Intro to Simple Harmonic Motion (Horizontal Springs) - Online Tutor, Practice Problems & Exam Prep

Simple harmonic motion (SHM) is exemplified by a mass-spring system, where the force is described by Hooke's law, F = kx. Key concepts include amplitude, period, frequency, and angular frequency, with relationships such as f = 1/T and ω = km. Velocity and acceleration vary sinusoidally, with maximum values at amplitude and equilibrium, essential for understanding oscillatory motion.

Intro to Simple Harmonic Motion

Hey, guys. So for this video, I'm going to introduce the motion of simple harmonic motion, the position velocity, and acceleration. I'm also going to be giving you guys some variables and definitions that'll chapter. So let's get to it. The most common type of simple harmonic motion that we'll study in this chapter, or sometimes you'll see as oscillation, is the mass-spring system. It’s where you attach some mass to some spring. So if I pull this thing all the way back here, right, as I pull this on a spring and I let it go, the only force that's going to be acting on that thing is the spring force, and that's equal to kx. I'm going to pull this thing back and I'm going to let it go. And basically, this thing is going to go all the way back to the equilibrium position, but it's going to overshoot it and then land on the other side. So it's going to go like something like that. And so basically, after that, it's just going to oscillate between those two points forever. Those maximum displacements here. So the displacement is maximum on the left side here and on the right side right here. And so at the equilibrium position, we know that the x is equal to 0. We call these two points on the outsides the amplitude. So we got plus a and minus a right here. The amplitude is just the maximum displacement that this object has. It’s always the initial push or pull that you apply against this object. So if you push it out to some distance and release it, that's the amplitude. What does the velocity look like when you do this? We're going to pull this thing all the way back, and then you're just going to let it go from rest. That means that at this point right here at the amplitude, the velocity is equal to 0. So then what happens is this thing starts speeding up faster and faster, goes through the equilibrium position, and when it gets to the other side right here, it has to stop and then turn around. When it gets to the other side, it has to stop and then turn around. That means that the velocity at this other amplitude is also equal to 0. And then what happens is as it's going back through here, the velocity is maximum here, and then it just keeps on doing that over and over again. We have maximum velocity at the center. And so it does this whole entire cycle over and over again. The two variables that are related to that are the period and the frequency. The period, which is, the letter t, is the time that it takes for you to complete one complete cycle, for you to finish one complete cycle. What's related to that variable is the frequency. These things are just related by inverses of each other. So you see that the frequency is just 1 divided by t. So if you ever have one of them, you can get to the other one. So for example, if I have a t that's equal to, like, 2 seconds, then the frequency is just going to be equal to 1 half of a cycle per second, and the units for that are going to be in hertz. Another variable that's related to the related to the frequency, but not the same thing, is the angular frequency. That's the letter omega. It’s the units are radians per second, and all omega is is just the frequency times 2 pi. We can also write it in a different way because the period and the frequency are inverses of each other. What do the forces look like at these three points? So when I pull it all the way back, the spring wants to push harder and harder against me. So at this maximum displacement here, that means the force is also going to be maximum. So take a look at kx. If x is maximum, then that means that F_s also has to be maximum. And so we know that the maximum forces are going to be here at the endpoints. And because of the equilibrium position, where the x is equal to 0, then that means that the force is equal to 0. Now what about the acceleration? Well, the acceleration and the force are related just by F = ma. So if F is max here, then that means the acceleration is max here and here. If F equals 0, then that means a is equal to 0. So it looks like x, F, and a are all related to each other, and they're all in sync. We’ve got x is 0, F is 0, and a is 0 at the middle, whereas the weird one is this velocity here. So that’s maximum when the other ones are 0. Let's take a look at an example here and see what we can find out.

Example

Hey, guys. Let's take a look at this example here. We've got a mass on a spring, and it's pulled 1 meter away from its equilibrium position. So we've got 1 meter, and we're just gonna draw a little box here. And then you release it from rest. This is a mass spring system. Right? So it's just gonna go and reach the other side where now the distance here, the displacement is gonna be negative 1 meter. And then it's just gonna go back and forth forever. So here in this first part, we are asked to calculate the amplitude, but we know that the amplitude is just the maximum displacements on either side. So that means that the amplitude of this is just 1 meter.

