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

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$0.25$

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(B) The potential energy $U$ of a simple harmonic oscillator is given by $U = \frac{1}{2}m\omega^2y^2$,where $y$ is the displacement.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 displacement $y = \frac{a}{2}$,we substitute this into the ratio:$\frac{U}{E} = \frac{(\frac{a}{2})^2}{a^2} = \frac{\frac{a^2}{4}}{a^2} = \frac{1}{4} = 0.25$.

When the displacement is half the amplitude,the ratio of potential energy to the total energy is

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