### 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 will then displace 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 given 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.