Simple Harmonic Motion of Vertical Springs - Online Tutor, Practice Problems & Exam Prep

In vertical mass-spring systems, the equilibrium position is where the restoring force equals the weight of the mass, represented by the equation $k\mathrm{\Delta L}=\mathrm{mg}$. The distance the spring stretches to reach equilibrium is $\mathrm{\Delta L}$, while any additional displacement is the amplitude. Understanding these concepts is crucial for analyzing oscillatory motion and energy conservation in spring systems.

Simple Harmonic Motion of Vertical Springs

We're going to see that vertical systems are very similar to horizontal, and the only difference has to do with the equilibrium position. But rather than tell you, I just want to show you using this example. So let's take a look. We've got a 0.5 meter spring, and it's hanging from the ceiling. So we've got that distance 0.5 meters, and that's going to be the original length of the spring. Then it stretches down by some distance after I attach a mass to it. Now, why does it stretch down and stop? Well, we've got an extra force to consider because now we have the object's weight that pulls it down, whereas, originally, we didn't have that before in horizontal systems. So as the thing is pulling the spring down, the restoring force gets higher and higher upwards. And so what ends up happening is that that restoring force ends up cancelling out with the gravity, and so it reaches a new equilibrium position. So in horizontal mass spring systems, we said that x equals 0 is where like no forces were acting on it. But in vertical mass spring systems, the equilibrium is where these forces will cancel out. That's the important part. And so we call that distance, that hanging distance until it reaches equilibrium delta l. So in this case, it is delta l equals to 0.2. So again, that delta l really represents the spring system's hanging deformation or hanging distance. It's the amount that you need to stretch the spring so that you reach equilibrium. And that equilibrium condition is where these two forces balance out. Well, the upward force is going to be k times delta l, and the downward force is going to be mg. So that means at equilibrium, we've got k δ l = m g . So let's look at the second part of this problem. 

After all of this happens, we're going to pull the spring system an additional 0.3 meters downward. So now, just like in horizontal spring systems, your additional push or pull was the amplitude. So that means that once you pull this thing downwards 0.3, now this thing is just going to go up and down between these two amplitudes. So this is going to be the positive and that's going to be the negative amplitude. So it's just going to oscillate up and down like that. And so it's important to remember that this amplitude is the additional push or pull, and it's not the delta l. So it's not the natural hanging distance or deformation; it's the additional push or pull. So in problems, what you'll see is that you're going to attach a mass, and it's going to stretch by some distance. What that represents is delta l, and then you're going to pull it down an additional something, and that is going to be the amplitude. So don't confuse those two. Alright. So now basically we have everything we need to solve the problem. This first part is asking us for the force constant, so it's asking us for k. So we're just going to use this new equilibrium equation that we have. So we've got k δ L = m g . Now I know what mg and delta L are. I just have to figure out what this k constant is. So I've got k, once I just move this to the other side, I've got k = 5 ⋅ 10 for gravity, and then the delta L is what? Well, the delta L we said, that the natural stretching distance was 0.2 meters. So that means I got a k constant of 250 newtons per meter, and that's it. So what about the second part here? The second part is now asking us at its maximum height, so it's oscillating. At maximum height, how far is that ceiling, from the block? Great. So let's check it out. So we've got this thing here, and so I'm going to represent this whole entire line here as this motion. The original distance, the original length of the spring was that black dot, and then you hang it down, and it reaches some equilibrium position. And then you're going to pull it down an additional 0.3 meters. So that's the amplitude, and it's just going to oscillate between these two points. Okay. So we're being asked for basically this distance right here. What is the distance between the ceiling and its maximum height? That's where it reaches that maximum amplitude. And so that distance here is going to be the letter d. Well, we're told that this thing has an original length of 0.5 meters, and the equilibrium position is when it stretches an additional 2. So that means that the length of equilibrium, I'll call this Leq, is equal to 0.7 meters. So now what happens is it goes 0.3 down and then 0.3 up. So at this bottom part here, this length here is 1.0, And at the top point right here, it's 0.7 minus 0.3. And so we say that at maximum height, the distance d away from the ceiling is equal to 0.4 meters, and that's it. Alright, guys. That's it for vertical oscillations.

