# Spring constants in series

#### chrom68

I have two springs arranged in series (remember circuit diagrams from physics class!).

One has a low stiffness constant (K1 = 5) and the other connected to it has a much higher contant (K2 = 100). According to the equation (as used in wikipedia):

$$\frac{1}{k}=\frac{1}{k_1}+\frac{1}{k_2}$$

http://en.wikipedia.org/wiki/Hooke%27s_law" [Broken]

Question 1)
Using this i get my equivalent spring constant to be K = 4.76, which is less than K1?

I don't understand why. I would expect the equivalent constant to be much higher (but less than K2=100).

Question2)
This formula doesn't consider the initial lengths of each spring. How could it do so?

Last edited by a moderator:
Related Other Physics Topics News on Phys.org

#### berkeman

Mentor
I have two springs arranged in series (remember circuit diagrams from physics class!).

One has a low stiffness constant (K1 = 5) and the other connected to it has a much higher contant (K2 = 100). According to the equation (as used in wikipedia):

$$\frac{1}{k}=\frac{1}{k_1}+\frac{1}{k_2}$$

http://en.wikipedia.org/wiki/Hooke%27s_law" [Broken]

Question 1)
Using this i get my equivalent spring constant to be K = 4.76, which is less than K1?

I don't understand why. I would expect the equivalent constant to be much higher (but less than K2=100).

Question2)
This formula doesn't consider the initial lengths of each spring. How could it do so?
Probably best to draw the springs in their initial state, and then compressed under some force. You should be able to derive the equation based on that.

The overall K will be lower than the weaker K, because your force "sees" the weaker K, plus some additional compression beyond what just the weaker K offers. Thus, the overall K appears a bit weaker than the weaker K. Make sense?

Last edited by a moderator:

#### mgb_phys

Homework Helper
I don't understand why. I would expect the equivalent constant to be much higher (but less than K2=100).
Imagine you have a very soft spring and a very hard spring in series.
You haven't done anything to make the soft spring any harder so the overall stiffness can't be any more than the soft spring (imagine attaching the soft spring to an infinite stiffness rod).
But you have added a strong spring which will give a little (however small) and so the overall system must have a little more give = overall stiffness must be less

#### chrom68

I understand that the soft spring doesn't become any harder therefore shouldn't the overall stiffness still be greater than K1 (by just a small amount)? Surely adding a stiffer spring in series wouldn't make the stiffness weaker overall.

#### berkeman

Mentor
I understand that the soft spring doesn't become any harder therefore shouldn't the overall stiffness still be greater than K1 (by just a small amount)? Surely adding a stiffer spring in series wouldn't make the stiffness weaker overall.
Draw the drawings I suggested in my post, and work through the numbers with some examples. What do you find?

Quiz Question -- what do you get for the composite K when you put two springs with identical K values in series? Why?

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving