Double Spring System with different K springs, where will weight rest?

[hellocubed]

in this system
http://imgur.com/oZsUW

The left and right springs are different springs with Different constant K.
Even though they are different, the mass in the middle will oscillate Symmetrically between them.... (exactly the same as if they were identical springs).
This is because, even though one has a higher constant K... it means that it pushes AND pulls harder... which "equilibrates" with the other spring... I think.



I was wondering, when the mass is no longer oscillating and is at rest, is it still at a symmetric distance between the walls? I would think that the weight would rest closer/further than the more powerful spring, considering the distance of the walls.....

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[Dasypus]

Durn it, I need to stop answering these things. Milski's right ↓ , solve with opposed forces.

 

[milski]

If there is no energy loss, it will oscillate in a simple harmonic motion. The equilibrium position will not be in the middle between the two walls but somewhat towards the spring with the lower k (provided the springs have equal length when relaxed).

If there is energy loss, they will eventually stop in equilibrium position, which is as already mentioned, closer to the spring with lower k.

For equilibrium:

k1d1=k2d2 or d1/d2=k2/k1.

 

[Dasypus]

So what happens if the springs have different resting lengths?

 

[milski]

So what happens if the springs have different resting lengths?

Just look at Morsetlis' answer below. I was trying to make an entry in the stupid comment of the week competition.

 

[Morsetlis]

In equilibrium, either both springs will be extended, or both springs will be compressed.

The change in length from full length of the weaker spring (lower k) will be GREATER than the change in length from full length of the stronger spring (higher k). Thus, the mass will come to rest (with energy loss) closer to the weaker spring if both springs are compressed, and closer to the stronger spring if both springs are extended.

However since spring lengths can vary, it's easier to think of the resting position of the object as displacement from full length of either spring. The stronger spring has less displacement, and the weaker spring has more displacement.