Twin spring system with a mass in between (X and Y direction)

In summary: Excel spreadsheet you have already created with this information?I have created an Excel spreadsheet with this information. Please see the link below. Excel File - Motion of Mass with Two Springs and Gravity Excel File - Motion of Mass with Two Springs and Gravity Excel File - Motion of Mass with Two Springs and Gravity Excel File - Motion of Mass with Two Springs and GravityIn summary, In order to solve for the motion of a mass with two springs and gravity, the user first draws the free body diagram of the forces acting on the mass at equilibrium, and then looks at how those forces change as the mass moves vertically and horizontally from that equilibrium position.
  • #1
MrNewton
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3
Hello my fellow scientists!

This is my first post on this forum, but most certainly not my last. I am relatively new to physics. I mean, i understand the basics. I would like to make it my hobby. You know, really find a passion for the physics. BTW, i am from the Netherlands, so please do not judge my grammer :)


1. Homework Statement

First, see the appendix.

If the springs are coupled with each other, Without the mass, the springs are at restposition. When the mass is applied between them the springs extend to a certain place.Thereafter a punch is applied to the mass in the X Direction.

I need to find the movement of the mass in both direction X and Y.

There is not much details given, i can fill this in for myself. Ofcourse i am assuming the simulation takes place on earth, with an acceleration g = 9,81m/s^2

Fs = Force Spring
Fg = Force Gravity

Homework Equations


I have done this for a single spring/mass system under a angle of approximatly 45 degrees where a plate is also applied under an angle of 45 degrees so the mass doesn't fall straight down.

i am not sure which equations i need to apply. In this case. I need a start on this project so i can finish it on my own.

The Attempt at a Solution



I am not sure where to begin in this project. I have done it before, but only with 1 spring and 1 mass under a 55 degree angle, with a bottom plate liniear to the spring and mass so the mass wouldn't fall straigt down.

But since there are 2 springs now its is a lot more complicated (in my opinion)
My question to you, can you please help me make a start on this project? I am not asking for you to solve this for me, but guide me trough it.

Thank you in advance!
 

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  • #2
Welcome to the PF. :smile:
MrNewton said:
i am not sure which equations i need to apply. In this case. I need a start on this project so i can finish it on my own.
Is your intention to write a numerical simulation of the motion of the mass, or to come up with a solution for the motion?

In either case, start by drawing the Free Body Diagram (FBD) of the forces on the mass at the equilibrium position, and then look at how those forces change as the mass moves vertically and horizontally from that equilibrium position. Can you make that start and show us your work?
 
  • #3
berkeman said:
Welcome to the PF. :smile:

Is your intention to write a numerical simulation of the motion of the mass, or to come up with a solution for the motion?

Thanks for your fast reply! It is indeed my intention to wire a simulation of the motion of the mass (i will do this in excel eventually).

I am working hard on a FBD, i will upload it when i have it figured out.
 
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  • #4
berkeman said:
Welcome to the PF. :smile:

In either case, start by drawing the Free Body Diagram (FBD) of the forces on the mass at the equilibrium position, and then look at how those forces change as the mass moves vertically and horizontally from that equilibrium position. Can you make that start and show us your work?

So here is my FBD. The two springs are the diagonal vectors(Fs2 & Fs1), and the gravity downwards is Fg
The constant of the springs are 10kg/m.
I divided the diagonal forces of the springs in there X component and the Y component. Fg has an Y component of 0 at the moment of rest.

So the resulting force in the Y direction is: Fry = Fg - (Fs2y+ Fs1y)

The resulting force in the x direction is: Frx = Fs2x -Fs1x

Fs1 = Force Spring 1
Fs2 = Force Spring 2
Fry = Force Resulting Y direction
Frx = Force Resulting X direction
Fs2x = Force Spring 2 X direction
Fs2y = Force Spring 2 Y direction
etc.Sorry for the metric units, i know you are used to the american units since you are in sillicon valley(?)
 

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  • #5
MrNewton said:
Sorry for the metric units, i know you are used to the american units since you are in sillicon valley(?)
Never be sorry for using SI units! :smile: (I don't even know how to use non-SI units...)

