Video: Hooke’s Law

Ed Burdette

In this video we learn Hooke’s Law describing the spring restoring force based on spring force constant and displacement from equilibrium.

09:07

Video Transcript

In this video, we’re going to learn about Hooke’s law. We’ll see what this law describes, how it works, and how to apply it practically.

To start out, consider all these different objects: a wristwatch, a pool diving board, shocks on a car, a mattress, a ballpoint pen, and a door lock. What do all these objects have in common? If you answered springs, you’re right. Many objects we encounter in everyday life use springs. And Hooke’s law tells us how those springs perform.

Imagine that there is a spring attached on one end to a wall and on the other end, you put your hand and begin to push. We know from experience that as we push the spring in and compress it, the spring resists that force. We also know that if we were to grab onto the spring and begin to pull it, the spring would resist that stretching. We understand then that in general springs resist being compressed or stretched. This is something that we understand through our experience.

And what Hooke did was take this experience and quantify it based on experiments. Setting up his experiment, Hooke observed that every spring has an equilibrium length — a length it goes to naturally when it’s not compressed or stretched. Hooke within displaced the spring from its equilibrium length — some distance we could call Δ𝑥 — by either stretching the spring or compressing the spring shorter than its natural length.

By carefully using known weights to stretch or compress the spring and by measuring the stretched or compressed length of the spring relative to its equilibrium length, he was able to derive a relationship for the force that the spring exerted. We can call it 𝐹 sub 𝑠. That force is equal to negative the change of the spring’s length times a constant 𝑘. As Hooke repeated his experiments with different types of springs: some short and fat, some long and thin, and some in between, he found that each individual spring had its own constant 𝑘.

Based on this, 𝑘 was given the name spring constant and it’s measured in units of newtons per metre. The Δ𝑥 in this equation which came to be known as Hooke’s law is the displacement of the spring from its equilibrium length. And the minus sign in the equation tells us that the spring resists whatever external forces acting on it — whether to compress or stretch. This is to say that the direction of the spring force is opposite the direction of the external force on the spring.

Bringing all this together, we can say that Hooke’s law tells us that the force a spring exerts is equal to the spring constant 𝑘 multiplied by the displacement of the spring from equilibrium whether being stretched or compressed and that the force direction is opposite the direction of the external force. In this equation, 𝐹 sub 𝑠 is in units of newtons, 𝑘 the spring constant is in units of newtons per metre, and Δ𝑥 the displacement is in units of metres.

Let’s get some practice using Hooke’s law in a couple of examples.

An object of mass 𝑚 equals 332 grams is suspended from a vertically hanging spring, causing the spring to extend by 4.89 centimetres. What extension of the spring would be produced by suspending an object with a mass of 505 grams from it?

We’re told in this example that if we start out with a spring hanging under its own weight and then attach a mass of value 332 grams to the end of the spring, the spring will extend a distance — we can call Δ𝑥 — of 4.89 centimetres beyond its natural length. What we then want to know is if we attached a different mass — call it 𝑚 sub two — of 505 grams to the same spring, then what extension of the spring — we can call it Δ𝑥 two — from its equilibrium length would be find?

So we want to solve for Δ𝑥 two. And to find it, we can recall Hooke’s law. Hooke’s law tells us that the restoring force exerted by a spring is equal to negative its spring constant times its displacement from equilibrium. If we consider the forces that are acting on the spring when the mass 𝑚 is hanging from the spring and it’s in equilibrium, we can say that the gravitational force is equal in magnitude to the spring force — negative 𝑘 times Δ𝑥.

If we were to then write a similar equation, but this time for the second mass 𝑚 two, it would read 𝑚 two times 𝑔 equals negative 𝑘 the same spring constant because it’s the same spring times Δ𝑥 two. If we then divide these equations by one another, we see that on the left-hand side, the factor of 𝑔 cancels out and on the right-hand side, negative 𝑘 the spring constant cancels.

We’re left with a ratio 𝑚 over 𝑚 two is equal to the ratio Δ𝑥 over Δ𝑥 two. Rearranging to solve for Δ𝑥 two, we see it is equal to Δ𝑥 times the ratio of masses 𝑚 two over 𝑚. We’ve been giving Δ𝑥, 𝑚, and 𝑚 two in our problem statement. So we’re ready to plug in and solve for Δ𝑥 two. When we enter these values on our calculator, we find Δ𝑥 two is 7.44 centimetres. That’s the displacement from equilibrium of the spring with the second mass on it.

Let’s do another example using Hooke’s law.

The springs of a pickup truck act like a single spring with a force constant of 1.30 times 10 to the fifth newton metres. By how much will the truck be depressed by its maximum load of 1.00 times 10 to the third kilograms? If the pickup truck has four identical springs, what is the force constant of each spring?

In the first part of this exercise, when we’re asked by how much the truck is depressed, we’re searching for a displacement which we can call Δ𝑥. In part two where we want to solve for the force constant of each of the four identical springs, we can call that force constant 𝑘 sub four. To start off on our solution, we can recall Hooke’s law for springs.

This law tells us that a spring’s restoring force is equal to negative its spring constant multiplied by its displacement from equilibrium Δ𝑥. In the case of our pickup truck that’s fully loaded, we can call the mass of that load 𝑚 given as 1.00 times 10 to the third kilograms. And we’re told that the overall spring constant of the truck’s suspension system, which we’ve called 𝑘, is 1.30 times 10 to the fifth newton metres.

Since the fully loaded truck is in equilibrium, we can write that the magnitude of the gravitational force on the full load of the truck is equal to the magnitude of the restoring spring force or 𝑚 times 𝑔 equals 𝑘 times Δ𝑥. And when we rearrange for Δ𝑥, we find it’s equal to 𝑚 times 𝑔 over the spring constant 𝑘. The acceleration due to gravity 𝑔 we’ll let be exactly 9.8 metres per second squared. When we plug in for 𝑚, 𝑔, and 𝑘 and enter this expression on our calculator, we find Δ𝑥 is 7.54 centimetres. That’s how much the springs of the truck’s suspension system are compressed under this load.

Next, we want to solve for the spring constant if rather than one collective spring constant for the suspension system, it was equally divided among the four wheels of the truck. Since there is an equal division, that means that 𝑘 sub four is equal to 𝑘 the spring constant of the truck collectively divided by four. Plugging in for 𝑘 to three significant figures, we find a value of 3.25 times 10 to the fourth newton metres. That’s the effective spring constant of each of the truck’s wheels suspension systems.

Let’s summarize what we’ve learnt so far about Hooke’s law. We’ve seen that Hooke’s law describes the restoring force of a stretched or compressed spring. As an equation, Hooke’s law says that the spring restoring force is equal to negative the spring constant 𝑘 multiplied by the spring’s displacement from equilibrium Δ𝑥. We’ve also seen that the spring constant 𝑘 differs from one spring to another and that this constant is given in units of newtons per metre. And finally, we’ve seen that the minus sign in Hooke’s law means that the restoring spring force points in the opposite direction of any external forces acting on a spring.

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