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Problem 7 .59, we're told we have a spring that does not obey hook's law, but rather has an extra quadratic term in its force equation.
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These are the particular spring constants we have.
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And what we would like to do first is to get an expression for the potential energy in this spring, with the initial condition, a constant of integration, or whatever you like, that it zero.
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Compression or extension, the potential energy is zero.
00:39
And then we imagine we have a 0 .9 kilogram mass that's pulled one meter away from its 2 -1 meters to the right of the equilibrium position and then let go.
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And then we want to know what the velocity of that mass will be when its exposition is half a meter.
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So, part a.
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Now, we know that the force in a given direction of the partial derivative of the potential in that direction, and through the fundamental theorem of calculus, basically, this means that we can write the potential as this, its value at zero minus the integral from zero to x, of x of what we'll call s our constant of integration or not constant entire variable of integration the s we're told we want this to be zero so is the negative integral from zero to x of negative alpha s minus beta s squared d s this is equal to one half alpha x squared which you would expect because it does have the linear behavior of a spring that's then modified with the quadratic behavior and then we have a one -third beta x cubed now in addition to that so for part b we want to use conservation of energy...