Lesson Video: Deformation of Springs | Nagwa Lesson Video: Deformation of Springs | Nagwa

Lesson Video: Deformation of Springs Physics

In this lesson, we will learn how to use the formula 𝐹 = 𝑘𝑥 to calculate the deformation of a spring, defining the spring constant as the resistance of a spring to deformation.


Video Transcript

Given a spring, say, one of the springs we see here, it’s possible to stretch the spring out to make it longer or to compress it to make it shorter. This is known as deforming the spring. But before talking about deformation, let’s consider what springs do naturally when there’s no force acting on them.

Say we have this spring right here attached to a wall. Right now, with nothing attached to its free end, no force pushing on it or pulling it, the spring has a length that we call the natural length of the spring. This length is just how long a spring is without being compressed or stretched at all. This is the length that the spring is always trying to get back to. We could call it the preferred length of the spring.

So, we have this spring sitting at its natural length, but then say that we grab onto the free end of it and we start to pull. As a result of this pull, the spring stretches out a distance that we can call 𝑥. By applying a net force to the spring, what we’ve done is we’ve changed its shape. We’ve deformed it. In order to keep the spring this long though, we know we’ll have to continue applying that same force, let’s call it 𝐹, to the end of the spring. If we were to suddenly let go and stop applying this force, the spring would return to its natural length.

But let’s imagine that rather than letting go of the spring, we actually pull harder on it. We double the force. If we pull twice as hard, then we’ll double the extension of the spring. We’ve added an extension of 𝑥 to it so that now the total combined extension of the spring from its equilibrium, or natural length, is two 𝑥. And so long as our spring is mechanically able to handle all this stretching, we can increase the force and increase the stretch. The relationship between these two quantities, the force we apply to a spring and how much it stretched or compressed from its natural length, is called Hook’s law.

Hook’s law tells us that the net force acting on a spring is directly proportional to the stretch or the compression of the spring 𝑥 from its natural length. We’re seeing evidence of Hook’s law as we double our force and get twice as much stretch in our spring. Now, the way that we’ve written this law 𝐹 is directly proportional to 𝑥 is one way to say it. But there’s another mathematically equivalent way to write this out. We could also say that the force acting on a spring is equal to a constant of proportionality, we’ll call it 𝑘, multiplied by 𝑥.

Let’s think for a moment about this constant and what it does physically. When we stretch this spring, when we make it longer than its natural length, the spring resists that stretch. If we’re holding the free end of the spring with our hand, we feel the spring pulling back on it. This constant of proportionality 𝑘 is a measure of just how much a particular spring resists stretching or compression. For that reason, the name that we normally give 𝑘 is the spring constant. The units of this spring constant, the units of 𝑘, are newtons per meter. That is units of force divided by units of distance.

The spring constant of a spring is essentially saying if we want to stretch or compress a given spring a distance of one meter, then here’s how much force in newtons we’ll need to apply to do that. Some springs are very stiff and have high spring constants. It takes a lot of force to stretch or compress them some distance. Other springs have very low spring constants. They offer very little resistance to deformation. But in all cases, regardless of the spring, the spring has a spring constant, which resists deformation from its natural length.

Now, here’s something interesting about spring constants. Let’s say that we bring our spring back to its natural length, the length it originally had. Starting from that equilibrium position, say that we then stretch our spring a distance of one meter. And let’s say further that the particular spring constant of the spring we’re working with is 100 newtons per meter. This means that in order to stretch our spring this one-meter distance, we needed to apply 100 newtons of force to it.

Then, let’s say we stretch our spring some further distance out so that now the spring looks like this. And let’s say to get the spring out to this length, that we’re now pulling on it with a force of 400 newtons. If we were to mark off an additional meter from this stretched length of our spring and then pull the spring yet further so it covers that gap. A question is, how much additional force does it take to make our spring extend this meter versus this original meter of stretch?

The spring constant 𝑘 tells us that it takes just as much additional force to stretch this meter as it did to stretch our original meter from equilibrium. Meaning, we’d now be pulling with a force of 400 plus 100, or 500 newtons, to get the spring this length. In other words, no matter how much we stretch or compress the spring, the spring constant is truly a constant.

This isn’t always the case. On rare occasions, we’ll encounter spring constants that change depending on how far the spring has been stretched or compressed. But unless we’re told otherwise, it’s always safe to assume that the spring constant is indeed a constant. No matter how far our spring is deformed from its equilibrium length, the spring constant gives us a set value for how much force it will take to move the spring yet another meter. Now, that we’ve considered the spring constant in Hook’s law equation, there are two other points about this expression that we ought to mention.

The first has to do with this variable 𝑥. It’s important to realize that 𝑥 is the displacement of a spring from its natural, or equilibrium, length, not the natural length itself. For example, considering this spring, the natural length of this spring would be this distance. Here, we can call that 𝐿. But say we put a mass on top of the spring, which compressed it down from its equilibrium length. It’s that distance that we would call 𝑥, rather than the original length 𝐿 of the spring.

