A mass is resting on a horizontal surface without friction. It is connected on both sides by slinky springs k1 and k2 (that is they have different spring constants, and the equilibrium position will not be in the center of the system). First I had to find the equation of motion after setting up the forces. I've seen an example of a mass connected on either side by a spring and k=k1+k2. This is because whichever side of equilibrium you place the mass, the forces act in the same direction, one pulls and one pushes, basically. In this however, the slinky springs naturally are of length L0, or they are close to zero before stretched. Therefore both springs are pulling away from the center no matter where you set the displacement. It is just that one is pulling more than the other, being displaced further. (This is my reasoning so far.) So if I displace to the right, I say sum of forces=ma=-k1x+k2x. That is the force of k1 pulls to the left while k2 to the right. So I have mx"+(k1-k2)x=0 or x"+((k1-k2)/m)x=0. Is this right? Or should I have done (K1+k2)?(adsbygoogle = window.adsbygoogle || []).push({});

Thanks, Josh

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# Homework Help: Slinky springs and eq of motion.

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