The potential energy of a simple harmonic osc | vedclass

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

Vedclass · Accepted Answer

$\frac{1}{4}E$

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(B) The potential energy $U$ of a simple harmonic oscillator at a displacement $y$ is given by $U = \frac{1}{2}m\omega^2y^2$.The total energy $E$ of the oscillator is given by $E = \frac{1}{2}m\omega^2a^2$,where $a$ is the amplitude.The ratio of potential energy to total energy is $\frac{U}{E} = \frac{\frac{1}{2}m\omega^2y^2}{\frac{1}{2}m\omega^2a^2} = \frac{y^2}{a^2}$.Given that the particle is half way to its end point,the displacement is $y = \frac{a}{2}$.Substituting this value into the ratio,we get $\frac{U}{E} = \frac{(\frac{a}{2})^2}{a^2} = \frac{a^2/4}{a^2} = \frac{1}{4}$.Therefore,$U = \frac{1}{4}E$.

The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where $E$ is the total energy)

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