Springs connected to a mass in series?

[flinnbella]

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

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.

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.

 

[erobz]

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?

 

[flinnbella]

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?

 

[erobz]

Delta y?

Think about the force applied to the second spring, what force is the second spring applying to the first spring?

 

[flinnbella]

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

 

[erobz]

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?

 

[flinnbella]

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?

 

[erobz]

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)?

 

[flinnbella]

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?

 

[erobz]

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 )

 

[erobz]

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?

 

[flinnbella]

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

 

[erobz]

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?

 

[flinnbella]

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?

 

[jack action]

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?

 

[erobz]

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.

 

[flinnbella]

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.

 

[flinnbella]

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

 

[jack action]

however, I don't see something similar for the ones in series

Yes, you already did part of the job earlier:

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:
$$F_1 = F_2 = F_{eq}\ (= mg)$$

 

[erobz]

This means:
$$ ... F_{eq}\ (= mg)$$

Just for clarity. When in static equilibrium

 

[flinnbella]

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.

 

[erobz]

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.

 

[jack action]

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 $\Delta y_{eq}$?

 

[flinnbella]

deltay1 + deltay2 is the total displacement, so deltayeq is the same as their sum

 

[erobz]

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.

 

[flinnbella]

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).

 

[erobz]

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.

 

[flinnbella]

I don't see why we are trying to find this F_eq.

 

[erobz]

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.

 

[jack action]

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.

 

[flinnbella]

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.

 

[erobz]

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.

 

[flinnbella]

I got k_1+k_2/(k_1 + k_2) = k_eq. So I have the forces in terms of k1, k2.

 

[erobz]

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)

 

[jack action]

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?

 

[flinnbella]

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)

k1*k2/(k1 + k2)

 

[erobz]

k1*k2/(k1 + k2)

Good. Now divide by numerator and denominator by $k_1 \cdot k_2$ and you will see the start of a more general form that should look familiar if you've taken circuits.

 

[Lnewqban]

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.

Note how much you can stretch two rubber bands (in parallel) at once.
Now, tie two ends of those bands (in series) and stretch both as one.

At the point that you are applying similar force as before, there will be much more stretching (x of balance) for the second case.

 

[flinnbella]

Note how much you can stretch two rubber bands (in parallel) at once.
Now, tie two ends of those bands (in series) and stretch both as one.

At the point that you are applying similar force as before, there will be much more stretching (x of balance) for the second case.

View attachment 332054

Oh this makes some more sense, not 100% understanding but it helps. How did you make the graphic?

 

[Lnewqban]

Oh this makes some more sense, not 100% understanding but it helps. How did you make the graphic?

What is still unclear?

Left graphic:
Balance load equals mg.
Both, the strong and the weak spring, must deflect the same in order to produce a combined reactive force equaling mg.
Reactive individual force of strong spring is greater than the reactive individual force of weak spring, but summation of both is the balance load mg.

Could you try describing the right graphic?

 

[SammyS]

k1*k2/(k1 + k2)

So, if this is k_eq , what do you get for 1/k_eq ?

To restate that:

If $\displaystyle \quad k_{eq}=\dfrac{k_1 \cdot k_2}{k_1 + k_2}$ ,

then what is $\displaystyle \quad \dfrac 1 {k_{eq} } \ ?$

 

[flinnbella]

So, if this is k_eq , what do you get for 1/k_eq ?

To restate that:

If $\displaystyle \quad k_{eq}=\dfrac{k_1 \cdot k_2}{k_1 + k_2}$ ,

then what is $\displaystyle \quad \dfrac 1 {k_{eq} } \ ?$

1/keq = (k1 + k2)/(k1*k2)

 

[flinnbella]

What is still unclear?

Left graphic:
Balance load equals mg.
Both, the strong and the weak spring, must deflect the same in order to produce a combined reactive force equaling mg.
Reactive individual force of strong spring is greater than the reactive individual force of weak spring, but summation of both is the balance load mg.

Could you try describing the right graphic?

I don't see why the springs both exert a force of the mass. Spring 2 exerts a force because it's attached, but spring 1 isn't directly connected to the mass so how can it apply a force to the mass? Makes no sense.

 

[erobz]

1/keq = (k1 + k2)/(k1*k2)

Yeah, but now you can turn that into something familiar (resistors in parrallel) by dividing each term in the numerator by the denominator.

I don't see why the springs both exert a force of the mass. Spring 2 exerts a force because it's attached, but spring 1 isn't directly connected to the mass so how can it apply a force to the mass? Makes no sense.

They don't both exert a force on the mass. The FBD of the mass only includes its weight and the force from sping 2. That force is transmitted through spring 2 and is applied to spring 1. Then that force is transmitted through spring one to the fixed structure.

 

[Lnewqban]

I don't see why the springs both exert a force of the mass. Spring 2 exerts a force because it's attached, but spring 1 isn't directly connected to the mass so how can it apply a force to the mass? Makes no sense.

What prevents spring 2 to fall down to the ground, together with the attached mass, is a force of magnitude mg pulling them up.
The net force on that spring 2 is then zero, and it remains in balance.
Exactly the same thing happens to spring 1 (which is attached to ground and to spring2-mass).

 

[SammyS]

1/keq = (k1 + k2)/(k1*k2)

Simplify that.

(k1 + k2)/(k1*k2) = k1/(k1*k2) + k2/(k1*k2) = ?

 

[jack action]

I don't see why the springs both exert a force of the mass. Spring 2 exerts a force because it's attached, but spring 1 isn't directly connected to the mass so how can it apply a force to the mass? Makes no sense.

The mass exerts a force on spring 2 and spring 2 must exert the same force on spring 1.

The fascinating thing is that since both spring displacements are added up such that the total displacement is always larger than any of the springs alone, the equivalent spring that would replace both springs is also always less stiff than the softest of the springs.

 

[flinnbella]

The mass exerts a force on spring 2 and spring 2 must exert the same force on spring 1.

The fascinating thing is that since both spring displacements are added up such that the total displacement is always larger than any of the springs alone, the equivalent spring that would replace both springs is also always less stiff than the softest of the springs.

Why does spring 2 exert same force on spring 1? I don't think those are force action-reaction pairs.

 

[erobz]

Why does spring 2 exert same force on spring 1? I don't think those are force action-reaction pairs.

The hand pulls on the spring with force $F$.

What force does the wall apply to the spring? What force does the spring apply to the wall? Remember, we are talking about "massless" springs. Free the spring from external bodies and sum external forces i.e. $\sum F_{ext}= \cancel{m}^0 a = 0$

 

[jack action]

Why does spring 2 exert same force on spring 1? I don't think those are force action-reaction pairs.

Yes, there are.

Doing the free body diagrams (FBD) for each component (the ceiling $c$, the spring $k_1$, the spring $k_2$ and the mass $m$):

By definition, from one FBD to the next:
$$F_c = F_{k1}$$
$$F_{k1} = F_{k2}$$
$$F_{k2} = F_m$$
mass FBD:
$$F_m = mg$$
Therefore:
$$F_c = F_{k1} =F_{k2} = F_m = mg$$
The force $mg$ goes through every component.