The intricacies of spring expansion

AlexIn summary, the problem involves two identical springs made from steel wire of equal length and loaded with equal masses. The first spring stretched to 1/10 of its initial length, and the question is what is the percent elongation of the second spring. Through analysis, it is determined that the second spring will have a different spring constant and will experience less elongation for the same amount of force. Using a proportion, the percent elongation of the second spring is found to be 20%.
  • #1
freemind
Howdy,

I don't know how to solve this problem:

2 identical pieces of steel wire of equal length were used to manufacture 2 springs. Diameter of the 1st spring coil was d, diameter of second was 2d. Both springs were then loaded with equal masses. As a result, the first spring stretched to 1/10 of its initial length.
What was the percent elongation of the 2nd spring?

I've found (through arc-length integration of two space-curves) that the two coils are in a length ratio of [tex]\frac{\sqrt{5}}{\sqrt{2}}[\tex]. Now what? I don't know how a change in coil length affects the spring constant. I'm quite sure that the spring constant is different for the double-diameter coil, but don't know how it differs. Any help would be greatly appreciated.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2


Hi there,

It seems like you've made some good progress in your analysis of the problem. To determine the percent elongation of the second spring, we need to consider a few factors. First, the spring constant is directly proportional to the cross-sectional area of the wire, so the double-diameter coil will have a different spring constant than the first spring. This means that the second spring will experience a different amount of force for the same amount of elongation as the first spring.

Secondly, the percent elongation of a spring is inversely proportional to its spring constant. This means that the second spring, with its higher spring constant, will experience less elongation for the same amount of force as the first spring.

Taking these factors into account, we can use the formula for percent elongation, which is: (change in length/original length) x 100%. Since the first spring stretched to 1/10 of its original length, its change in length is 1/10 of its original length. We can then set up a proportion with the percent elongation of the second spring as the unknown:

(1/10 of original length)/(original length) = (percent elongation of second spring)/100%

Solving for the unknown, we get a percent elongation of 20% for the second spring. This means that the second spring will experience 20% elongation for the same amount of force as the first spring.

I hope this helps! Let me know if you have any further questions or if you need clarification on anything. Keep up the good work!


 
  • #3


Hi there,

It seems like you are trying to solve a problem related to the expansion of springs and the relationship between their length and spring constant. I'm not an expert in this area, but I can offer some general guidance that may help you solve this problem.

Firstly, let's clarify the concept of spring expansion. When a spring is loaded with a mass, it stretches or expands due to the force applied. This expansion is directly proportional to the applied force, which is known as Hooke's Law. In other words, the more force applied, the more the spring will expand.

Now, in your problem, you have two identical springs with different coil diameters. The first spring, with a diameter of d, is loaded with a mass and stretches to 1/10 of its initial length. The second spring, with a diameter of 2d, is also loaded with the same mass. You are trying to find the percent elongation of the second spring.

To solve this problem, you need to understand the relationship between the spring's length and its spring constant. The spring constant, denoted by k, is a measure of the stiffness of a spring. It is a constant value that relates the force applied to the amount of stretch or compression of the spring. In other words, the higher the spring constant, the stiffer the spring, and the less it will stretch for a given force.

Now, in your problem, you have two springs of equal length and material, but with different coil diameters. This means that their spring constants will be different. The spring constant is inversely proportional to the coil diameter, so the spring with a diameter of 2d will have a lower spring constant compared to the spring with a diameter of d.

To find the percent elongation of the second spring, you will need to use the formula for Hooke's Law: F = -kx, where F is the force applied, k is the spring constant, and x is the displacement or expansion of the spring. Since the first spring expands to 1/10 of its initial length, we can say that x = 1/10L, where L is the initial length of the spring.

Now, to find the spring constant of the second spring, we can use the ratio you found, [tex]\frac{\sqrt{5}}{\sqrt{2}}[\tex]. This ratio represents the relationship between the lengths of the two springs. So,
 

1. What causes springs to expand during springtime?

The main cause of spring expansion is the increase in temperature during the spring season. As the weather warms up, the molecules in the metal of the spring begin to move faster, causing the spring to expand.

2. Is the expansion of springs a uniform process?

No, the expansion of springs is not a uniform process. The rate and amount of expansion can vary depending on factors such as the type of metal used, the design of the spring, and the surrounding temperature and environment.

3. Can the expansion of springs be controlled?

Yes, the expansion of springs can be controlled through various methods such as using different types of metals, adjusting the design and size of the spring, and regulating the temperature and environment in which the spring is used.

4. How does the expansion of springs affect their performance?

The expansion of springs can affect their performance in several ways. If the expansion is too great, it can cause the spring to lose its elasticity and become weaker. On the other hand, if the expansion is not enough, the spring may not function properly and may not provide the necessary force or support.

5. Are there any potential issues or problems caused by spring expansion?

Yes, there can be potential issues or problems caused by spring expansion. For example, if the expansion is not taken into account when designing a machine or structure, the spring may not fit properly and can cause malfunctions. Additionally, extreme expansion can cause stress on the material and lead to breakage or failure of the spring.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
335
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
469
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
24
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
449
  • Introductory Physics Homework Help
Replies
3
Views
367
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
719
  • Introductory Physics Homework Help
Replies
10
Views
1K
Back
Top