So the second question now, we're asked to find the period. So that is going to be that letter T. So what does the period look like? Well, we're told that the mass takes 2 seconds to reach the maximum displacement on the other side. So if you release it over here, then this time that it takes for you to go all the way over to the other side was equal to 2 seconds. Now the question is, is that the whole period? No. It's not. Because we said that the period is equal to the time that it takes you to complete one whole entire cycle here. So this 2 seconds really only represents a half period. This thing has to go all the way back to the other side for another 2 seconds, and that's going to be another half period, which means that these points in between here are actually quarter periods. Right? So this is a quarter. This is a quarter. These are all quarter periods. And this is the smallest sort of division that you can make. So you've got half periods and quarter periods. So we're told that it's 2 seconds to get to the other side, which means that the full period of this motion is going to be 4 seconds.

Now we've got, in this third part here, the angular frequency of the motion. So how do we relate angular frequency to the period? Well, we've got an equation up here that can do that. So if we're looking for the angular omega, we can use either 2 pi times the frequency, or we can use 2 pi divided by the period, which we actually know. So I'm just going to go ahead and use that. I've got omega equals 2 pi divided by the full period of the cycle, which is 4. And so what you should get is 1.57, and that's going to be radians per second. So that's it for this one. Let's keep going with some more examples.

ω = 2 π T

A mass-spring system with an angular frequency ω = 8π rad/s oscillates back and forth. (a) Assuming it starts from rest, how much time passes before the mass has a speed of 0 again? (b) How many full cycles does the system complete in 60s?

• A. 1/4 s; 480 cycles
• B. 1/8 s; 480 cycles
• C. 1/8 s; 240 cycles
• D. 1/4 s; 240 cycles

