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 metres 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 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 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 our 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 that 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 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 metre. 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.
