# Homework Help: Correct Formal Equation (Force due to a Spring)?

1. Mar 20, 2009

### PFStudent

Hey,

1. The problem statement, all variables and given/known data.
What is the correct formal equation for the Force due to a Spring?
$${\vec{{F}_{s}}} = {{-k}{\vec{r}}}$$
OR
$${\vec{{F}_{s}}} = {{-k}{\Delta}{\vec{r}}}$$

The reason this question arose was because I noticed in my textbook (The Fundamentals of Physics (6th Edition)) the spring force is defined as,
$${\vec{{F}_{s}}} = {{-k}{\vec{d}}}$$
where ${\vec{d}}$ is explicitly referred to as the displacement.

So that's when I recalled that if we consider two distance vectors (one for initial distance and one for final distance) in three-dimensions,
$${{\vec{r}}_{i}}$$
and
$${{\vec{r}}_{f}}$$
then we can write their displacement as follows,
$${{\Delta}{\vec{r}}} = {{{\vec{r}}_{f}}-{{\vec{r}}_{i}}}$$

Upon noting the above this confused me as to which equation,
$${\vec{{F}_{s}}} = {{-k}{\vec{r}}}$$
OR
$${\vec{{F}_{s}}} = {{-k}{\Delta}{\vec{r}}}$$
was the correct formal equation for the force due to a spring.

That's when I thought of trying to derive the Work due to a spring from each and see if it came out to the same result, however that didn't quite work out so well and I've explained below why.

2. Relevant equations.
Knowledge of Classical Mechanics and Calculus.

3. The attempt at a solution.
Derivation 1.
$${\vec{{F}_{s}}} = {{-k}{\vec{r}}}$$

$${{d}{{W}_{s}}} = {\vec{{F}_{s}}\cdot{d}{\vec{r}}}$$

$${{W}_{s}} = {{\int_{{r}_{i}}^{{r}_{f}}}\vec{{F}_{s}}\cdot{d}\vec{r}}$$

$${{W}_{s}} = {{\int_{{r}_{i}}^{{r}_{f}}}{\left({-k}\vec{r}}\right)\cdot{d}\vec{r}}$$

$${{W}_{s}} = {{-k}{\int_{{r}_{i}}^{{r}_{f}}}\vec{r}\cdot{d}{\vec{r}}}$$

$${{W}_{s}} = {\left{-k}\left[\frac{{\left(\vec{r}\right)}^{2}}{2}\right]\right|_{{r}_{i}}^{{r}_{f}}}$$

$${{W}_{s}} = {\left\frac{-k}{2}\Biggl[\vec{r}\cdot\vec{r}\Biggr]\right|_{{r}_{i}}^{{r}_{f}}}$$

$${{W}_{s}} = {{\frac{-k}{2}}{\Biggl[{\left({\vec{r}}_{f}\cdot\vec{r}_{f}}\right)-\left({\vec{r}}_{i}\cdot{\vec{r}}_{i}\right)\Biggr]}}$$

$${{W}_{s}} = {{\frac{-k}{2}}{\Biggl[{{\left|{\vec{r}}_{f}\right|}{\left|{\vec{r}}_{f}\right|}{\cos{{\phi}_{f}}}}-{{\left|{\vec{r}}_{i}\right|}{\left|{\vec{r}}_{i}\right|}{\cos{{\phi}_{i}}}}\Biggr]}}{,}{\,}{\,}{\,}{\,}{{\phi}_{i}} = {{\phi}_{f}} = {0{\,}{\,}\text{rad.}}$$

$${{W}_{s}} = {\frac{k}{2}\Biggl({{{{r}_{i}}^{2}}{-}{{{r}_{f}}^{2}}}\Biggr)}$$

Derivation 2.
$${\vec{{F}_{s}}} = {{-k}\Delta{\vec{r}}}$$

$${{d}{{W}_{s}}} = {\vec{{F}_{s}}\cdot{d}{\vec{r}}}$$

$${{W}_{s}} = {{\int_{{r}_{i}}^{{r}_{f}}}\vec{{F}_{s}}\cdot{d}\vec{r}}$$

$${{W}_{s}} = {{\int_{{r}_{i}}^{{r}_{f}}}{\left({-k}\Delta\vec{r}}\right)\cdot{d}\vec{r}}$$

$${{W}_{s}} = {{-k}{\int_{{r}_{i}}^{{r}_{f}}}\Delta\vec{r}\cdot{d}{\vec{r}}}$$

$${{W}_{s}} = {\left{-k}\left[\frac{{\left(\Delta\vec{r}\right)}^{2}}{2}\right]\right|_{{r}_{i}}^{{r}_{f}}}$$

$${{W}_{s}} = {\left\frac{-k}{2}\Biggl[\Delta\vec{r}\cdot\Delta\vec{r}\Biggr]\right|_{{r}_{i}}^{{r}_{f}}}$$

$${\left\frac{-k}{2}\Biggl[\Delta\vec{r}\cdot\Delta\vec{r}\Biggr]\right|_{{r}_{i}}^{{r}_{f}}} = {\text{???}}$$

So, from the first derivation I could derive the Work due to a spring pretty easily. However, from the second derivation I was confused on how to evaluate the last step since technically the substituion of the limits was supposed to lead to ${{\Delta}{\vec{r}}}$, however I already had ${{\Delta}{\vec{r}}}$ there. Which derivation is correct and why?