Equations of Simple Harmonic Motion

Hey, guys. So for this video, I'll be giving you the solid equations that we'll need for mass spring systems and simple harmonic motions. Let's check it out. So remember that as the spring is moving back and forth along the spring, the acceleration is always changing. And because it's never constant, we can't use the kinematic equations from early on in physics. Now the equations that we've seen so far are the force and the acceleration, but both of these things depend on x. So what if we wanted a new set of equations that we can actually depend on t on time for? So I'm not going to derive them, but they're right down here. So let's take a look at the subtle differences between them. They're all sinusoidal and the beginnings have \( a \omega \) and \( a \omega^2 \). Those are the front terms. Now, you have to be careful because the sign changes between these. We've got positive, negative, and negative, and we've also got changes between cosines and sines. So for those of you who've taken calculus, you'll be able to relate these functions using derivatives. There are 2 important things about these functions. The first is that these are functions of time. So what these questions will look like is they'll give you a specific time. And if you have \( a \) and \( \omega \), you can plug that into these equations to figure out what the position, velocity, and acceleration are. The second thing is that because we're using cosines and sines, just make sure that your calculator is in radians mode. Okay. So these were equations of time, whereas the equations that we've been dealing with so far, the old equations, are functions of position. So these are asking us for the forces and acceleration at a specific position. So now what if I wanted to know what the maximum values of these things are? Well, I can relate this back to my simple harmonic motion diagram. I know that at the endpoints, the acceleration and the maximum and the forces are maximum. So that means that the for the old equations, these things are maximized when x is equal to either amplitude. So because I've got these equations on the left, I can figure out their maximum forms. So I've got \( \pm k a \), and the acceleration is \( \pm \frac{k}{m} \times a \). So what about these bottom equations then? When are those things maximum? Remember that one's position and one's, for one's time. Well, if we take a look here, we've got cosines and sines, and cosines and sines always oscillate between positive one and negative one, and they do that forever. So that means that these things are going to be maximized when the sine or cosine graphs are equal to their endpoints, which is positive or negative one. And so a good way to remember where these things are maximum and what those are, these are just going to be whatever's on the front terms here. So what that means is that \( x_{\text{max}} \) is just going to be \( \pm a \), which we actually know. The maximum displacement is the amplitude. The maximum velocity is \( \pm a \omega \) and the maximum acceleration is \( a \omega^2 \). Okay. So we've gotten all the maximum terms for these. One other thing that we can do is we've gotten two expressions for the maximum acceleration. One is a function of position and one of time. So we combine those two things together. They represent the same variable, and we can use that to figure out what this \( \omega \) term is. Now up until now, the \( \omega \), which is the angular frequency, we've only been able to relate to the linear frequency and the period. But if you combine those two equations and solve for \( \omega \), what we'll get is that \( \omega \) is equal to the square root of \( \frac{k}{m} \). This is arguably the most important variable in this whole chapter, so make sure you remember that. \( \omega \) is always \( \sqrt{\frac{k}{m}} \). So to see how all of this stuff works, let's go ahead and do an example. So in this first example, I've got a 4 kilogram mass and it's at rest and it's attached to a spring. So I've got 4 kilograms. I'm given the k constant, and I'm told that it's pulled 2 meters. So now I'm supposed to find out what the angular frequency is. So I'm supposed to find out what \( \omega \) is. Let's write out my equation for \( \omega \). I've got \( \omega = 2\pi \times \text{frequency} \) and I'm not given any information about the frequency or the period. So what I'm going to use is this new \( \omega \) that gives me 7.07. And I know the units for that are radians per second. So now that I've got the angular frequency, now for this part b, I'm asked for when \( t = 0.5 \) seconds, what is the velocity? So I'm given a \( t \) and I'm asked to find out what the velocity is. So I can use my velocity as a function of t. And so let's go ahead and look up there for a second. So I'm told that the velocity of \( t \) as a function of t is that equation. So I've got \( -a \omega \) and then I've got \( \sin(\omega \times t) \). So I know what my \( \omega \) is, I just figured that out. I'm going to plug in that 0.5 seconds for \( t \). Now all I need to do is find the amplitude. So I'm told in this case that the mass has pulled 2 meters. So that means that I've got an amplitude that's equal to 2 meters. So it means that the velocity as a function of time is equal to \( -2 \times 7.07 \times \sin(7.07 \times 0.5) \). Now if you do this and you make sure that your calculator is in radians mode, we'll get a velocity when \( t = 0.5 \) seconds. That's equal to 5.42 meters per second. So all this means, notice how I didn't get a negative sign. This positive sign means the velocity at this particular time points to the right. So let's move on to part c now. Part c is similar to part b, except now it's giving us a position. It's telling us the position is equal to 0.5, and now we're supposed to find the acceleration. So now we're going to use the position formulas, the position functions. If I want to find out the acceleration as a function of position, then I'm just going to use \( -\frac{k}{m} \times x \). So the acceleration when the position is equal to 0.5 is going to be equal to \( -\frac{200}{4} \times 0.5 \). So I'm getting an acceleration of negative 25 meters per second squared. So that is the acceleration. And all that means is that at this particular instance, the now for this last one here, I'm supposed to figure out what the period of oscillation is. So now for this last one here, I'm supposed to figure out the period. So for part d, I'm supposed to figure out the period. So \( t \). So let's go ahead and look up in my equations and figure out which one has \( t \). So I've got this, big \( \omega \) equation in here and this \( \omega \) contains that variable \( t \) in there. So I'm going to write that out. So \( \omega = 2\pi \times \text{frequency} = \frac{2\pi}{t} = \sqrt{\frac{k}{m}} \). So all of these things are equal to each other and I've got I've already got what my \( \omega \) term is, and I'm just trying to figure out what this \( t \) term is. So I can just go ahead and use that relationship directly. So I've got, \( \omega = \frac{2\pi}{t} \), and now all I can do What I can do is just, just trade places. So the \( t \) will come up and then the \( \omega \) will take its place. So they'll just trade places there. So I've got \( t = \frac{2\pi}{\omega} \). So \( t = \frac{2\pi}{7.07} \). So what I get is a period of 0.89 seconds. Alright, guys. So that's it for this one. Let's keep going.

A 4-kg mass on a spring is released 5 m away from equilibrium position and takes 1.5 s to reach its equilibrium position. (a) Find the spring's force constant. (b) Find the object's max speed.

• A. 3.28 N/m; 4.52 m/s
• B. 4.39 N/m; 5.25 m/s
• C. 17.54 N/m; 10.47 m/s
• D. 2.01 N/m; 3.54 m/s

