# Conceptual Question About Vertical Spring System

• lvslugger36
In summary, the conversation discusses the motion of a spring with a mass attached to it when released from rest. It is observed that when the mass passes through equilibrium, the force mg is equal to the force kx, where acceleration is zero. However, when considering the energy equation, it is found that 2mg = kx, which seems contradictory. It is clarified that this is because the equations are being mixed from different starting points. It is suggested to use the energy equations from the starting point y0, and it is noted that the stretch will be the same regardless of how the mass is lowered.
lvslugger36
Suppose there is a spring with a mass m attached to it, and it is released from rest. It will oscillate, starting from a certain point y0, going back to equilibrium and then down to let's say y2. Let's say in the first case, I call y1 my x = 0 point for the spring. When it passes equilibrium, is it not true that mg = kx since the acceleration is zero? However, let's repeat the case for which i will call the equilibrium point the x=0 for the spring. Now, from energy we have: mgx = kx^2/2. This means that 2mg = kx. How can this be possible?

lvslugger36 said:
Suppose there is a spring with a mass m attached to it, and it is released from rest. It will oscillate, starting from a certain point y0, going back to equilibrium and then down to let's say y2. Let's say in the first case, I call y1 my x = 0 point for the spring. When it passes equilibrium, is it not true that mg = kx since the acceleration is zero?
yes
However, let's repeat the case for which i will call the equilibrium point the x=0 for the spring. Now, from energy we have: mgx = kx^2/2. This means that 2mg = kx. How can this be possible?
You are mixing apples and oranges. For the 2nd case, the mass has kinetic energy at x= 0 (point y1). You are better off using the energy equations from the starting point y0. Then from the 2 'no speed' positions (at release and at the end point when the mass comes to a temporary stop), then mg(2x) = 1/2(k(2x)^2 , k = mg/x. Note that the stretch is double the stretch if the mass was slowly lowered to its final at rest position y1.

Last edited:
Thanks for your help; that's brilliant. However, wouldn't the stretch be the same even if it was lowered? For the first equation: x = mg/k. For the one slowly lowered, since the forces are balanced once it reaches equilibrium, shouldn't the stretch be x = mg/k also?

lvslugger36 said:
However, wouldn't the stretch be the same even if it was lowered? For the first equation: x = mg/k. For the one slowly lowered, since the forces are balanced once it reaches equilibrium, shouldn't the stretch be x = mg/k also?
Yes. When it is slowly lowered with your hand applying an external force, then you release it once it reaches it's equilibrium position, the stretch is mg/k. Or, if it is damped due to friction, it will ultimately, after oscillating back and forth a few times, settle to that same point, x =mg/k.

It is important to note that the equations mg = kx and mgx = kx^2/2 represent two different scenarios and cannot be compared in this way. In the first case, we are considering the forces acting on the mass at a specific point in time, when it is passing through equilibrium. At this point, the acceleration is zero and the forces are balanced, so mg = kx. However, in the second case, we are looking at the total energy of the system, which includes the potential energy stored in the spring. This energy is dependent on the displacement of the mass from the equilibrium point, hence the equation mgx = kx^2/2. It is not valid to equate these two equations as they represent different aspects of the system. The value of kx in the second equation represents the maximum displacement of the mass, not its position at a specific point in time. Therefore, it is possible for 2mg to equal kx in the second case, as it is a different scenario with different variables.

## 1. How does a vertical spring system work?

A vertical spring system works by using the force of gravity and the elasticity of a spring to create a back-and-forth motion. When the spring is stretched, it stores potential energy. When released, this potential energy is converted into kinetic energy, causing the spring to compress and expand repeatedly until the energy is dissipated.

## 2. What factors affect the behavior of a vertical spring system?

The behavior of a vertical spring system is affected by several factors, including the mass of the object attached to the spring, the stiffness of the spring, and the force of gravity. The amount of potential energy stored in the spring also plays a role in determining how high or low the object will bounce.

## 3. How does the length of the spring affect the motion of the system?

The length of the spring affects the motion of the system by determining the equilibrium position, or the point at which the forces of gravity and the spring are balanced. A longer spring will have a lower equilibrium position, while a shorter spring will have a higher equilibrium position. This affects how high or low the object attached to the spring will bounce.

## 4. Can a vertical spring system exhibit simple harmonic motion?

Yes, a vertical spring system can exhibit simple harmonic motion under certain conditions. In order to exhibit simple harmonic motion, the restoring force (in this case, the force of the spring) must be directly proportional to the displacement from the equilibrium position. If this condition is met, the motion will be periodic and can be described using mathematical equations.

## 5. How can a vertical spring system be used in real-world applications?

Vertical spring systems have a variety of real-world applications, including in shock absorbers for vehicles, pogo sticks, and trampolines. They are also commonly used in engineering and construction to absorb and dissipate energy, such as in earthquake-resistant buildings. Additionally, vertical spring systems are used in scientific experiments to study the behavior of springs and their relationship with other factors, such as gravity and mass.

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