Springs connected to a mass in series?

  • #1
flinnbella
26
4
Homework Statement
Find the frequency of oscillation of mass m suspended by two
springs having constants k1 and k2, in each of the configurations
shown.
Relevant Equations
x = A sinωt + B cosωt
Screenshot_20.png


I found the answer for the springs in parallel, but not for the ones in series. I believe I don't understand how the forces are interacting properly.

Screenshot_21.png

Here's a force diagram I drew. Everytime I try to make equations from this though my answer dosen't make sense. The mass m has a gravititoanl force applied to it, and spring 2 applies a restoring force to mass m. On spring 1, the mass m applies a gravitational force, and spring 2 applies a restoring force onto spring 1 towards itself. Spring 1 applies a restoring force towards itself.

I think there's something wrong with this interpretation, but I don't understand what.
 
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  • #2
If there is one spring and you applied a force ##F## you would expect deflection ##\Delta y##. If you use a second identical spring to apply the force ##F## to the first spring what deflection would you expect?
 
  • #3
erobz said:
If there is one spring and you applied a force ##F## you would expect deflection ##\Delta y##. If you use a second identical spring to apply the force ##F## to the first spring what deflection would you expect?
Delta y?
 
  • #4
flinnbella said:
Delta y?
Think about the force applied to the second spring, what force is the second spring applying to the first spring?
 
  • #5
erobz said:
Think about the force applied to the second spring, what force is the second spring applying to the first spring?
It is applying k_2*delta_y to the first spring
 
  • #6
flinnbella said:
It is applying k_2*delta_y to the first spring
While it may be true, I’m think it’s missing the point. ##F## is applied to the second spring. What is force is applied to the first? You can think of this like two ropes if you like. If you apply ##F## to the second rope, what is the tension developed in each rope?
 
  • #7
erobz said:
While it may be true, I’m think it’s missing the point. ##F## is applied to the second spring. What is force is applied to the first? You can think of this like two ropes if you like. If you apply ##F## to the second rope, what is the tension developed in each rope?
If it was two ropes, the force would be F? If the ropes are in equilibrium then T = F, so the force from rope 2 to rope 1 is F?
 
  • #8
flinnbella said:
If it was two ropes, the force would be F? If the ropes are in equilibrium then T = F, so the force from rope 2 to rope 1 is F?
Correct. ##F## is applied to both ropes. Now, put the springs back. Each spring is acting under applied load ##F##. Meaning the total deflection the end of the second spring sees relative to the origin is ( assuming identical springs)?
 
  • #9
erobz said:
Correct. ##F## is applied to both ropes. Now, put the springs back. Each spring is acting under applied load ##F##. Meaning the total deflection the end of the second spring sees relative to the origin is ( assuming identical springs)?
The displacement of spring 1 and of spring 2 by the mass m?
 
  • #10
flinnbella said:
The displacement of spring 1 and of spring 2 by the mass m?
Forget the mass for a minute. We are applying a force ##F## with our hands. What is the total deflection of our hand? ( Remember a force ##F## acting on either particular spring causes deflection ##\Delta y## of that particular spring )
 
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  • #11
Maybe my line of questions isn’t working. Let’s shift gears. Would you write Newton’s Second law for the mass (as you currently believe it to be) so we can examine it?
 
  • #12
erobz said:
Maybe my line of questions isn’t working. Let’s shift gears. Would you write Newton’s Second law for the mass (as you currently believe it to be) so we can examine it?
Netwon second law is that the product of the mass and acceleration of an object is equal to the sum of all the forces applied to it
 
  • #13
flinnbella said:
Netwon second law is that the product of the mass and acceleration of an object is equal to the sum of all the forces applied to it
On third thought this is going to make it worse I think.

They want you to find the ##k_{eff}## value of a single spring that could effectively replace the system of springs to solve your problem. What did you find was that ##k_{eff}## value for the parallel configuration?
 
  • #14
erobz said:
On third thought this is going to make it worse I think.

They want you to find the ##k_{eff}## value of a single spring that could effectively replace the system of springs to solve your problem. What did you find was that ##k_{eff}## value for the parallel configuration?
Is there a single spring that can replace the two springs? Isn't there a force between the springs as well, so how can a single spring account for that?
 
  • #15
With the parallel setup, the displacement is the same for each spring - thus also the same for an equivalent spring replacing both - but the forces from each spring are added together to be applied to the equivalent spring. So:
$$\Delta y_1 = \Delta y_2 = \Delta y_{eq}$$
$$F_1 + F_2 = F_{eq}$$
And, for each spring:
$$F_n = k_n \Delta y_n$$
From these, you found the value of ##k_{eq}## in terms of ##k_1## and ##k_2##. These are basically 6 equations with 6 unknowns, namely: ##F_1,\ F_2,\ k_{eq},\ \Delta y_1,\ \Delta y_2## and ##\Delta y_{eq}##. (We also know that ##F_{eq} = mg##)

Find now the same equations that relate the displacements and the forces together when the springs are in series. Which variables are the same and which ones are added together?
 
  • #16
flinnbella said:
Is there a single spring that can replace the two springs? Isn't there a force between the springs as well, so how can a single spring account for that?
Newton’s third law. That’s the idea I was trying to convey. The force of spring 2 on spring 1 is a third law partner with the force of spring 1 on spring 2 ( together forming a third law pair).

There is an effective spring constant in ether case. You just have to figure out how to apply the constraints.
 
