total energy is kinetic?

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