A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is then pulled down 6.0 cm and released. If the ball makes 30 oscillations in 20 s, what are its (a) mass and (b) maximum speed?

• A. .17 kg; .56 m/s
• B. .63 kg; .29 m/s
• C. .17 kg; 1.12 m/s
• D. .63 kg; .15 m/s

Example

Hey, guys. Let's check out this problem. So we've got an elastic cord, and we're told that it's 65 centimeters long when some weight hangs from it. And then we attach more weight from it, and it stretches down even farther. So I'm just gonna go ahead and draw a little picture for that. So in this case, for this first box, let's go ahead and draw out the forces. We know that there's a weight acting on this box and that's \( m_1 g \). But \( m_1 g \) is the weight of this object, and we're told that that's equal to 75. Now when that happens, it's in equilibrium with the elastic cord which you can kind of just think about as a spring. And the equilibrium condition is that we have \( k \times \Delta l_1 \). Right? So we have the restoring force that acts upwards. Now for the second situation, it's the exact same thing. So in this case, we've got \( m_2 g \) and we know that that's equal to 180. And the restoring force acting in this direction is now \( k \times \Delta l_2 \) because this spring has now gotten longer now that \(\Delta l\) is even farther. It's sagging even farther. So we've got 2, like, equilibrium conditions to consider here. We are always gonna start with our equilibrium condition that \( k \times \Delta l \) is equal to \( mg \). So for the first box, \( k \times \Delta l_1 \) is equal to \( m_1 g \). And now for the second situation, we've got \( k \times \Delta l_2 \) equals \( m_2 g \). Now let's look at all of our variables. What are we looking for? We're looking for the spring constant \( k \). So really what we're gonna be solving in this problem is \( k \). And the \( k \) is the same for both of the elastic cords because it's the same thing. Okay. So I've got \( m_1 g \) is equal to 75, and I've got \( m_2 g \) is equal to 180. Now, what about this \( l_1 \) and \( l_2 \)? So what about these distances that I'm told? Is the 65 centimeters the \( l_1 \) and the 85 centimeters the \( \Delta l_2 \) or the \( l_2 \)? Well, no, we can't say that because the \(\Delta l\)s represent the change in distance. So remember that \(\Delta l\) so I'm gonna write this out. That \(\Delta l\) is actually just the length final minus length initial. That's what \(\Delta l\) represents. So really what they're actually giving us is the final length of the spring. This is the number that they're giving us and that is equal to 75. For the second one, we have \( k \times (l_{f2} - l_i) \) is equal to 180. Okay. So we've got now we've got 2 variables to consider. We We don't know what the \(\Delta l\) is because we don't know what the initial length of the spring is. So these \( k \)s and these \( l_i \)s are unknowns. What I am given though is that \( l_{f1} \) is equal to 0.65, 65 centimeters and \( l_{f2} \) is equal to 0.85. So really what this becomes is an equation like this. So \( k \times (0.65 - l_i) \) equals 75 and then \( k \times (0.85 - l_i) \) is equal to 180. Okay. So here we've gotten to a situation where I have 2 unknowns. Right? I've got this \( k \) that's unknown and the \( l_i \) that's unknown, \( k \) and \( l_i \), and the rest are just numbers. So what have we done in the past to sort of deal with the situation? We always either do substitution or we add the two equations together. But first what I'm gonna do is sort of expand this and I'm just gonna multiply this \( k \) straight throughout. So I get \( k \times 0.65 - k \times l_i = 75 \) in this formula, and Then I've got \( 0.85 k - k l_i = 180 \). That's just distributing the \( k \)s into both of these terms. Okay. So now what I'm gonna do is if I'm gonna do substitution or I'm just gonna stack these equations on top of each other, let's see what I get. So I'm gonna bring these 2 together. So I've got \( 0.85 k - k l_i = 180 \), and then we've got \( 0.65 k - k l_i = 75 \). So what do I do with these two equations to get rid of those 2 unknowns? Just have to subtract those 2 equations. So if you subtract these two equations, what's gonna happen is that these two things, a minus, and a minus are gonna cancel out and then you just have to subtract straight down. So \( 0.85 - 0.65 = 0.20 k \) and then \( 180 - 75 = 105 \). So now \( k \) is just equal to 525 newtons per meter. So that is our force constant. So let me know if you guys have any questions for this one. I know this is kind of a weird problem.

