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The force on the centroid is NOT = mg. It is m(g - T) where T is the tension in the spring at the position of the centroid. I think that a more rigorous treatment will find that the topmost coil experiences F = ma = mg since T has a purely downward component. The way to sort this out is to conduct an experiment comparing the fall of a slinky and a ...


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what you need to do is replace the springs with equal and opposite forces on both ends and do a Free Body Diagram. The answer comes straight out of this.


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Hard to answer this without just suggesting an answer. Consider the forces on the mass. There is gravity that is downwards, but I guess the mass is not moving so think about the forces on the mass - you could also think about the forces the point where the two springs joint together. Hope this is helpful.


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Take a look at the junction of the two springs. If you displace the block and hold it there, the junction will also be at rest. At the junction, two forces are pulling it, $k_1 x_1$ to the left and $k_2 x_2$ to the right. Since the junction is at rest, the two forces must balance. Therefore, $$k_1 x_1 = k_2 x_2$$ EDIT 1: Another calculus-based ...


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simple. the box is confined in space by two springs. there are only two forces acting on the box, $F_1$ and $F_2$, exerted by those two springs. Well, the two forces must balance each other out by Newton's third law, or the law of action-reaction pairs. so we get $$ F_1 = -F_2 = -k_1 x_1 = k_2 (-x_2) --> k_1 x_1 = k_2 x_2 $$ (note that $x_1$ and $x_2$ ...


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You actually do not need to solve the equation at all in order to determine that, whatever the answer, it must be the same for A and B. The only difference between A and B is the direction of k1. However if you look at $F=-kx$ is always a restoring force whose magnitude is proportional to the magnitude of the displacement from equilibrium. So the mass ...


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I'll get you started. It's not as bad as you might think at first glance. I think it's easiest to see how to proceed using vector notation. Let $\mathbf r_1, \mathbf r_2, \mathbf r_3$ denote the positions of the three masses in the plane containing the hoop. Let $\ell_{ij}$ denote the length of the spring attaching mass $i$ to mass $j$, then ...


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The are no means to hold kinetic energy in MASSLESS spring. So, the potential energy of extension, will instantly converted to infinite speed of contracting spring. If spring is absolutely rigid itself, it will contract to equivalent minimal length and will start to extending back. Once it extends to initial length, it will start to contract back, and so ...


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I don't think this is the correct answer to the problem. The correct answer is pretty straightforward and simple (it's a sine function times a constant). Check out Kleppner and Kolenkow (Intro to Mechanics, page 128) for the correct solution. Also, I've done this problem using the Lagrangian, but I've never seen Laplace Transforms being used for this problem ...


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I haven't combed through your math, but I suspect that you made some mistakes along the way. For instance, in your first set of equations $\ddot{x}_1(t)=-(x_1-x_2)$ should actually be $\ddot{x}_1(t)=(x_2-x_1-L)$. You can check yourself on this fact because the force ought to be $0$ when $d=L$. Furthermore, you need to be careful about your minus signs ...



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