And as we think about it, this makes sense. If a spring is already at its natural length, then it won’t experience any force to restore it to its natural length since it’s already there. It’s only by deforming the spring, by stretching or compressing it, that we bring in a restoring force 𝐹. So, that’s one thing to remember about this equation.

The other thing to remember is that sometimes we’ll see a minus sign in front of the right-hand side, in front of the 𝑘 times 𝑥. The reason for that is that this force 𝐹 is a restoring force. It’s a force exerted by the spring to return itself to its natural, or equilibrium, length. When we stretch a spring out, as we have in this case, the force we’re exerting on the spring is in the direction of the displacement 𝑥. But the force exerted by the spring is opposite that direction. In this case, the force exerted on the spring is to the right. But the force the spring exerts is to the left.

Because force on a spring is in the direction of displacement, while the force of a spring points the opposite way, Hook’s law is written with a plus or minus sign depending on which of the two equal but opposite forces we’re considering. So, when we see a minus sign in Hook’s law, it means we’re talking about the force of the spring, which is opposite displacement. Let’s now put these ideas into practice through a couple of examples.

Which of the following formulas correctly shows the relation between the force applied to a spring, the change in the spring’s length Δ𝑥, and the spring constant, also known as stiffness of the spring, 𝑘? A) 𝐹 is equal to 𝑥 divided by Δ𝑘. B) 𝐹 is equal to 𝑘 divided by Δ𝑥. C) 𝐹 is equal to one-half 𝑘 times 𝑥 squared. D) 𝐹 is equal to 𝑘 times Δ𝑥. E) 𝐹 is equal to one divided by 𝑘 times Δ𝑥.

O𝑘ay, so, in this exercise we’re looking for a mathematical relationship that correctly connects three quantities, the force applied to a spring, the change in the spring’s length Δ𝑥, and the spring constant called 𝑘. Let’s start out by writing down those three variables. We’ll call the force 𝐹, the change in the spring’s length is Δ𝑥, and then the spring constant is 𝑘.

If we consider the units of each one of these three terms, putting these square braces around the term means that we’re talking about the units of that variable. Then, we can say that within the SI system, the base unit of force is the newton, the base unit of a change in length, or length at all for that matter, is the meter, and the units of the spring constant 𝑘 are newtons per meter.

Taking a look now at our five answer options, we see that on the left-hand side of each one, we have the force 𝐹 by itself. In other words, regardless of which answer option we choose, the left-hand side of that option will have units of newtons. Physically then, we know that in order for the equation to hold true, whichever equation we end up choosing the units of the right-hand side must also equal newtons. Only answer options which meet this criterion can possibly be correct. So, starting with answer option A, let’s look at what the units on the right-hand side of these expressions boil down to.

For choice A, we have 𝑥, a distance, in the numerator. So, that has units of meters. And then, that distance is divided by a spring constant, which itself has units of newtons per meter. To clean this up a bit, we can multiply both top and bottom by the units of meters, which will lead to the cancellation of that term in our denominator. And this leaves us with the overall units of meters squared per newton. Since this is not equal to the base units of force, newtons. We know that answer option A can’t be our choice.

Moving on to answer option B, here, we have a spring constant in units of newtons per meter divided by a change in length with the units of meters. If we divide both the numerator and denominator of this expression by meters, then we’ll wind up with final units of newtons per meter squared. This also is not a match simply for units of newtons. So, answer option B isn’t our choice either.

Then, we get to option C. Here the one-half has no units; that’s just a constant. So, we multiply the units of the spring constant 𝑘 newtons per meter by the units of the spring’s length, meters squared. A newton multiplied by a meter squared divided by a meter gives us final units of newton meters. Once again, this is not simply units of newtons. So, option C isn’t correct either.

Moving to option D, here, we have a spring constant with the units of newtons per meter multiplied by a change in spring length with units of meters. Here, we see that the meters in denominator and numerator cancel out, leaving us simply with units of newtons. This is a very good sign. It means that option D is a candidate for the correct relationship between force, spring constant, and spring length. To see if it’s the only candidate. Let’s move on to option E and evaluate that.

The units of the right-hand side of answer option E are one over a newton per meter multiplied by a meter. Once again, our units of meters cancel out. But this time we don’t wind up with units of newtons ultimately, but rather units of one over newtons. This isn’t a match for the units of the left-hand side of our expression. So, answer option E isn’t a candidate. This tells us that option D is our correct choice. Force is equal to 𝑘 times Δ𝑥. This is the only answer option for which the units on the left-hand side are equal to the units on the right-hand side.

Let’s look now at a second example involving spring deformation.

The spring shown in the diagram has a constant of 50 newtons per meter. What is the mass of the object attached to it? Answer to the nearest gram.