A chair of mass 30 kg on top of a spring oscillates with a period of 2s. (a) Find the spring's force constant. You place an object on top of the chair, and it now oscillates with a period of 3s. (b) Find the object's mass.

• A. 296 N/m; 1286.9 kg
• B. 188 N/m; 42.9 kg
• C. 188 N/m; 817.4 kg
• D. 296 N/m; 38.1 kg

Here’s what students ask on this topic:

What is the equilibrium position in a vertical mass-spring system?

In a vertical mass-spring system, the equilibrium position is where the restoring force of the spring equals the weight of the attached mass. This is represented by the equation:

$k\mathrm{\Delta L}=mg$

Here, $k$ is the spring constant, $\mathrm{\Delta L}$ is the distance the spring stretches to reach equilibrium, $m$ is the mass, and $g$ is the acceleration due to gravity. At this position, the forces are balanced, and the system is in a state of rest.

How do you calculate the spring constant in a vertical mass-spring system?

To calculate the spring constant $k$ in a vertical mass-spring system, you can use the equilibrium condition where the restoring force equals the weight of the mass:

$k\mathrm{\Delta L}=mg$

Rearranging for $k$, we get:

$k=\frac{mg}{\mathrm{\Delta L}}$

Here, $m$ is the mass, $g$ is the acceleration due to gravity, and $\mathrm{\Delta L}$ is the distance the spring stretches to reach equilibrium. By substituting the known values, you can solve for $k$.

What is the difference between the equilibrium position and amplitude in a vertical spring system?

The equilibrium position in a vertical spring system is the point where the restoring force of the spring equals the weight of the attached mass, represented by $k\mathrm{\Delta L}=mg$. This is the natural hanging distance of the spring.

Amplitude, on the other hand, is the additional displacement from the equilibrium position when the spring is pulled or pushed. It represents the maximum extent of oscillation on either side of the equilibrium position. Unlike $\mathrm{\Delta L}$, which is the static stretch due to the mass, amplitude is the dynamic displacement during oscillation.

How do you determine the maximum height of a block in a vertical spring system?

To determine the maximum height of a block in a vertical spring system, you need to consider the original length of the spring, the equilibrium position, and the amplitude of oscillation. The equilibrium position is given by:

$k\mathrm{\Delta L}=mg$

Let's denote the original length of the spring as $L$ and the equilibrium length as $L$_eq. The maximum height is reached when the block oscillates to its highest point, which is $L$_eq minus the amplitude $A$. Therefore, the distance from the ceiling to the block at maximum height is:

$d=L$_eq-A

What is the role of the spring constant in vertical oscillations?

The spring constant $k$ plays a crucial role in vertical oscillations as it determines the stiffness of the spring. A higher spring constant means a stiffer spring, which requires more force to stretch or compress. In the context of vertical oscillations, the spring constant is used to calculate the equilibrium position using the equation:

$k\mathrm{\Delta L}=mg$

It also affects the frequency of oscillation, with a stiffer spring resulting in a higher frequency. The spring constant is essential for understanding the dynamics of the system and predicting its behavior under various conditions.