Seems like a good start. The directions and magnitudes of the spring forces will change with the (x,y) position of the mass, so that will all get factored into the simulation. I like that you are going to do it in Excel (at least at first), since I often do such initial simulations with an Excel spreadsheet. You will need to decide on a good time step value to use to give you good resolution (base it on an initial estimate of the resonant frequency of the system, and some target accuracy).

Can you say what the columns will be in your spreadsheet? :smile:
 
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  • #6
What is the significance of the 60 cm horizontal distance as shown in the figure? Is it fair to consider the following picture: There is a single spring of relaxed length 60 cm which is attached to two points on the ceiling. Then a mass is attached at its mid point (30 cm) mark and the mass is allowed to descend to its equilibrium position. Then a sharp impulse is delivered to the mass in the same plane as the plane defined by the extended spring. I am asking because I used google translate (it's getting better) and that's the impression I got. It also seems that you are asked to provide a simulation and not necessarily a solution. Is that correct?
 
  • #7
MrNewton said:
So here is my FBD.
It's not easy to read - too dark - but it looks like an FBD for the original position. You need an FBD for an arbitrary time later. There will be x and y displacements from that original position, leading to different angles and spring lengths. You will also need variables for x and y velocities and accelerations.

Next step is to write the ΣF=ma equations for the x and y directions.
 
  • #8
kuruman said:
What is the significance of the 60 cm horizontal distance as shown in the figure? Is it fair to consider the following picture: There is a single spring of relaxed length 60 cm which is attached to two points on the ceiling. Then a mass is attached at its mid point (30 cm) mark and the mass is allowed to descend to its equilibrium position. Then a sharp impulse is delivered to the mass in the same plane as the plane defined by the extended spring. I am asking because I used google translate (it's getting better) and that's the impression I got. It also seems that you are asked to provide a simulation and not necessarily a solution. Is that correct?

Good job on the translation! You are almost right. Except that there is not a single spring of 60 cm connected to the ceiling, but the 2 springs are connected to each other, without the mass. So in my opinion by the weight of the springs they are probably not parallel to the ceiling but the point where they are connected is a little towards the ground. I am not sure, it is a little vague. I don't believe that this factor is that important?

Translation of the text from the first picture:
When the springs are connected to each other without the mass they are at their equilibrium position.(relaxed position).
If a mass is applied the springs will be stretched to a certain point. Thereafter a punch is applied.
 
  • #9
berkeman said:
Never be sorry for using SI units! :smile: (I don't even know how to use non-SI units...)

Seems like a good start. The directions and magnitudes of the spring forces will change with the (x,y) position of the mass, so that will all get factored into the simulation. I like that you are going to do it in Excel (at least at first), since I often do such initial simulations with an Excel spreadsheet. You will need to decide on a good time step value to use to give you good resolution (base it on an initial estimate of the resonant frequency of the system, and some target accuracy).

Can you say what the columns will be in your spreadsheet? :smile:

Absolutely, timestep is essential. I am not sure which timestep i will use. I will probably find out the best one by trying. If you have suggestions please tell me :).

I didnt find enough time tonight to look further into the simulation, but my colums will be:
Timestep = Timestap of the event
Fg = Force gravity
Fx = Force in X direction
Fy = Force in Y direction
Fxy = Force in X & Y direction combined(pythagoras)
Ax = acceleration in the X direction
Ay = acceleration in the Y direction
X= x position of the mass
Y = Y position of the mass

Am i missing any critical columns?
 
  • #10
haruspex said:
It's not easy to read - too dark - but it looks like an FBD for the original position. You need an FBD for an arbitrary time later. There will be x and y displacements from that original position, leading to different angles and spring lengths. You will also need variables for x and y velocities and accelerations.

Next step is to write the ΣF=ma equations for the x and y directions.
Thank you kindly for your reply. Bij F=ma, F is force, m = mass, a = acceleration i assume?
 
  • #11
MrNewton said:
Thank you kindly for your reply. Bij F=ma, F is force, m = mass, a = acceleration i assume?
Yes, and ΣF means the sum of all forces, i.e. the net force.
 
  • #12
MrNewton said:
Except that there is not a single spring of 60 cm connected to the ceiling, but the 2 springs are connected to each other,
Apart from the effective spring constant, it's the same as if it were a single spring.
MrNewton said:
the weight of the springs they are probably not parallel to the ceiling but the point where they are connected is a little towards the ground
You are supposed to treat the springs as massless.
 
