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(I) $(a)$ At what displacement of a $S H O$ is the energy half kinetic and half potential? (b) What fraction of the total energy of a SHO is kinetic and what fraction potential when the displacement is one third the amplitude?

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a. the displacement is $x=\frac{A}{\sqrt{2}}$b. since the total energy is the sum of kinetic and potential energies, the kinetic energy will be $\frac{8}{9}$times that of the total energy.

Physics 101 Mechanics

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So at what displacement would with the, um with the potential energy and with the potential energy be half of the total energy. Ah, making, of course, the kinetic energy. Also equaling half of the total energy would be the position of a simple harmonic oscillator. Given that this occurs. So we can say that for party the, um, potential energy of the spring should be equal to 1/2 times the total energy. And we know that the total energy is equaling 1/2 K a squared, and this must be equal to, um, the potential energy of a spring. So essentially half of k x squared. So we can say that, um, 1/2 k x squared would be equal to 1/4 K a squared. And so we find that X is gonna be equal to a divided by radical, too. So this would be the position such that half of the total energy is potential, while the other half of the total energy is kinetic. And then for part B of A. And this is only for a simple harmonic oscillator. And then for part B, we see that exes equaling the amplitude divided by three Therefore, the ah potential injury over the total energy would be equal to 1/2 k x squared, divided by 1/2 K a squared, and so this would be equal to a over three squared times, one over a squared, and this is equally one over nine. Therefore, the potential energy is gonna be, um, 1/9 of the total energy. So we can say that potential energy equals total energy over nine. And then the kinetic energy is, of course, eight times the total amount of energy divide by nine. So this would be the two answers for part B. And again that that answer for party would be ex equaling the amplitude divided by radical, too. That is the end of the solution. Thank you for watching.

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