Video Transcript
In this video, we’re going to learn
about oscillating springs. We’ll see how these springs move,
how to mathematically describe their motion, and we’ll learn about simple harmonic
motion.
To get started, imagine that in
celebration of International Pi Day on March 14th of every year, your physics
teacher has constructed a gigantic oscillating spring with a height of 10𝜋 meters
tall. What’s special about this
oscillating spring, besides the fact that it’s absolutely massive, is that it is
equipped with a motor which helps to drive its motion.
The motor is designed though so
that it only supplies enough energy to the spring to counteract the energy lost
through friction as the mass on the end of the spring moves up and down up and
down. In other words, the motion of this
spring doesn’t decrease over time, but rather it stays constant with the power
that’s added in by the motor as needed.
To help your class understand
spring motion better, your teacher has issued a challenge. Given the initial position of the
mass on the end of the spring, what your teacher wants to know will be the position
of the mass exactly one week from that time. In order to correctly predict the
mass’s position one week after release, we’ll want to know something about
oscillating springs.
One thing we know about springs is
that if we have a spring just by itself unstressed, the spring will move to what’s
called its natural length. That’s the length that a spring has
when it’s not being compressed or stretched. If we then put a mass on the end of
the spring and, say, we stretch it out some distance, then we know that the spring
under this tension will have a tendency to compress itself back to its natural
length.
And in particular, if we calculated
the force that acts on our mass 𝑚 to pull it backward, Hooke’s law tells us that
that force is equal in magnitude to what’s called the spring constant, or the force
constant, which is a property of the spring naturally, and the distance that spring
has been displaced from its equilibrium length.
So, the more we displace a spring,
either by stretching or compressing it, the more force the spring experiences to try
to be restored to its natural length. Whenever a system experiences a
force like this, which is a restoring force, it tries to restore equilibrium, that’s
proportional to the displacement of the system from equilibrium, then that system is
said to exhibit simple harmonic motion. Simple harmonic motion, sometimes
abbreviated SHM, is the way that an object which is subject to Hooke’s law
moves.
To talk a little bit more about
this type of motion, let’s imagine that instead of a horizontally oriented spring
system, we have a vertically oriented system. Once again, we have our spring,
which has a natural length all by itself. And if we then hang a mass from
that spring, the spring will stretch a little bit and the system overall will find a
new natural length. We’ll call the position of the
middle of the mass, where 𝑥 is equal to zero.
And now, we can create a plot where
we have position on the vertical axis and time on the horizontal axis. What we’re going to do is set this
mass in motion by stretching it and then releasing it. And as the mass moves up and down,
we’re gonna chart its motion on this plot. If we stretch this mass out and
then say we release it to start moving, we know that because of Hooke’s law the mass
will start to move upward. And let’s say that instead of
starting to track its motion at the moment we release the mass, we wait until the
mass is back at its equilibrium position.
When the mass has reached this
equilibrium position which we’ve called where 𝑡 is equal to zero, it’s moving
upward with its maximum speed. As the mass moves from its
equilibrium position to this highest point, we plot its motion on this graph. At this maximum amplitude because
the spring is compressed and again wants to return to its natural length, there is a
restoring force that pushes the mass downward now. Once the mass has been restored to
its equilibrium position, it now has a significant speed in the downward
direction.
As the mass moves to its maximum
spring extension, we continue to plot its motion. The mass then moves back up and
then back down and then back up again. And then, the mass will continue
moving. But we’ll stop plotting it and when
it’s back at its equilibrium position just because our graph has run out of
room.
Because this mass that was
oscillating on the spring was subject to Hooke’s law, that means that the plot we’ve
made is a picture of what simple harmonic motion looks like. The first thing we may notice
looking at this curve is that it looks like a sine curve. And that’s correct. That’s actually one of the
hallmarks of simple harmonic motion. We can go further in describing
this curve though. We could actually get to a point of
creating a function, the position of our mass as a function of time.
To start figuring out what this
function 𝑥 of 𝑡 might be, let’s consider the maximum value that our function
reaches. That’s called the amplitude of this
wave, which we can symbolize with a capital 𝐴. And just as we get to positive 𝐴
on one side, we reach negative 𝐴 on the other.
Imagine that if instead of our
functions having this sine shape, it was just a flat horizontal line at 𝑥 equals
𝐴. In other words, it was just a
straight level line. In that case, for the function 𝑥
as a function of 𝑡, it would be a constant. It would simply equal 𝐴. But we know that really we want to
modulate this value capital 𝐴 with a sine function. After all, our function goes from
positive 𝐴 to negative 𝐴 back and forth back and forth.
So, let’s say in our function, we
multiply this amplitude 𝐴 by the sine of the variable on our horizontal axis, the
time 𝑡. This function for 𝑥 is actually
looking pretty good because it gives us our maximum and minimum function values. And it follows the overall shape of
our curve, a sine curve. There is only one thing that’s a
little bit off about this function and that’s what’s inside the argument of the sine
function. Right now, it’s just the time
𝑡. To see why we might still be
missing some information in our function 𝑥 as a function of 𝑡, let’s look at the
sine function itself.