  • #13
MrNewton said:
Am i missing any critical columns?
Yes, as I posted, you will need x and y velocities.
 
  • #14
haruspex said:
Apart from the effective spring constant, it's the same as if it were a single spring.

You are supposed to treat the springs as massless.

Thank you. I will try to convert this into an simulation and come back to you guys! (/girls)
 
  • #15
Since the mass is in its equilibrium position, the sum of all the forces must be 0. So Fs1x = Fs2x
And the same goes in the Y direction. But look! Fg must be equal to the sum of Fs1 &Fs2.
But when i calculate Fs1 +Fs2 the answer is, 14,14 kg/m. (The springs, Fs1 & Fs2 are the same, 10kg/m and the angle is 45 degrees)
What does this mean for Fz? How do i calculate Fz to make the answer 14,14?
In my opinion Fg = m(ass) * a(cceleration) = 2kg * 9.81 = 19,62. Am i doing something wrong?

EDIT: I just realized something when i was in the shower. The only value given in the assignment was the distance between the springs (60cm). The constant for the spring(10kg/m), the mass(2kg) and the angle of the spring(45 degrees), i declared them myself. But they probably must all be in proportion to each other. So for example i can declare the spring constant and the mass myself, but a specific angle comes with them. In that case i cannot declare the angle myself am i right? Another example, if a declare the spring constant, and the angle of the spring, a spicific mass comes with them. I cannot declare the mass myself?

So this means that F should equal (Fs1+Fs2).
That gives the next equation: Fg = m * a
14,14 = m * 9,81
m = 14,14 / 9,81
m = 1,44kg
 

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  • #16
Are you interested in the full, nonlinear description applicable for large motions, or are you looking for a linearized, small motion description?
 
  • #17
MrNewton said:
Since the mass is in its equilibrium position, the sum of all the forces must be 0. So Fs1x = Fs2x
And the same goes in the Y direction. But look! Fg must be equal to the sum of Fs1 &Fs2.
But when i calculate Fs1 +Fs2 the answer is, 14,14 kg/m. (The springs, Fs1 & Fs2 are the same, 10kg/m and the angle is 45 degrees)
What does this mean for Fz? How do i calculate Fz to make the answer 14,14?
In my opinion Fg = m(ass) * a(cceleration) = 2kg * 9.81 = 19,62. Am i doing something wrong?

EDIT: I just realized something when i was in the shower. The only value given in the assignment was the distance between the springs (60cm). The constant for the spring(10kg/m), the mass(2kg) and the angle of the spring(45 degrees), i declared them myself. But they probably must all be in proportion to each other. So for example i can declare the spring constant and the mass myself, but a specific angle comes with them. In that case i cannot declare the angle myself am i right? Another example, if a declare the spring constant, and the angle of the spring, a spicific mass comes with them. I cannot declare the mass myself?

So this means that F should equal (Fs1+Fs2).
That gives the next equation: Fg = m * a
14,14 = m * 9,81
m = 14,14 / 9,81
m = 1,44kg
It's not clear to me which values are given. I see the spring constant and relaxed length are given, but I don't see a given mass or angle. I wouid assume that at some point a mass will be supplied and the angle should be computed from that, not the other way about.
But the more interesting work is the simulation. You could just leave all the initial conditions as unknowns for now and work on that.
 
  • #18
haruspex said:
It's not clear to me which values are given. I see the spring constant and relaxed length are given, but I don't see a given mass or angle. I wouid assume that at some point a mass will be supplied and the angle should be computed from that, not the other way about.
But the more interesting work is the simulation. You could just leave all the initial conditions as unknowns for now and work on that.

The only value given is the distance between the springs (60cm). I do not know if the drawing is in scale or not (i assume not). It is either that, or i can declare the springconstant and mass myself.

You are right, but i want to calculate the forces also in its equilibrium position.
 
  • #19
Okay so I'm just an undergrad but like how about you place the ball at an arbitrary position, let's say (x0, y0) with the spring being streched, and then compute the net force in terms of the angles(which could be further expressed in terms of x and y coordinates of the ball) and then have two differential equations corresponding to the x and y coordinate ?
 