Example

Hey, guys. Let's take a look at this example. So we're told that a 4-kilogram mass is on a spring. It's oscillating at 2 hertz and it's moving at 10 meters per second once it crosses the equilibrium position. So on a mass spring system, when it's oscillating back and forth, as it crosses the equilibrium position right here, we know that the speed is maximum. So that's what they're telling us. They're telling us that v_{\text{max}} = 10 \text{ m/s}. So I'm going to start writing everything out. I know that the mass is equal to 4. I've got the frequency is equal to 2, and I've got v_{\text{max}} = 10. And what I'm supposed to find is I'm supposed to find the time it takes to get from equilibrium all the way out to its maximum distance. So in other words, how long does it take to get from equilibrium all the way out to its maximum distance? We know that that is 1 quarter of the period. One full period is the whole entire cycle, so we're just looking for that quarter. So if we're looking for t/4, we might as well just find what t is. So let's just use our equations to find out what time is. Well, I've got the big omega equation down here, but I also know that t and the frequency are inverses of each other. And I have what the frequency is. So that means that the period is just one half of a second. But that's not what I'm looking for, I'm looking for a quarter. So if I got t/4, that's just going to be 1 quarter of 1 half of a second, and so that equals 1 eighth of a second. So that's the answer to part a. So what does part b ask us? Part b asks us to find out what the amplitude is. Let me write that down here. So we're supposed to figure out what A is. Let's look at all of our equations and figure out where A is. Well, A is kind of present in all of them, so let's rule out the ones that we can't. We don't know anything about the mass. We don't know anything about the maximum acceleration. We don't know anything about time, so we can't use these guys. So basically, I'm just going to have to use this equation, all my max equations. So I've got I don't know what x_{\text{max}} is because otherwise that would be the amplitude, and I don't know what the acceleration max is either. But I do know what the v_{\text{max}} is. So let me go ahead and use that equation for v_{\text{max}} because that's the one that I know most about. So I've got v_{\text{max}} = A \times \omega. So if I rearrange for this, I've got that A = v_{\text{max}} / \omega. So now I have what v_{\text{max}} is. I don't know what omega is, so let me go ahead and find that. So let me check that. I've got v_{\text{max}}. Now I just have to go over here and find out what \omega is. So let's use my big omega equation. Omega is equal to 2\pi frequency. Do I know the frequency? Yes, I do. So that means 2\omega = 2\pi \times f. So \omega f is just equal to 2. The frequency is 2 hertz. So the omega is equal to 4\pi. So I'm just going to stick that right back in there. So that means that the amplitude is just 10 meters per second divided by, right, that's the V_{\text{max}}, and then I've got 4\pi. So I can just go ahead and simplify that and say that it's \frac{5}{2\pi}, and I'm going to go ahead and highlight that and box it so you guys see it. So we've got \frac{5}{2\pi}. Now this last one is asking me to find the maximum acceleration. So now we're actually going to go ahead and solve for a_{\text{max}}. So which one are we going to use? We've got this a_{\text{max}} over here, but I'm going to need to know k, and I don't have the spring constant. So instead, I'm going to use not this a_{\text{max}}. I'm going to use this a_{\text{max}}. So I'm going to use the a \omega^2. So a_{\text{max}} is the amplitude which is \frac{5}{2\pi}, and then I've got omega which is 4\pi. So if I square this guy, it's just going to be 16 \times \pi^2. Right? So this is 4\pi. So \omega^2 = 16\pi^2. So now what happens is I've got a \pi on the bottom in the denominator. Got a \pi in the numerator so they cancel. And then I've got a 16 in the numerator and a 2 in the denominator. So all of that stuff simplifies. So I've got a_{\text{max}} = 5 \times 8 \times \pi. So that means a_{\text{max}} = 40 \times \pi, and that's the answer. So you should definitely become familiar with all these pies popping up all over the places. Let me know if you have any questions. Let's keep moving on.

What is the equation for the position of a mass moving on the end of a spring which is stretched 8.8cm from equilibrium and then released from rest, and whose period is 0.66s? What will be the object's position after 1.4s?

• A. .88cos(19.04t); .79 m
• B. .088cos(19.04t); .08 m
• C. .88cos(9.52t); .86 m
• D. .088cos(9.52t); .064 m

