Equivalent Spring Constant in Series and Parallel Configurations

[Abhishekdas]

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...

 

[Andrew Mason]

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

 

[Abhishekdas]

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...

 

[ehild]

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

 

[Andrew Mason]

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

 

[Abhishekdas]

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...

 

[Abhishekdas]

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...

 

[Abhishekdas]

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

 

[Andrew Mason]

Both derivations are correct.

AM

 

[Abhishekdas]

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

 

[SammyS]

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.

 

[Abhishekdas]

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

 

[SammyS]

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

Yes, because to get 1/k_parallel = 1/k₁ + 1/k₂ , you need Δx₁ = Δx₂ .

 

[Abhishekdas]

Yes, because to get 1/k_parallel = 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...

 

[SammyS]

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.

 

[Abhishekdas]

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