Lastly, technically shouldn't the limits in the formal equation for the Work due to a spring actually be vectors like below,
$${{W}_{s}} = {{\int_{{\vec{r}}_{i}}^{{\vec{r}}_{f}}}\vec{{F}_{s}}\cdot{d}\vec{r}}$$

as opposed to (not being vectors),

$${{W}_{s}} = {{\int_{{r}_{i}}^{{r}_{f}}}\vec{{F}_{s}}\cdot{d}\vec{r}}$$

Thanks,

-https://www.physicsforums.com/member.php?u=79507"

Last edited by a moderator: Apr 24, 2017
2. Mar 20, 2009

### LowlyPion

I think it is confusing to represent Hooks Law as

F = -kΔr

because the F is a linear expression of the displacement from the relaxed position, not any arbitrary difference in displacement.

In taking the integral to determine the work/PE relationship of the spring, then you are necessarily evaluating it over a force equation that is linear in displacement. (Corresponding to your first derivation result.)

You can think of displacement as Δr, but ri is necessarily ro which is the relaxed point of the spring.

3. Mar 21, 2009

### PFStudent

Hey,
Ahh, that makes sense. Since we always evaluate the force due to a spring through a linear displacement where the initial distance is the relaxed position (and because so is necessarily zero) and the final distance is the distance from the relaxed position.

However, the above leads to an interesting question - if the displacement is always the final distance from the relaxed position, where the relaxed position is the initial distance and because so is necessarily zero (because we are measuring the final distance from the initial distance (aka. relaxed position)) then in the following equation,
$${{\vec{F}}_{s}} = {{-k}{\vec{r}}}$$

We can let,
$${\vec{r}} = {{\vec{r}}_{f}}$$

So that we have,
$${{\vec{F}}_{s}} = {{-k}{{\vec{r}}_{f}}}$$

However, if we choose to derive the Work due to a spring from the above notation we run in to a problem (explained below).

Derivation 3.
$${{\vec{F}}_{s}} = {{-k}{{\vec{r}}_{f}}}$$

$${{d}{{W}_{s}}} = {\vec{{F}_{s}}\cdot{d}{\vec{r}}}$$

$${{W}_{s}} = {{\int_{{r}_{i}}^{{r}_{f}}}\vec{{F}_{s}}\cdot{d}\vec{r}}$$

$${{W}_{s}} = {{\int_{{r}_{i}}^{{r}_{f}}}{\left({-k}{\vec{r}}_{f}}\right)\cdot{d}\vec{r}}$$

$${{W}_{s}} = {{-k}{\int_{{r}_{i}}^{{r}_{f}}}{\vec{r}}_{f}\cdot{d}{\vec{r}}}$$

$${{W}_{s}} = {\left{-k}\left[\frac{{\left({\vec{r}}_{f}\right)}^{2}}{2}\right]\right|_{{r}_{i}}^{{r}_{f}}}$$

$${{W}_{s}} = {\left\frac{-k}{2}\Biggl[{\vec{r}}_{f}\cdot{\vec{r}}_{f}\Biggr]\right|_{{r}_{i}}^{{r}_{f}}}$$

$${\left\frac{-k}{2}\Biggl[{\vec{r}}_{f}\cdot{\vec{r}}_{f}\Biggr]\right|_{{r}_{i}}^{{r}_{f}}} = {\text{???}}$$

The problem in this third derivation is that my limits are: ${r}_{i}$ and ${r}_{f}$, however my variable is already ${\vec{r}}_{f}$.

So, why is the above derivation wrong?

Thanks,

-https://www.physicsforums.com/member.php?u=79507"

Last edited by a moderator: Apr 24, 2017
4. Mar 21, 2009

### LowlyPion

I don't see any problem.

ri is necessarily 0, meaning that d ≡ Δr.

Your integrals can only be evaluated at 0 to rf.

5. Mar 21, 2009

### PFStudent

Hey,
In that case how would one ever derive,
$${{W}_{s}} = {\frac{k}{2}\Biggl({{{{r}_{i}}^{2}}{-}{{{r}_{f}}^{2}}}\Biggr)}$$
if we could only evaluate our integral, to find the Work done by a spring, from ${0}$ to ${{\vec{r}}_{f}}$?

Instead of the above, if we are limited to evaluating from ${0}$ to ${{\vec{r}}_{f}}$, we would have,
$${{W}_{s}} = {{\frac{-k}{2}}{{\left({\vec{r}}_{f}\right)}^{2}}}$$

Which is not correct since the correct formal equation for the Work done by the Spring is,
$${{W}_{s}} = {\frac{k}{2}\Biggl({{{{r}_{i}}^{2}}{-}{{{r}_{f}}^{2}}}\Biggr)}$$

Thanks,

-https://www.physicsforums.com/member.php?u=79507"

Last edited by a moderator: Apr 24, 2017
6. Mar 21, 2009

### LowlyPion

I think you are confusing work done to a spring and work done by a spring.

