Bead attached to a spring and moving along a horizontal wire

AI Thread Summary
The discussion revolves around finding the potential energy function U(x) for a bead attached to a spring on a frictionless wire. The book's solution states that U(x) = (1/2)k((h²+x²)^(1/2) - l0)², which the user initially disputes, arguing that only the x-component of the spring's force should be considered. However, a counterpoint is raised regarding energy conservation, suggesting that the book's calculation accurately reflects the energy stored in the spring when released. Ultimately, both solutions are shown to differ only by an additive constant, indicating that they are both valid representations of the potential energy. The discussion highlights the importance of considering all components of force and energy in such problems.
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I'm self-studying an introductory book on mathematical methods and models and came across the following problem:

1. A bead of mass m is threaded onto a frictionless horizontal wire. The bead is attached to a model spring of stiffness k and natural length l0, whose other end is fixed to a point A at a vertical distance h from the wire (where h > l0). The position x of the bead is measured from the point on the wire closest to A. Find the potential energy function U(x).

Homework Equations



I'm rather puzzled by the solution given in the book, which claims that since the length of the spring is (h2+x2)1/2 and its extension is (h2+x2)1/2 - l0, then U(x) = (1/2)k((h2+x2)1/2 - l0)2.

The Attempt at a Solution



I think that's incorrect because only the x-component of the force exerted by the spring on the bead is relevant to the calculation of U(x). The x-component is -k(l-l0)cos\theta, where l = (h2+x2)1/2 and cos\theta = x/(h2+x2)1/2. Hence, U(x) is -\int(-kx + kl0x/(h2+x2)1/2)dx. Is this correct?
 
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Humm. Interesting question. I'm not sure I can answer it. At first thought you are correct, but then ...

With the spring stretched the energy stored in the spring will be that calculated in the book. What happens to the energy when the spring is released? Where does it go if you apply conservation of energy? The wire is frictionless so as far as I can see all the energy ends up in the bead (eg not some fraction of that due to the Cosθ issue you raise).

So I conclude the book is correct but I'm willing to be convinced you are right and the book is wrong.
 
See if you can show that both expressions are correct, although they might differ by an additive constant.
 
The integral should evaluate to

kx2/2 - kl0(h2 + x2)1/2 + C.

Hence, the difference between my solution and the book's is:

kx2/2 - kl0(h2 + x2)1/2 + C - (k(h2 + x2)1/2l0 + (1/2)kh2 + (1/2)kx2 - (1/2)kl02)
= C - (1/2)kh2 + (1/2)kl02,

so the two solutions do appear to differ by a constant...
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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