Here, we’re gonna plot sine as a
function of 𝑥. We know that when 𝑥 is equal to
zero, so is the sin of zero. We know that as 𝑥 increases, the
sine function has a maximum value of one and then a minimum value of negative one
and that when 𝑥 is exactly equal to two 𝜋, the sine function comes back to its
original value it started at, at 𝑥 equals zero. We’ve just mapped out one complete
wavelength of the sine curve. But we know that in general the
wavelength of a real-life physical system might not exactly match this form.
What if the wavelength of this
curve we’ve plotted out is a little bit shorter or a little bit longer or much
shorter or much longer? To allow for this possibility,
we’ll want to insert one more variable in the argument of our sine function. The variable we’ll insert is
𝜔. And that’s not an accidental
choice. Recall that 𝜔 is an angular speed
of our system. Let’s look back over at our
oscillating spring to see why we made this choice of the angular speed for that
variable.
If we consider this whole system
from an energy perspective, we know that because energy is conserved, it will
involve an exchange of potential energy and kinetic energy. When our mass is maximally extended
and is not moving, this system has no kinetic energy, but has plenty of potential
energy due to the stretched spring. We can recall that the potential
energy of this mass at this point is equal to one-half the spring constant 𝑘 times
its displacement from equilibrium, which is its amplitude squared.
If we then release the mass, that
potential spring energy will be converted into kinetic energy, energy of motion of
the mass, as it returns back to its equilibrium position. When the mass is in this position,
it no longer has potential energy due to the spring, but it does have kinetic
energy. That kinetic energy is equal to
one-half the mass times the speed of the mass squared. And because energy is conserved,
that’s equal to the spring potential energy one-half 𝑘𝐴 squared.
Cancelling out the common factors
of one-half, if we then divide both sides of this equation by the mass on the end of
the spring and also divide both sides by the amplitude squared, we find that the
spring constant 𝑘 over the mass on the spring 𝑚 is equal to the maximum speed the
spring has divided by its amplitude quantity squared. If we now take the square root of
both sides of this equation, see that we have a speed over a distance, speed over
amplitude. This distance capital 𝐴 is
effectively a radial distance.
And we can use the fact that linear
speed 𝑣 is equal to radial distance 𝑟 multiplied by angular speed 𝜔. This means that 𝑣 over 𝑟 is equal
to 𝜔. And on the right-hand side of our
expression, we have a speed over an effective radial distance. So, we make the substitution. And we find that the angular speed
of our oscillating spring system is equal to the square root of the spring constant
over the mass on the end of the spring. This means that the equation we’ve
chosen to show the position of our mass as a function of time is directly connected
to physical parameters of the scenario.
𝜔 is rooted in the spring constant
and the mass on the end of the spring. And those factors are what
determine the angular frequency of our system and, therefore, its period of
oscillation.
Speaking of our way of period, we
can remember that that period capital 𝑇 is equal to the inverse of the frequency of
oscillation. And since the frequency 𝑓 is equal
to the angular speed 𝜔 over two 𝜋, that means that period 𝑇 is also equal to two
𝜋 divided by 𝜔. Let’s get some practice now working
with an oscillating spring system through an example.
A type of clock keeps time by
the oscillation of a small object bouncing on a spring. What force constant of a spring
is needed to produce a period of 0.370 seconds for an object of mass 0.0210
kilograms?
We’ll label this force constant
we want to solve for 𝑘. And we can start on our
solution by recalling that the period 𝑇 is equal to two 𝜋 divided by an
angular speed 𝜔 and that 𝜔 for an oscillating spring system is equal to the
square root of the spring, or force constant 𝑘, divided by the mass of the
system 𝑚.
In the problem statement, we’re
told the values for the period 𝑇 and the mass 𝑚. So, we can combine these two
equations to help us solve for the force constant 𝑘. 𝑇 is equal to two 𝜋 times one
over 𝜔, which is equal to two 𝜋 times the square root of 𝑚 over 𝑘. When we rearrange this
expression to solve for that force constant 𝑘, we find it’s equal to four 𝜋
squared times the mass 𝑚 all divided by the period squared.
Plugging in for 𝑚 and capital
𝑇, when we enter this expression on our calculator, we find that, to three
significant figures, 𝑘 is 6.06 newtons per meter. That’s the force constant
needed in this oscillating spring to produce this period for this mass.
Let’s summarize what we’ve learned
so far about oscillating springs. We’ve seen that a mass on a spring
that moves according to Hooke’s law displays simple harmonic motion. This means that its motion is
described by a sinusoidal equation. We’ve also seen that a mass moving
in simple harmonic motion has a position described by the function position as a
function of time is equal to 𝐴 times the sin of 𝜔𝑡. In this equation, capital 𝐴 is the
amplitude of the oscillation and 𝜔 is the angular speed.
And finally, we saw that for an
oscillating spring system, by energy conservation, that angular speed 𝜔 is equal to
the square root of the force constant of the spring divided by the mass on it. Which means, we can rewrite our
position as a function of time to be equal to the amplitude of the wave times the
sin of the square root of 𝑘 over 𝑚 all multiplied by the time 𝑡.