When you are compressing a spring, there is work done by the pusher. Since the spring is resisting the direction of motion it is doing negative work, and is said to be storing potential energy.

When it comes time to release the spring, the its force is in the direction of motion and as it gives up potential energy it results in positive work.

7. Mar 21, 2009

### PFStudent

Hey,
I see. Ok, so let me see if I understand this correctly.

When considering the formal equation for the Force due to a spring,
$${{\vec{F}}_{s}} = {{-k}{\vec{r}}}$$
it is understood that,
$${\vec{r}}$$
is defined as the displacement and,
$${\vec{r}} = {{\Delta}{\vec{r}}}$$
where,
$${{\Delta}{\vec{r}}} = {{\vec{r}}_{f}-{\vec{r}}_{i}}$$

However, within the context of considering the Force due to a spring we are only interested in the displacement measured from the position of the relaxed state of the spring - in other words, since we are only interested in this particular displacement we have that,
$${{\vec{r}}_{i}} = {0}$$
and therefore,
$${{\Delta}{\vec{r}}} = {{\vec{r}}_{f}}$$
Where now,
$${\vec{r}} = {{\vec{r}}_{f}}$$

In the interest of clarity in our notation let,
$${\vec{r}} = {{\vec{r}}_{s}}$$
where we have defined ${{\vec{r}}_{s}}$ as the spring's displacement, specifically the displacement of the free end from its position when the spring is in the relaxed state.

So now the formal equation for the Force due to a spring is,
$${{\vec{F}}_{s}} = {{-k}{{\vec{r}}_{s}}}$$

This new notation helps clear up the misconception that when we are trying to find the Work done by a spring force,
$${{W}_{s}} = {{-k}{\int_{{r}_{i}}^{{r}_{f}}}\vec{r}\cdot{d}{\vec{r}}}$$
we can now see that rather than considering the spring's displacement ${{\vec{r}}_{s}}$ (as we have defined it within the context of the formal equation for the Force due to a spring), due to the nature of the integration we instead consider some variable-displacement ${\vec{r}}$ which represents the displacement as the object moves from ${{\vec{r}}_{i}}$ to ${{\vec{r}}_{f}}$ (of the Work done by the spring force).

This new perspective helps clear up the confusion that the displacement used in finding the Force due to a spring is NOT the same displacement used when finding the Work done by a spring, because due to the nature of the integration we instead consider a variable-displacement that represents the object's displacement as it moves from ${{\vec{r}}_{i}}$ to ${{\vec{r}}_{i}}$.

Is all of the above correct?

Thanks,

-https://www.physicsforums.com/member.php?u=79507"

Last edited by a moderator: Apr 24, 2017
8. Mar 22, 2009

### kbaumen

After a quick glance, isn't $${{W}_{s}} = {{\frac{-k}{2}}{{\left({\vec{r}}_{f}\right)}^{2}}}$$

the same as

$${{W}_{s}} = {\frac{k}{2}\Biggl({{{{r}_{i}}^{2}}{-}{{{r}_{f}}^{2}}}\Biggr)}$$

Because $r_{i}$ = 0. I mean, if the spring is at rest, than the displacement is equal to 0. Isn't that so?

Last edited by a moderator: Apr 24, 2017
9. Mar 22, 2009

### PFStudent

Hey,
Not quite, from my understanding, the displacement ${{\vec{r}}_{s}}$ is NOT the same as the displacement ${\vec{r}}$ used when calculating the Work done by the spring.

The reason for this is because in,
$${{\vec{F}}_{s}} = {{-k}{{\vec{r}}_{s}}}$$
the spring's displacement ${{\vec{r}}_{s}}$ is defined as the displacement of the free end from its position when the spring is in the relaxed state.

Where in,
$${{W}_{s}} = {{-k}{\int_{{r}_{i}}^{{r}_{f}}}\vec{r}\cdot{d}{\vec{r}}}$$
the displacement here refers to the object's displacement as it moves from ${{\vec{r}}_{i}}$ to ${{\vec{r}}_{f}}$ as Work is done by the spring on the object.

Thanks,

-https://www.physicsforums.com/member.php?u=79507"

Last edited by a moderator: Apr 24, 2017
10. Mar 22, 2009

### kbaumen

Well, the end of the spring moves the same distance from it's relaxed state, as does the object connected to it's end. If we talk about work, then there are two kinds of work done. A positive $W_{f}$ done by a force that is pulling the spring and a negative work done by the spring itself (negative because the force is acting in the opposite direction of the movement).

Regarding your original question, I'd say that it doesn't really matter, both notations are fine.

$$\Delta \vec{r}$$

refers to a difference between the initial and final displacement vectors from your axis' origin and

$$\vec{r}$$

refers to the displacement of the object (or the end of the spring) where your origin of the axis is in the initial position of the object. In a sense, these two refer to the same vector.

IMO, if you're writing any papers regarding this, then use whichever notation you like, just stick to it. On the other hand, I'm no professional, so this is just my opinion.

Last edited by a moderator: Apr 24, 2017