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Springs in parallel? 
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#1
Jan1313, 09:38 AM

P: 8

Hi. I have a mass between 2 springs. See attachment. Both springs are compressed.
I'm trying to convince myself that the springs act in parallel. I've done the calculations and it checks out ok, but I'm struggling to visualise this. In my mind, if an external force is applied to the mass to move it to the right, then the left hand spring is assisting this motion which goes against the idea that the springs are in parallel. Can anyone help me get my head around this? Thanks. 


#2
Jan1313, 09:59 AM

P: 11

I don't quite agree that the left hand spring is assisting the motion. It is in fact opposing it. Just think of it, it would certainly be easier to displace the body if that spring weren't there.
As far as why they're considered to be in parallel, you can see it from the math if you don't feel it intuitively. You can see that each spring is displaced by half of what it would have been without the other spring, as is the case in parallel springs. 


#3
Jan1313, 10:03 AM

P: 208

For parallel springs, each spring is displaced by the same amount as it would be if the other one weren't there. If you don't like to picture it with one spring on each side, pretend they're both on the same side and you can see why the displacements are equal. Either way, Tik is absolutely incorrect in his statement.



#4
Jan1313, 10:08 AM

P: 8

Springs in parallel?
Thanks for your replies.
If the springs were not precompressed, then yes, I can completely see that the left hand spring would be opposing the motion. Moving the mass to the right would compress the right hand spring and extend the left hand spring, both opposing the motion. But with precompressed springs I'm finding it difficult to understand why the left hand spring isn't assisting the motion. 


#5
Jan1313, 10:13 AM

P: 11

I'm sorry, I didn't notice the fact that the springs were precompressed.



#6
Jan1313, 10:35 AM

P: 504

You are correct that with an initial push to the right, the spring on the left is still helping push it to the right. However, the initial compression of the spring on the right makes up for this.
I think it will make more sense to you if you set k=1 and graph the force of each spring and the net force as well. 


#7
Jan1313, 10:36 AM

P: 504

You may be getting confused by setting up the spring on the right. If you do it rigorously, you actually have three negative signs multiplying together



#8
Jan1313, 10:43 AM

P: 504

Spring force
Fs=kXs The extension of the spring Xs is actually the opposite of the coordinate x Xs=X The force Fs Points in the x direction F = Fs Putting it together F = Fs F = (kXs) F = (k(X)) F = kX 


#9
Jan1313, 04:32 PM

Sci Advisor
PF Gold
P: 11,889

One would have an effective k of k1 +k2 (side by side, with the forces adding because the lengths are the same) one would have a k of 1/ (1/k1 +1/k2) (end to end where the force is the same for both springs and the one which interests you would, I think, have an effective spring constant of k1k2. So I think the electrical resistor analogue of your set up would be more like a three terminal arrangement of two resistors in series with the measurement point at the centre and the end points at two different potentials. 


#10
Jan1413, 09:18 AM

P: 8

Sophiecentaur, thanks for your reply, but you statement 'k1k2' doesn't make sense. If the 2 springs have the same spring constant k, then you are saying that the overall spring constant is zero!
This is exactly what I'm struggling with. My feeling is that the springs should 'cancel themselves out', but I know that maths says the springs act in parallel. 


#11
Jan1413, 09:30 AM

Sci Advisor
PF Gold
P: 11,889

The mass will be in in a potential well when in equilibrium. The force when displaced will be x(k1k2), won't it? I think that means the effective k, for displacements from equilibrium, has to be k1k2. This, to me, seems like a star of resistors, in some way, as the neutral position corresponds to the divided potential between the supply terminals and current needs to be added or subtracted from the node in order to change its potential.
[Not sure the k1k2 thing is a simple as that, now I've thought some more. I must think some more, some more!] 


#12
Jan1413, 11:07 AM

Sci Advisor
PF Gold
P: 11,889

Yep. I've thought some more and, of course, there's less force from the left spring and more force from the right spring and the changes are xk1 and xk2  in the same direction.
The red herring that was fooling me was the fact that the forces are cancelling out at equilibrium  but that has nothing to do with what happens to the forces as you displace the mass. I should have gone further with my electrical analogy. It's obvious that you can decouple a point in a circuit by connecting capacitors either to 'rail' or to ground  or to both. They appear in parallel. 


#13
Jan1513, 07:34 AM

P: 8

Thanks for everyone's response, but unfortunately I still haven't got my head around this. I can't understand how the left spring can be adding to the overall stiffness when it is effectively assisting the motion to the right.
If the springs are the same, then we are saying that the overall stiffness is 2k. If I remove the left spring then the stiffness is k. I can't understand how the stiffness is halved by removing an assisting spring!! The more I think about it the less I'm convinced the springs are acting in parallel ... 


#14
Jan1513, 08:39 AM

P: 907

The left hand spring is _assisting_ _less_ as the object is displaced further right.
The right hand spring is _resisting_ _more_ as the object is displaced further right. The delta is in the same direction in both cases. The two effects add. 


#15
Jan1513, 08:46 AM

Mentor
P: 41,301

Imagine that the compressed springs in their initial positions each exert a force F_{0} on the mass. The net force is zero, thus the mass is in equilibrium. Now let's move that mass a distance Δx to the right of equilibrium and figure out how much force the springs exert in that position. The right spring must be compressed further, so the force that it now exerts on the mass is F_{0} + ΔF to the left, where ΔF = kΔx. Now consider the left spring. It used to exert a force of F_{0} to the right, but now it is not as compressed. Thus the force it exerts on the mass is now only F_{0}  ΔF to the right. So what's the net force on the mass in its displaced position? ƩF = (F_{0}  ΔF) (F_{0} + ΔF) = 2ΔF = 2kΔx. Thus the restoring force from the two springs is equivalent to a single spring of twice the stiffness. 


#16
Jan1513, 08:47 AM

P: 8

Ah yes, brilliant. That makes sense. Thank you.



#17
Jan1513, 08:51 AM

P: 8

thanks doc al, that's further clarified things. I'm now fully convinced the springs act in parallel. Thanks again to everyone.



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