Taking a look at our diagram, we see in the first case our unstretched spring, the spring with no mass attached to it. This spring has an equilibrium length of 15 centimeters. But then, when we attach a mass to the end of our spring and we can call that mass 𝑚, we see the spring stretches out and that stretch is given as a distance of 2.5 centimeters. From this information, we want to solve for the mass of the object attached to the spring.

As we start doing this, it’s helpful to realize that once the mass is attached to the spring and the spring is stretched out, then this mass-spring system is in equilibrium. This means that the force upward on the mass is equal to the force downward on it. And to see just what those forces are, let’s sketch out a free body diagram of this mass. Just like any mass of object, this mass is subject to the force of gravity. And we know that that force is equal to the mass of our object multiplied by the acceleration due to gravity.

Now, if this were the only vertical force acting on our mass, then it would be accelerating downward. But it’s not the only force, there’s the restoring force of the spring that acts on the mass as well. And to know what this force is, it will be helpful to recall Hook’s law. This law says that the restoring force provided by a stretched or compressed spring is equal to the spring constant 𝑘 multiplied by the displacement of the spring from equilibrium.

So, when it comes to the upward force on our mass, that’s equal to the force provided by the spring. It’s equal to the spring constant times 𝑥, the displacement of the spring from its natural length. And just like we said, this mass is stationary as it hangs off the end of the spring. That means these two forces are equal in magnitude to one another. In other words, we can write that 𝑘 times 𝑥 is equal to 𝑚 times 𝑔. If this weren’t true, if these two forces weren’t equal in magnitude to each other, then the mass would be accelerating. But we know that it’s not. So, 𝑘 times 𝑥 is equal to 𝑚 times 𝑔. And it’s 𝑚, the mass of our object, that we want to solve for.

To do that, let’s divide both sides of the equation by the acceleration due to gravity 𝑔, causing that term to cancel out on the right. And what we find is that the object’s mass is equal to 𝑘 times 𝑥 divided by 𝑔. At this point, we can recall that the acceleration due to gravity is approximately 9.8 meters per second squared. And then, looking up at our problem statement, we see that the spring constant 𝑘 of our spring is 50 newtons per meter. The last thing we want to solve for and then plug into our equation is 𝑥.

And at this point, we’ll need to be a bit careful. Notice that over in our diagram, we’re given this 15-centimeter distance. That’s this distance here, the length of our unstretched spring. But the 𝑥 in Hook’s law, and the 𝑥 we’re using in our equation, is not the natural length of a spring. Rather, it’s the stretch or compression of that spring from equilibrium. In other words, is the displacement of the spring. And that is given as a distance of 2.5 centimeters. So, the 𝑥 in our equation is 2.5 centimeters. 𝑘 is 50 newtons per meter. And 𝑔 is 9.8 meters per second squared. At this point, with all our numbers plugged in, we’re almost ready to calculate the mass 𝑚.

There are two things we should keep in mind though. First, our displacement is in units of centimeters, whereas the distances in the other terms in our expression are in units of meters. This means that the terms in our expression aren’t all on equal footing. And we’ll want to make sure they are before we combine them. To do that, let’s convert 2.5 centimeters to that same distance in meters. Since 100 centimeters is equal to one meter, that tells us if we shift the decimal place two spots to the left in this distance, we’ll get that same distance in meters. It’s 0.025 meters. And now, we’re ready to multiply all these values together.

When we do, we get this answer 0.12755 and so on kilograms. But we know that we want to give our answer to the nearest gram. Since there are 1000 grams in one kilogram, that means to convert this answer to some number of grams, we’ll want to shift the decimal place three spots to the right. Doing so, we get 127.55 dot dot dot grams. And lastly, we want to round this answer to the nearest gram. That means we look at this value right to the right of the decimal point, and since it’s equal to or greater than five, that indicates that to round this answer to the nearest gram, we’ll round this number up by one. In other words, our answer is 128 grams. That’s the mass of the object attached to the spring to the nearest gram.

Let’s take a moment now to summarize what we’ve learned in this lesson. First off, we saw that an uncompressed, an unstretched, spring, like this one here, is at what’s called its natural, or equilibrium, length. We learned that when a spring is deformed, that is stretched or compressed, the spring exerts a restoring force to try to return to its natural length. That restoring force 𝐹 is equal to what’s called the spring constant 𝑘 multiplied by the displacement of the spring from equilibrium. For example, if this spring was stretched this distance, then that distance beyond the equilibrium point would be 𝑥.

Furthermore, we saw that the spring constant 𝑘, which has units of newtons per meter, indicates the resistance of a given spring to being deformed. The higher 𝑘 is, the harder it is to stretch or compress a spring. And lastly, we saw that for a given spring, 𝑘, the spring constant, can be assumed to be a constant value. In other words, no matter how much a spring is deformed, it will still require the same amount of force to stretch or compress the spring an additional meter.

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