# Two spring system

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The ends of two identical springs are connected. Their unstretched lengths l are negligibly small and each has spring constant k. After being connected, both springs are stretched an amount L and their free ends are anchored at y = 0 and x = +/- L as shown (see attachment). The point where the springs are connected to each other is now pulled to the position (x , y). Assume that (x , y) lies in the first quadrant.

A. What is the potential energy of the two-spring system after the point of connection has been moved to position (x , y)? Keep in mind that the unstretched length l of each spring is much less than L and can be ignored (i.e., l << L).

Express the potential in terms of k, x, y, and L.

B. Find the force F on the junction point, the point where the two springs are attached to each other.

Express F as a vector in terms of the unit vectors x and y.

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##### Solution Summary

The potential energy of the two-spring systems are examined. The force on the junction point are determined.

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Point C is the (x,y) point. We have to determine the lengths of the two springs in the given configuration.

________

Length of spring 1 = BC = √(L-x)2+y2

_________

Length of spring 2 = CA = √(L+x)2+y2

Part A : Potential energy stored in a spring = ½ kd2 where d is the extension/compression of spring (change in length from its normal length). In the present case as the normal length is negligible as compared to the extended length, d ...

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