Equivalent Spring Constant in Series and Parallel Configurations

In summary: You are correct as far as the derivation goes. You need to get the relation between the extension of the total length (x1+x2) and the external force mg. The natural lengths are irrelevant.
  • #1
Abhishekdas
208
0
Equivalent spring constant...

Homework Statement


This is a general doubt i have...when two springs of spring constants k1 and k2 are connected in series the equivalent k is given by 1/k=1/k1+1/k2 and when they are in parallel it is k=k1+k2...Now in these cases do we assume that the natural length of the springs are same?


Homework Equations





The Attempt at a Solution


I was trying to derive these relations...

The one for parrallel seems simple...If we hang a mass the force acting on it is (k1+k2)x = mg (i have assumed the initial lengths of springs are same so the extansion is also same for both)
so so a spring having k=k1+k2 acts identical to having these two in parallel...So tell me if this was correct...

And i am not getting how to derive it for the series connection...
I was going lke this...correct me all the way...
I hang a weight to this setup...now springs are mass less so at the point of intersection the force due to upper spring=force due to lower spring ...
so k1*x1=k2*x2
x1 and x2 are the respective extensions...
and kx2 supports mg so kx2 = mg...this was as far as i got ...

So please help me out with these derivations and answer my earlier doubt(abt the lengths being equal)...

Thank you...
 
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  • #2


Abhishekdas said:

Homework Statement


This is a general doubt i have...when two springs of spring constants k1 and k2 are connected in series the equivalent k is given by 1/k=1/k1+1/k2 and when they are in parallel it is k=k1+k2...Now in these cases do we assume that the natural length of the springs are same?

I hang a weight to this setup...now springs are mass less so at the point of intersection the force due to upper spring=force due to lower spring ...
so k1*x1=k2*x2
x1 and x2 are the respective extensions...
and kx2 supports mg so kx2 = mg...this was as far as i got ...

So please help me out with these derivations and answer my earlier doubt(abt the lengths being equal)...
You are trying to find the effective spring constant k such that:

mg = k(x1+x2)

So write out the equations for x1 and x2 in terms of mg and k1, k2.

AM
 
  • #3


Andrew Mason said:
You are trying to find the effective spring constant k such that:

mg = k(x1+x2)

So write out the equations for x1 and x2 in terms of mg and k1, k2.

AM

Are you talking about the parallel or the series connection?

Is my derivation for parallele correct...and you i showed what i have done for series...didnt get far...
 
  • #4


Abhishekdas said:
And i am not getting how to derive it for the series connection...
I was going like this...correct me all the way...
I hang a weight to this setup...now springs are mass less so at the point of intersection the force due to upper spring=force due to lower spring ...
so k1*x1=k2*x2
x1 and x2 are the respective extensions...
and kx2 supports mg so kx2 = mg...this was as far as i got ...

So please help me out with these derivations and answer my earlier doubt(abt the lengths being equal)...

Thank you...

It is correct as far. You need to get the relation between the extension of the total length (x1+x2) and the external force mg. The natural lengths are irrelevant.

ehild
 
  • #5


Abhishekdas said:
Are you talking about the parallel or the series connection?
I was showing what you have to do to solve for the effective spring constant for two springs in series.

AM
 
  • #6


Abhishekdas said:
And i am not getting how to derive it for the series connection...
I was going lke this...correct me all the way...
I hang a weight to this setup...now springs are mass less so at the point of intersection the force due to upper spring=force due to lower spring ...
so k1*x1=k2*x2
x1 and x2 are the respective extensions...
and kx2 supports mg so kx2 = mg...this was as far as i got ...

Here i made a small mistake (typing error)...in the last line i meant k2x2 supports mg so k2x2=mg...

Anyway thanks a lot Andrew mason and ehild...for your replies...
And you I have come to something so please read my next post...
 
  • #7


Taking up your suggestions i did this...

I had got k2x2 = mg... (1)

now k2x2=k1x1 so k1x1 = mg... (2)

now you guys said get a relation between k(x1+x2) and mg...

So i got x1=mg/k1 (from eq 1) and x2 =mg/k2

adding we get x1+x2= mg(1/k1+1/k2) (3)

So now k(x1+x2)=mg k=equivalent spring constant...

subtituting (x1+x2) from eq 3
kmg(1/k1+1/k2)=mg...

Simplifying we get 1/k=1/k1+1/k2 which is the required relation...So am i correct here?

And thanks ehild for saying that it is independent of their natural lengths...that was bothering me...And after deriving it i realize that...
 
  • #8


And you is the derivation for parallel correct? the one i gave in the question...?And for parallel the natural lengths are equal?
 
  • #9


Both derivations are correct.

AM
 
  • #10


Ok...thanks...and abt my other question...The natural lengths in case of series don't matter but what about the parallel connection?
 
  • #11


Abhishekdas said:
OK...thanks...and about my other question...The natural lengths in case of series don't matter but what about the parallel connection?

For parallel, what's important is that the springs stretch the same amount. That's what's assumed in the derivation of the effective spring constant (for parallel). If the springs don't have the same unstretched length, you can attach a length of rope or cable or string or thread or …, to give them the same unstreched length.
 
  • #12


Sa SammyS ...basically in normal case the unstreched length of the springs are same right? Thanks...
 
  • #13


Abhishekdas said:
Sa SammyS ...basically in normal case the unstreched length of the springs are same right? Thanks...
Yes, because to get 1/kparallel = 1/k1 + 1/k2 , you need Δx1 = Δx2 .
 
  • #14


SammyS said:
Yes, because to get 1/kparallel = 1/k1 + 1/k2 , you need Δx1 = Δx2 .

But that is for series right? and we are talking about parallel...series the natural lengths don't matter...
 
  • #15


Abhishekdas said:
But that is for series right? and we are talking about parallel...series the natural lengths don't matter...

Yes, you're right. I just grabbed a formula w/o thinking.

But the result for parallel does require that the two springs stretch the same amount. Usually, that's most easily done by having their natural lengths be equal.
 
  • #16


Ok...thanks ...
that settles it...good...both parallel and series sorted...thanks guys...
 

FAQ: Equivalent Spring Constant in Series and Parallel Configurations

What is an equivalent spring constant?

An equivalent spring constant is a value that represents the stiffness of a spring system that behaves similarly to a given set of springs. It takes into account the number of springs, their individual spring constants, and their arrangement to calculate a single value that can be used to model the system as a single spring.

How is the equivalent spring constant calculated?

The equivalent spring constant is calculated by adding the reciprocals of each individual spring constant and then taking the reciprocal of that sum. This value is then adjusted based on the arrangement of the springs (series or parallel) to give the final equivalent spring constant.

Why is the equivalent spring constant important?

The equivalent spring constant allows for a more simplified and manageable model of a complex spring system. It can be used to predict the behavior of the system and make calculations more efficient. It also allows for easier comparison between different spring systems.

What factors can affect the equivalent spring constant?

The equivalent spring constant is affected by the number of springs in the system, their individual spring constants, and their arrangement (series or parallel). It can also be affected by external factors such as temperature, material properties, and the presence of any external forces.

How can the equivalent spring constant be experimentally determined?

The equivalent spring constant can be experimentally determined by measuring the displacement and applied force of the spring system. By plotting this data and calculating the slope of the resulting curve, the equivalent spring constant can be determined. This can also be done by using a force sensor and a motion sensor to collect data in real-time.

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