  • #20
MrNewton said:
i can declare the springconstant and mass myself.
Then let's start with unknowns to represent these, k and m, and write 2L for the spring length.
In terms of those, can you obtain an expression for the rest height of the mass?
Next, can you write the differential equations for the general position?
Then we can move onto turning those equations into a simulation.
 
  • #21
haruspex said:
Then let's start with unknowns to represent these, k and m, and write 2L for the spring length.
In terms of those, can you obtain an expression for the rest height of the mass?

Im sure i can, but i doubt i understand you correct. By "Rest height of the mass" you mean the distance between the mass and the ceiling? And K is the spring constant? I am not sure what you mean with the 2L for the spring lengthy. Twice the length of the spring probably because there are 2 springs but I am not sure how to draw this into the fbd.
 
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  • #22
MrNewton said:
Then we can move onto turning those equations into a simulation.

Im sorry, i hope i am not boring you with my stupidness, and all my questions :)
Perhaps you can make an drawing of what you mean :smile::smile:
 
  • #23
I have a really silly question, but :
1. Are we assuming the mass to be a point mass?
2. Are the springs allowed to "turn" or "twist" ?
 
  • #24
MrNewton said:
By "Rest height of the mass" you mean the distance between the mass and the ceiling? And K is the spring constant? I am not sure what you mean with the 2L for the spring lengthy. Twice the length of the spring probably because there are 2 springs
Yes. Each spring length L.
 
  • #25
haruspex said:
Yes. Each spring length L.
I am very sorry but i do not understand how i should draw this setup or what you are suggesting. I really need a push on this one :smile:
Ive been trying since 4:00 PM and its now 9:30 PM over here. I am not making a great progress (because of the lack of values i guess :frown:)
 
  • #26
MrNewton said:
I am very sorry but i do not understand how i should draw this setup or what you are suggesting. I really need a push on this one :smile:
Ive been trying since 4:00 PM and its now 9:30 PM over here. I am not making a great progress (because of the lack of values i guess :frown:)
Try to answer these questions in sequence:

If the relaxed spring lengths are L and the mass hangs at rest a distance h below the ceiling, how long are the springs?
What is the extension of the springs?
What is the tension in the springs?
What equation does that give for the vertical force balance?
 
  • #27
haruspex said:
Try to answer these questions in sequence:

If the relaxed spring lengths are L and the mass hangs at rest a distance h below the ceiling, how long are the springs?
The relaxed springs are 30 cm each. (Without the mass, this is given in the assigment!) In the relaxed position the are parallel to the ceiling (since the springs are weigthless)
I have printed the assigment, but offcourse the printed version is not in scale. But on my paper the 30 cm on the drawing is 6,7CM
The length of the spring drawn on my paper is 7.8 cm. This means that the theoretical length is 35 cm=0.35m. (30/6.7)*7.8

haruspex said:
Try to answer these questions in sequence:
What is the tension in the springs?

Using Hooke's law:

I calculated the spring constant by K = M*G/Lstretched - Lrest (L stands for length)
This gives me k = (2*9,81) * (0.35-0.3) = 400N/M
With this i can calculate the force in the spring: Fspring = -k*x = -400*0.35 = 140N

I doubt this is correct. Seems like a lot though
 
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  • #28
MrNewton said:
The relaxed springs are 30 cm each. (
I know, but I don't want you to use any numbers. Just use the symbols m, g, k, L, h, and write the equations.
 
  • #29
TachyonLord said:
I have a really silly question, but :
1. Are we assuming the mass to be a point mass?
2. Are the springs allowed to "turn" or "twist" ?

What do you mean with a point mass?

No, the springs are just as they are. Dont think to far
 
  • #30
haruspex said:
I know, but I don't want you to use any numbers. Just use the symbols m, g, k, L, h, and write the equations.
Height = √(Lspring2-302)
Mg = m*g
k = (m*g)-(Lstretched-Lrest)

Vertical force balance equasion: Fg = F1+2 Sum of forces in Y direction is 0 @ equilibrium(not sure if this is what you mean)
 
  • #31
Your simulation is not interested in equilibrium, so just define that position as your origin. Do as @haruspex advises and write out ##\vec F = m\vec a## equations in x and y
 
  • #32
MrNewton said:
Height = √(Lspring2-302)
Mg = m*g
k = (m*g)-(Lstretched-Lrest)

Vertical force balance equasion: Fg = F1+2 Sum of forces in Y direction is 0 @ equilibrium(not sure if this is what you mean)
A few errors there.
I defined L as the unstretched length, so use that, no numbers!
So the stretched length is √(L2+h2)
So how much is the extension? What is the tension? What is the vertical component of the tension?
 
