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(A) The potential energy $U$ of a simple harmonic oscillator at a displacement $y$ is given by $U = \frac{1}{2} m \omega^2 y^2$.The total energy $E$ o

Vedclass · Accepted Answer

$\frac{1}{4} E$

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(A) The potential energy $U$ of a simple harmonic oscillator at a displacement $y$ is given by $U = \frac{1}{2} m \omega^2 y^2$.The total energy $E$ of the oscillator is given by $E = \frac{1}{2} m \omega^2 A^2$,where $A$ is the amplitude.When the particle is halfway to its endpoint,the displacement $y$ is half of the amplitude,i.e.,$y = \frac{A}{2}$.Substituting this value into the potential energy formula:$U = \frac{1}{2} m \omega^2 (\frac{A}{2})^2$$U = \frac{1}{2} m \omega^2 (\frac{A^2}{4})$$U = \frac{1}{4} (\frac{1}{2} m \omega^2 A^2)$Since $E = \frac{1}{2} m \omega^2 A^2$,we get:$U = \frac{1}{4} E$.

The potential energy of a simple harmonic oscillator,when the particle is halfway to its endpoint,is

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