- #1

MilenaMath

- 8

- 0

Determine the total spring constant of the parallel springs.

Can anyone explain me the solution to this.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter MilenaMath
- Start date

- #1

MilenaMath

- 8

- 0

Determine the total spring constant of the parallel springs.

Can anyone explain me the solution to this.

- #2

DrChicken

- 3

- 0

Determine the total spring constant of the parallel springs.

Can anyone explain me the solution to this.

Since the springs are in parallel, they will both have the same displacement. Knowing this and examining the equation for the restoring force of the spring you can find that:

F = kx

F = k

Keeping the second equation in the same "form" as the first,

k = k

This is more of an introductory physics question. Make sure to read the stickies before posting.

- #3

MilenaMath

- 8

- 0

But the correct answer is 4K,I did the same thing,but the professor says 4K,I cant find the mistake

- #4

- 12,143

- 165

What is the spring constant of half of the original spring?

- #5

MilenaMath

- 8

- 0

But he says my mistake is that I cant say ##k_1= F/(x/2)## but the rest is ok

- #6

PhanthomJay

Science Advisor

Homework Helper

Gold Member

- 7,179

- 512

Your answer is correct but so is the professor's comment. You need to explain more clearly, in terms of the spring's unstretched lengths, not its displacement, why cutting a spring in half doubles its stiffness.

But he says my mistake is that I cant say ##k_1= F/(x/2)## but the rest is ok

- #7

MilenaMath

- 8

- 0

- #8

PhanthomJay

Science Advisor

Homework Helper

Gold Member

- 7,179

- 512

The deflection of the spring is directly proportional to its length and force, and the amount of its deflection is dependent on its geometric and elastic properties. Since you halve its length, you halve its deflection under a given force, and thus double its stiffness k. Then you put the cut springs in parallel, so each sees just half the force, which effectively quadruples the effective equivalent stiffness.

Share:

- Last Post

- Replies
- 16

- Views
- 7K

- Last Post

- Replies
- 6

- Views
- 3K

- Last Post

- Replies
- 4

- Views
- 4K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 6K

- Last Post

- Replies
- 11

- Views
- 4K

- Last Post

- Replies
- 3

- Views
- 7K

- Last Post

- Replies
- 2

- Views
- 6K

- Last Post

- Replies
- 4

- Views
- 5K