  • #33
haruspex said:
So how much is the extension? What is the tension? What is the vertical component of the tension?
But h is unknown? In know h of 1 spring in vertical component is 9,81 m/s2 because total vertical component is 19,62 but i cannot use this to calculate the length of the stretched spring can i? Since thet have different units
 
  • #34
MrNewton said:
But h is unknown?
That's why we write equations containing unknowns, so that we can find out what they are. Just follow the steps in post #32. Don't worry for now about which are known and which unknown.
 
  • #35
haruspex said:
That's why we write equations containing unknowns, so that we can find out what they are. Just follow the steps in post #32. Don't worry for now about which are known and which unknown.

Alright:

Extension = Stretched length - Length @ rest
Tension = -k * Extention
Vertical component of tension = SIN(theta)*Tension OR √(Heigth2+length@rest2)

With Theta i mean the angle between the spring and the ceiling.
 
<h2>1. What is a twin spring system with a mass in between?</h2><p>A twin spring system with a mass in between is a mechanical system that consists of two springs connected in parallel with a mass in between them. The system is designed to provide motion in both the X and Y directions.</p><h2>2. How does a twin spring system with a mass in between work?</h2><p>The twin spring system with a mass in between works by utilizing the properties of springs, such as their elasticity and ability to store and release energy. When a force is applied to the mass, the springs compress or stretch, storing potential energy. This energy is then released as the springs return to their original state, causing the mass to move in the opposite direction.</p><h2>3. What are the applications of a twin spring system with a mass in between?</h2><p>A twin spring system with a mass in between has various applications in different fields. It is commonly used in mechanical engineering for shock absorption and vibration isolation. It is also used in structures such as bridges and buildings to reduce the effects of earthquakes. Additionally, it can be used in sports equipment, such as trampolines, to provide a bouncing effect.</p><h2>4. What are the advantages of using a twin spring system with a mass in between?</h2><p>The advantages of using a twin spring system with a mass in between include its ability to provide motion in both the X and Y directions, making it versatile for various applications. It also has a high energy absorption capacity, making it useful for shock absorption. Additionally, it is a simple and cost-effective design that can be easily implemented in various systems.</p><h2>5. Are there any limitations to using a twin spring system with a mass in between?</h2><p>While a twin spring system with a mass in between has many advantages, there are also some limitations to consider. One limitation is that the system's motion is dependent on the force applied, so it may not be suitable for applications where a constant motion is required. Additionally, the system's performance may be affected by factors such as temperature and wear and tear of the springs over time.</p>

1. What is a twin spring system with a mass in between?

A twin spring system with a mass in between is a mechanical system that consists of two springs connected in parallel with a mass in between them. The system is designed to provide motion in both the X and Y directions.

2. How does a twin spring system with a mass in between work?

The twin spring system with a mass in between works by utilizing the properties of springs, such as their elasticity and ability to store and release energy. When a force is applied to the mass, the springs compress or stretch, storing potential energy. This energy is then released as the springs return to their original state, causing the mass to move in the opposite direction.

3. What are the applications of a twin spring system with a mass in between?

A twin spring system with a mass in between has various applications in different fields. It is commonly used in mechanical engineering for shock absorption and vibration isolation. It is also used in structures such as bridges and buildings to reduce the effects of earthquakes. Additionally, it can be used in sports equipment, such as trampolines, to provide a bouncing effect.

4. What are the advantages of using a twin spring system with a mass in between?

The advantages of using a twin spring system with a mass in between include its ability to provide motion in both the X and Y directions, making it versatile for various applications. It also has a high energy absorption capacity, making it useful for shock absorption. Additionally, it is a simple and cost-effective design that can be easily implemented in various systems.

5. Are there any limitations to using a twin spring system with a mass in between?

While a twin spring system with a mass in between has many advantages, there are also some limitations to consider. One limitation is that the system's motion is dependent on the force applied, so it may not be suitable for applications where a constant motion is required. Additionally, the system's performance may be affected by factors such as temperature and wear and tear of the springs over time.

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