What should the spring constant k of a spring be

Learn more about this course. Share this post. Q1 E : If a mass of normalsize 10 kg on a heavy industrial spring of normalsize cm creates an extension of normalsize 3 mm, then how much extension would be caused by normalsize 15 kg? How much mass would we have to load onto the original spring to get an extension of normalsize 5 mm? But behind this law is another law which is a fine example of an inverse proportionality.

What happens if we cut this spring into two equally sized pieces? More generally, the spring constant of a spring is inversely proportional to the length of the spring, assuming we are talking about a spring of a particular material and thickness. So suppose we cut the spring in the example above exactly in two, creating two shorter springs each of length normalsize 3 cm.

One of the smaller springs will have a spring constant which is twice the original. That is because the spring constant and the length of the spring are inversely proportional. That means that the original mass of normalsize 30 gm will only yield a stretch of normalsize 1 mm on the shorter spring. The larger the spring constant, the smaller the extension that a given force creates. Hopefully this makes intuitive sense — it should not be a surprise.

If we think of the original spring as being two shorter springs attached together, then the mass of normalsize 30 is stretching both smaller springs by normalsize 1 mm, giving a total stretch of normalsize 2 mm. If you pull a weight on a spring down and let it go, then it will oscillate around its mean position in what is called harmonic motion. If you look at these two examples, hopefully you will see the similarities in motions.

To get an extension of normalsize 5 mm, we would need a mass of m, where. In this experiment, we will investigate the forces exerted on and by a spring. If you have a large value of k, that means more force is required to stretch it a certain length than you would need to stretch a less stiff spring the same length. Once you have determined the spring constant of a spring, you can use that k value for all future calculations, unless the spring is damaged in some way.

The negative sign is in the equation because force is a vector quantity. The negative sign tells us that the direction of the elastic force is always opposite to the direction of the stretching motion. In other words, if you stretch a spring downward, you feel the spring pull upward. If you want to stretch the spring out and hold it in place, you must apply the same amount of force the spring is, but in the opposite direction. That is, to stretch a spring with spring constant k a distance x and hold it there in equilibrium, you must apply a constant force with a size given by.

This force that you are applying exactly balances the opposite force exerted by the spring, to achieve an equilibrium situation.

In this experiment, we will exploit this fact to find out the values of the spring constants of two springs. Note that you have two springs for this lab, color-coded green and blue. The green spring should already be hanging next to the meter stick. The spring is hanging from a device called a force sensor, which is hooked to the computer.

The sensor will tell the computer exactly how much force measured in newtons the spring is feeling. We will not use the sensor for this part of the lab. Just pull a little on each spring do not pull either to its limits! Question 1: Which spring is more difficult to stretch? Which spring do you think will have the higher spring constant? Question 2: When you pull the spring out and hold it, you should feel a force being exerted on you by the spring.

How does that force compare to the force you are applying? It only takes a minute to sign up. Connect and share knowledge within a single location that is structured and easy to search. Is this true? It seems like if it's an inherent property of the spring it shouldn't change, so if it does, why?

They're different springs than their large parent so they may have different values of an "inherent property": if a pizza is divided to 4 smaller pieces, the inherent property "mass" of the smaller pizzas is also different than the mass of the large one. You may have meant that it is an "intensive" property like a density or temperature which wouldn't change after the cutting of a big spring, but you have offered no evidence that it's "intensive" in this sense.

No surprise, this statement is incorrect as I'm going to show. One may calculate the right answer in many ways. For example, we may consider the energy of the spring. We may also imagine that the big spring is a collection of 4 equal smaller strings attached to each other. You could get the same result via forces, too. It's harder to change the length of the shorter spring because it's short to start with, so you need a 4 times larger force which is why the spring constant of the small spring is 4 times higher.

Now consider what happens when we divide the spring. We change only the length of the spring, whilst keeping A same cross-section area and E same spring, same material the same. When we make the spring four times shorter we essentially have the following:. Note, that this is assuming a rubber band like set-up, where we assume that the spring can be modelled by a uniform bar of elastic material.

A more rigorous proof of the dependence of spring constant and the length of the spring would involve the geometry of the spring and various torques on the spring elements when it is under load. However, all this complication just brings additional pre-factors to the spring constant, which are independent of the length of the spring.

All bonds between atoms can be thought as tiny springs obeying Hooke's law in case of small displacements. Now, what about connecting the strings in parallel? There are only two ways of combining strings in parallel or in series , hence the overall formula for force must be of a form bellow:.

So with a very simple thinking and some basic knowledge of the energy conservation, we could recover the law I assumed in my first part of explanation. EDIT : I noticed that there were some errors in the second part of my explanation, hence a complete overhaul. Also, I hope I clarified the first part of the explanation. In other words, the definition of the ideal spring is that it applies the force proportional to its deformation length at both endings of course.