General relationship for springs connected in series

[Vandetah]

Homework Statement

What is the effective force constant of 2 springs which are connected in series? What is the general relationship for n springs connected in series?

Homework Equations

im not sure if i should use

or is the formula for a series circuit also relevant?

The Attempt at a Solution

 

[Infinitum]

Hi Vandetah!

When massless springs are connected in series, the forces exerted by each is the same. Can you derive a relation using this fact?

 

[Vandetah]

so if there's two springs then it would be
k1 = k2?

 

[Infinitum]

so if there's two springs then it would be
k1 = k2?

Not quite. Even though their forces are same, their extension will be different. For two strings, you have,

[tex]x_1 = \frac{F}{k_1}[/tex]

[tex]x_2 = \frac{F}{k_2}[/tex]

And, [itex]x= x_1 + x_2[/itex]

Do you see a relation between the spring constants now?

 

[Vandetah]

u might laugh at this but, does it mean that two constant springs connected to each other create a signle force constant and therefore act as if it were a single spring?

 

[Infinitum]

u might laugh at this but, does it mean that two constant springs connected to each other create a signle force constant and therefore act as if it were a single spring?

No, they produce a single force. Not the same force constant. That should be clear from the above equations

 

[Vandetah]

No, they produce a single force. Not the same force constant. That should be clear from the above equations

so that is the general relationship, but what about the effective force constant? Can i use the equation i mentioned in the op?

 

[Infinitum]

so that is the general relationship, but what about the effective force constant? Can i use the equation i mentioned in the op?

Noo...Use the equations I wrote in my earlier posts to derive the relation for the effective force constant!

 

[Vandetah]

is there a difference between just a force constant and an effective one?

 

[Infinitum]

is there a difference between just a force constant and an effective one?

Well, force constant is for a simple, single spring. The effective constant is the effective spring you replace a given setup with, so that you get the same force exerted by that single spring which the whole setup would exert.

 

[Vandetah]

so if i derive the equations u mentioned i will get

k1 = [itex]\frac{F}{x1}[/itex]

and

k2 = [itex]\frac{F}{x2}[/itex]


so k = k1+k2??

 

[Infinitum]

so if i derive the equations u mentioned i will get

k1 = [itex]\frac{F}{x1}[/itex]

and

k2 = [itex]\frac{F}{x2}[/itex]


so k = k1+k2??

Noo

Use x = x1 + x2.

 

[Vandetah]

Noo

Use x = x1 + x2.

could it be?

x = [itex]\frac{F K_{2}+ F K_{1}}{K_{1}K_{2}}[/itex]

 

[Infinitum]

could it be?

x = [itex]\frac{F K_{2}+ F K_{1}}{K_{1}K_{2}}[/itex]

Yes, but you need to find the effective constant, and not the final extension

Take x = F/k.

 

[ehild]

Imagine that you measure the extension of both springs applying the same force to each. You get ΔL₁ with the first spring and ΔL₂ with the other spring. If you connect the springs, the same force acts on each - you pull the second one with F, it stretches by ΔL₂, the second spring pulls the first one with the same force, it extends by ΔL₁, so the connected springs stretch by the sum ΔL₁+ΔL₂.

You put the connected springs in a black box to hide the point where they are connected. You say it is a spring inside and ask somebody to measure the spring constant, applying force F. The person measures the change of length ΔL, and calculates the spring constant k=F/ΔL. That is the effective spring constant of the device inside the black box.

You know that ΔL=ΔL₁+ΔL₂, and also that ΔL₁=F/k1, ΔL₂=F/k2, so you can express the effective spring constant with the individual ones.

ehild

 

[Vandetah]

lol that post above got me more confused :(

 

[Infinitum]

lol that post above got me more confused :(

Ehild put all I was saying in one post.

Just use x=F/k in your last equation, where k is the effective constant.

 

[azizlwl]

1. Let 2 bodies connected with inextensible string.
If 1st body is pulled by a force and remained at rest, EQUAL force also exerted on 2nd object.

Thus spring#1 and spring#2 extended with equal force, x1 and x2

F=-k1x1=-k2x2 ...(1)

2. Total extension is x1+x2 due to F.
F=Keff(x1+x2)

Subt. (1) in (2)
-k2x2=-Keff((k2/k1)x2+x2)
k2=Keff((k2/k1)+1)

Keff =k1k2/k1+k2 like resistors in parallel

 

[ehild]

Keff =k1k2/k1+k2 like resistors in parallel

You mean that Keff=k1k2/k1+k2, that is Keff=2k2? It is wrong.
Or you just forgot the parentheses...

ehild