  • #17
erobz said:
Newton’s third law. That’s the idea I was trying to convey. The force of spring 2 on spring 1 is a third law partner with the force of spring 1 on spring 2 ( together forming a third law pair).

There is an effective spring constant in ether case. You just have to figure out how to apply the constraints.
Does that mean the restoring force of the spring have to equal the force of the weight? But the equal and opposite reaction for the weight is gravity pulling on it. I don't know what the equal and opposite force for the restoring force of the spring is though.
 
  • #18
jack action said:
With the parallel setup, the displacement is the same for each spring - thus also the same for an equivalent spring replacing both - but the forces from each spring are added together to be applied to the equivalent spring. So:
$$\Delta y_1 = \Delta y_2 = \Delta y_{eq}$$
$$F_1 + F_2 = F_{eq}$$
And, for each spring:
$$F_n = k_n \Delta y_n$$
From these, you found the value of ##k_{eq}## in terms of ##k_1## and ##k_2##. These are basically 6 equations with 6 unknowns, namely: ##F_1,\ F_2,\ k_{eq},\ \Delta y_1,\ \Delta y_2## and ##\Delta y_{eq}##. (We also know that ##F_{eq} = mg##)

Find now the same equations that relate the displacements and the forces together when the springs are in series. Which variables are the same and which ones are added together?
I solved for the parallel case a while ago, I understand how the displacements are the same, and thus the net force is the same as the sum of all the spring forces; however, I don't see something similar for the ones in series
 
  • #19
flinnbella said:
however, I don't see something similar for the ones in series
Yes, you already did part of the job earlier:
erobz said:
flinnbella said:
If it was two ropes, the force would be F? If the ropes are in equilibrium then T = F, so the force from rope 2 to rope 1 is F?
Correct. ##F## is applied to both ropes.
This means:
$$F_1 = F_2 = F_{eq}\ (= mg)$$
 
  • #20
jack action said:
This means:
$$ ... F_{eq}\ (= mg)$$
Just for clarity. When in static equilibrium
 
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  • #21
jack action said:
Yes, you already did part of the job earlier:

This means:
$$F_1 = F_2 = F_{eq}\ (= mg)$$
Ok. So the restoring force of each spring is equal. Then what? We don't know the displacement of each spring.
 
  • #22
flinnbella said:
Ok. So the restoring force of each spring is equal. Then what? We don't know the displacement of each spring.
But you do know the deflection of each( edit: in terms of the other), because you know the force applied to each and the spring constants.
 
  • #23
flinnbella said:
Ok. So the restoring force of each spring is equal. Then what? We don't know the displacement of each spring.
What is the relationship between ##\Delta y_1##, ##\Delta y_2## and ##\Delta y_{eq}##?
 
  • #24
deltay1 + deltay2 is the total displacement, so deltayeq is the same as their sum
 
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  • #25
flinnbella said:
deltay1 + deltay2 is the total displacement, so deltayeq is the same as their sum
Now use the relationship in #19 to relate the equivalent spring to either of the other springs.
 
  • #26
erobz said:
Now use the relationship in #19 to relate the equivalent spring to either of the other springs.
Ok, so k_1*delta_x = k_2*delta_y = k_eq*(delta_x + delta_y).
 
  • #27
flinnbella said:
Ok, so k_1*delta_x = k_2*delta_y = k_eq*(delta_x + delta_y).
Yeah, pick one (for the moment) of the spring forces to relate to the effective spring force.
 
  • #28
I don't see why we are trying to find this F_eq.
 
  • #29
flinnbella said:
I don't see why we are trying to find this F_eq.
It is a placeholder. We are trying to find the effective spring.
 
  • #30
flinnbella said:
Ok, so k_1*delta_x = k_2*delta_y = k_eq*(delta_x + delta_y).
Put those equations (##F_n = k_n\Delta y_n##) in the displacement equation instead.
 
  • #31
erobz said:
It is a placeholder. We are trying to find the effective spring
Ok I get that the total displacement of the mass is delta_x + delta_y, and this can be represented by a single spring instead of two. However, I don't see a relationship between the spring constants.
 
  • #32
flinnbella said:
Ok I get that the total displacement of the mass is delta_x + delta_y, and this can be represented by a single spring instead of two. However, I don't see a relationship between the spring constants.
Just continue on with #26. You are so close. We can talk about what it means as you go.

You have that ##A = B = C##, we want to pick ##A## or ##B## to relate to ##C##. Then you will use ##A## to relate to ##B##. In the end you will find the spring with constant ##k_{eq}##, that behaves as though it were identical to the series springs.
 
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  • #33
I got k_1+k_2/(k_1 + k_2) = k_eq. So I have the forces in terms of k1, k2.
 
  • #34
flinnbella said:
I got k_1+k_2/(k_1 + k_2) = k_eq. So I have the forces in terms of k1, k2.
I think you messed up the algebra or the application of the equations ( I can't tell because you don't shot the steps you took)
 
Last edited:
  • #35
This is what you did for the springs in parallel:
$$F_1 + F_2 = F_{eq}$$
$$k_1\Delta y_1 + k_2\Delta y_2 = k_{eq}\Delta y_{eq}$$
And because ##\Delta y_1 = \Delta y_2 = \Delta y_{eq}##:
$$k_1 + k_2 = k_{eq}$$
So what are you going to do with ##\Delta y_1 + \Delta y_2 = \Delta y_{eq}## in our case with springs in series?
 

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