# Find the effective force constant ke

1. Jul 31, 2009

### Pupil

This isn't really a homework problem, but I've been going over these Yale problem sets for physics 200 (http://oyc.yale.edu/physics/fundamentals-of-physics/content/resources/problem_set_3.pdf" [Broken]), and I'm having trouble with one of the questions.

1. The problem statement, all variables and given/known data
Two springs have the same unstretched length but different spring constants k1 and
k2 . Find the effective force constant ke if they are connected in series and in parallel.
(To ¯nd ke, imagine stretching the combination by an amount x and keeping track
of the force needed.)

2. Relevant equations
f = ma
fs = -kx
Pupil = awesome

3. The attempt at a solution
So I figured the parallel spring constant out fairly easily, but I'm having trouble with the series one. The way I imagined it is we tie spring with constant $$k_1$$ to the wall, and tie spring with constant $$k_2$$ to spring with constant $$k_1$$, and finally pull $$k_1$$ by some amount A. Spring with $$k_1$$ will stretch by some amount A1 and spring with $$k_2$$ will stretch by some amount $$A_2$$. Summing up, we have $$k_2A_2 + k_1A_1 = F$$. The problem is I don't know what $$A_1$$ and $$A_2$$ are. I know the springs won't stretch the same amount, so I'm stuck with this ugly expression. Help me get unstuck (don't tempt me to look at the answers)! Thanks!

EDIT: The title was supposed to be Spring Constant in Series...Don't know what happened.

Last edited by a moderator: May 4, 2017
2. Jul 31, 2009

### ideasrule

Re: Spring

Actually
$$k_2A_2 + k_1A_1 = F$$

isn't right. Think about the point where the two springs connect. The first spring applies k1A1 to the second, and the second applies k2A2 to the first, so F=k1A1=k2A2. The problem should be easy from there.

3. Jul 31, 2009

### Redbelly98

Staff Emeritus
Re: Spring

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That's not quite right. Springs in series experience the same force, so

k1A1 = k2A2 = F

However, the displacements add up, so we can also say

Anet = A1 + A2

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EDIT: Hey ideasrule, great minds think alike

Last edited by a moderator: May 4, 2017
4. Jul 31, 2009

### rock.freak667

Re: Spring

x1 is the extension produced by the force F in spring 1, similar for x2

The force on the first spring=force on the second spring
so we have

$$F=k_1x_1=k_2x_2$$

Now the effective spring constant relates to F by this

F=kex (where x=x1+x2)

i.e. we have $F=k_e(x_1+x_2)$

looks a bit unsettling right? BUT we know that $F=k_1x_1=k_2x_2$

you can put either 'kx' equal (replace F) to the formula above, and replace the 'other' x using the relation in the last formula. You will understand what I mean when you do it.

For parallel:

Each spring will stretch by the same amount, so finding ke is easier here.

EDIT: seems ideasrule and Redbelly98 can type faster than me

5. Jul 31, 2009

### Pupil

Re: Spring

Holy crap! The force of the second on the first is equal to the force of the first on the second. I forgot Newton's third law! *Face palm* Thanks guys. You rock!