Average energy mass on a spring

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To find the mean potential energy of a mass on a spring, the maximum potential energy is given by the formula 1/2*k*x0^2. The mean potential energy is indeed half of the maximum potential energy, but the calculation involves considering the potential energy at any instant, expressed as 1/2*k*x^2 = 1/2*k*x0^2*cos^2(ωt+ø). The average value of cos^2(ωt+ø) over a complete cycle is 1/2. Thus, the mean potential energy is confirmed to be half of the maximum potential energy.
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Homework Statement



Hi, I hope this isn't a silly question. I am looking to find the mean potential energy of a mass on a spring with spring constant k and maximum displacement x0.

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The Attempt at a Solution


I know the maximum energy is 1/2*kx0^2 so would the mean potential energy be half that?
 
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Yes, the mean PE will be half of the maximum , but not just like that.
PE at any instant is 1/2 kx2 = 1/2kx02cos2(ωt+ø) ,
avg. of cos2(ωt+ø) over a time period is 1/2.
 
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ah okay thanks!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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