Example

Hey, guys. Let's take a look at this example. So we're told that the velocity of a particle is given by this equation, and that's the only thing that we're given, and we're supposed to find out what the frequency of motion is. So what they're asking us is to figure out what f is. Let's take a look at my equation list. The only equation that involves frequency is the big omega equation right here. The problem is I don't have any of those other variables. I don't have k over m, I don't have the period, and I don't know what the omega is either. The only thing I am given is just the velocity equation. Let's take a look at the actual velocity equation with all the variables involved. So I've got the velocity is equal to negative 6 times the sine of 3 pi times t. But the form of the equation is I have negative a omega, and I've got sine of omega times t. So if you just look at this, what is this telling us? Well, we've got this negative 6 out here, and I've got a negative a omega. So that means that these things are equal to each other. So that means I have that a omega is equal to 6. And if you look inside the parentheses, I've got this omega over here that's equal to this 3 pi. So those two things are the same. So that means that omega, what they're telling me, is equal to 3 pi. Now let's take a look at my frequency equation. I don't have the period and I don't have the square root of k over m, but now I do have the omega. So let's relate that equation. So I've got omega is equal to 2 pi times the frequency, but omega, I just figured out was 3 pi. So I'm gonna set these 2 things equal to each other. So 2 pi times the frequency equals 3 pi, and now the frequency is just 3 pi over 2 pi, and what'll happen is the pi's will cancel, and so I get frequency like, the actual form of the equation to figure out what those things are. Let's look at part b. Part b is now asking us for the amplitude. So great. How do we find out the amplitude? We've got a bunch bunch of equations that involve the amplitude. So let's take a look. I'm not told the x max, v max, or a max. I'm never told any of those things, and I don't know what the a max or the restoring force is, so I can't use that. So I'm gonna have to again look at this equation right here. And from this equation, I figured out that A omega was equal to 6. So let's write that out. If A omega is equal to 6, then a is just going to be 6 divided by omega. But I figured out what omega was as well. Remember that omega was just equal to 3 times pi right over here. So that means that the amplitude is just 2 over pi once you go ahead and solve that. So I've got that's the amplitude, 2 over pi. Now for this last one, I'm supposed to find out what the velocity is at a specific time. So this is a question where they're giving me t and I've asked for v, so I'm just going to basically plug them into my formula. So the velocity when t is equal to 0.5 seconds is gonna be negative 6 times the sine of 3 pi times 0.5. Make sure that you're in radians mode, and you should get a velocity that's equal to 6 meters per second. And that's it. Let me know if you guys have any questions, and if not, let's keep going.

Here’s what students ask on this topic:

What is simple harmonic motion in a mass-spring system?

Simple harmonic motion (SHM) in a mass-spring system occurs when a mass attached to a spring oscillates back and forth around an equilibrium position. The restoring force is described by Hooke's law, $F=kx$, where $k$ is the spring constant and $x$ is the displacement from equilibrium. Key parameters include amplitude (maximum displacement), period (time for one complete cycle), frequency (cycles per second), and angular frequency (related to frequency by $\omega =2\pi f$). The motion is sinusoidal, with velocity and acceleration varying periodically.

How do you calculate the period and frequency of a mass-spring system?

The period $T$ of a mass-spring system is the time it takes to complete one full oscillation. It is given by $T=\frac{2}{\pi }\sqrt{\frac{m}{k}}$, where $m$ is the mass and $k$ is the spring constant. The frequency $f$ is the number of oscillations per second and is the reciprocal of the period: $f=\frac{1}{T}$. The units of frequency are hertz (Hz).

What is the relationship between angular frequency and the spring constant in SHM?

In simple harmonic motion (SHM), the angular frequency $\omega$ is related to the spring constant $k$ and the mass $m$ by the equation $\omega =\sqrt{\frac{k}{m}}$. This relationship shows that the angular frequency increases with a stiffer spring (higher $k$) and decreases with a larger mass. Angular frequency is measured in radians per second (rad/s).

How do you determine the maximum velocity and acceleration in a mass-spring system?

The maximum velocity $v$_max in a mass-spring system occurs at the equilibrium position and is given by $v$_max=Aω, where $A$ is the amplitude and $\omega$ is the angular frequency. The maximum acceleration $a$_max occurs at the maximum displacement (amplitude) and is given by $a$_max=Aω². These relationships highlight the sinusoidal nature of SHM, where velocity and acceleration vary periodically.

What are the key differences between position, velocity, and acceleration equations in SHM?

In simple harmonic motion (SHM), the position, velocity, and acceleration equations are sinusoidal functions of time. The position $x$ is given by $x=A\cos \left(\omega t\right)$. The velocity $v$ is the derivative of position and is given by $v=-A\omega \sin \left(\omega t\right)$. The acceleration $a$ is the derivative of velocity and is given by $a=-A\omega 2\cos \left(\omega t\right)$. These equations show how position, velocity, and acceleration are phase-shifted relative to each other.