Damped harmonic oscillation

[NTesla]

Homework Statement

I'm studying Damped harmonic oscillation from Kleppner and Kolenkow. However I'm stuck at one point. Lets take the case of a mass at one end of the spring, which is oscillating (left and right) with some damping. Lets say, at some instant, the mass is moving towards the equilibrium position from the right side. Then the restoring force would be towards the left. The damping force would be towards the right(since the velocity is towards left). Now kindly see the screenshot attached. In that page, value of gamma would be negative, which would mean that value of alpha would be negative, which in turn would mean that the amplitude in equation 11.10 would INCREASE exponentially. Now, this should not be happening since we are studying damped harmonic oscillation.

P.S: I'm typing this on mobile, so not able to write equation may kindly be glossed over.

Relevant Equations

Equation is in the screenshot

 

[Orodruin]

No, the value of ##\gamma## would not be negative, why do you think so?

 

[NTesla]

No, the value of ##\gamma## would not be negative, why do you think so?

In the situation that I've described : F = - kx + bv.
In such a case, if we write equation 11.8(shown in the screenshot), then gamma will have to be negative.

 

[Orodruin]

In the situation that I've described : F = - kx + bv.

No, it is not. It is still ##F = -kx - bv##.

Lets say, at some instant, the mass is moving towards the equilibrium position from the right side.

Ok, so ##x > 0## and ##\dot x < 0##.

Then the restoring force would be towards the left.

The restoring force is ##-kx## with ##k, x > 0## so ##-kx < 0## is to the left, yes.

The damping force would be towards the right(since the velocity is towards left).

The damping force is ##-bv = -b\dot x## with and ##\dot x < 0## that means that ##-bv > 0## if ##b > 0## resulting in ##\gamma > 0## as it should be.

In such a case, if we write equation 11.8(shown in the screenshot), then gamma will have to be negative.

No, see above.

 

[haruspex]

In the situation that I've described : F = - kx + bv.

How so? In that situation, v is negative and the damping force acts to the right, so it is -bv.

 

[NTesla]

No, it is not. It is still ##F = -kx - bv##.



Ok, so ##x > 0## and ##\dot x < 0##.



The restoring force is ##-kx## with ##k, x > 0## so ##-kx < 0## is to the left, yes.



The damping force is ##-bv = -b\dot x## with and ##\dot x < 0## that means that ##-bv > 0## if ##b > 0## resulting in ##\gamma > 0## as it should be.


No, see above.

Kindly help. I'm sure I'm not seeing things right. But where am I going wrong, that I'm not able to figure out.

 

[Orodruin]

You have drawn ##b\vec v## to the right and ##\vec v## to the left. If ##b## is positive, they should be in the same direction. What you have drawn is the actual force ##-b\vec v##.

 

[NTesla]

You have drawn ##b\vec v## to the right and ##\vec v## to the left. If ##b## is positive, they should be in the same direction. What you have drawn is the actual force ##-b\vec v##.

I've considered the moment when the mass is going to the left side, then velocity will be towards left. So the damping force would be towards right. In the equation I've written ##bv\hat i##, and restoring force as ##- kx \hat i##. How is that wrong ?

 

[Steve4Physics]

I've considered the moment when the mass is going to the left side, then velocity will be towards left. So the damping force would be towards right. In the equation I've written ##bv\hat i##, and restoring force as ##- kx \hat i##. How is that wrong ?

Maybe this will help.

F = -kx – bv where k, b >0.
This is valid at all positions. We can check-out all the possibilities...

Call the equilibrium position ‘O’. There are 4 cases to consider:

1. The mass is to the right of O, moving right and slowing down. In this case:
x is positive, so kx is positive, so -kx is negative.
v is positive, so bv is positive, so –bv is negative.

2. The mass is to the right of O, moving left and speeding up. In this case:
x is positive, so kx is positive, so -kx is negative.
v is negative, so bv is negative, so – bv is positive.

You can work out cases 3 and 4 (mass to the left of O) for yourself if it’s helpful.

 

[Orodruin]

I've considered the moment when the mass is going to the left side, then velocity will be towards left. So the damping force would be towards right. In the equation I've written ##bv\hat i##, and restoring force as ##- kx \hat i##. How is that wrong ?

It is wrong because, as you have drawn it, b is negative (or it would be pointing to the left!). And sure, you can use F = -kx + bv if you use negative instead of positive b. It just changes the definition of gamma to -b/m so gamma is still positive.

 

[haruspex]

I've considered the moment when the mass is going to the left side, then velocity will be towards left. So the damping force would be towards right. In the equation I've written ##bv\hat i##, and restoring force as ##- kx \hat i##. How is that wrong ?

Presumably, in ##bv\hat i##, v stands for the magnitude of the velocity. If so, ##b\vec v=-bv\hat i##.

 

[NTesla]

Maybe this will help.

F = -kx – bv where k, b >0.
This is valid at all positions. We can check-out all the possibilities...

Call the equilibrium position ‘O’. There are 4 cases to consider:

1. The mass is to the right of O, moving right and slowing down. In this case:
x is positive, so kx is positive, so -kx is negative.
v is positive, so bv is positive, so –bv is negative.

2. The mass is to the right of O, moving left and speeding up. In this case:
x is positive, so kx is positive, so -kx is negative.
v is negative, so bv is negative, so – bv is positive.

You can work out cases 3 and 4 (mass to the left of O) for yourself if it’s helpful.

In case 2: the restoring force is = ##-kx\hat i##.
The damping force in this case is: ##bv\hat i##.
So, the equation becomes: ##\vec F## = ##-kx\hat i## + ##bv\hat i##.
Would you agree to this equation..?

 

[NTesla]

 

[Orodruin]

View attachment 345400

I mean, now you have defined ##v## to be positive in the negative x-direction so ##\dot x = -v## not ##v## …

 

[NTesla]

I mean, now you have defined ##v## to be positive in the negative x-direction so ##\dot x = -v## not ##v## …

This is the way I've defined velocity: ##v## is only the magnitude of the velocity. Since the mass is going in the left direction so the ##\vec {velocity} ## = -##v\hat i##.

 

[Steve4Physics]

In case 2: the restoring force is = ##-kx\hat i##.

Yes, but note that the restoring force is ##-kx\hat i## in all 4 cases.

The damping force in this case is: ##bv\hat i##.

No. The damping force is always ##-bv\hat i##.

In case 2 the mass is moving left, so ##v## is negative. The damping force is:

##\vec F_{damping} = -bv\hat i##

##= - \text {(a positive quantity} \times \text {(a negative quantity)} ~\hat i##

##= \text {(a positive quantity)} ~\hat i##

So in case 2, ##\vec F_{damping}## acts to the right, as required,

So, the equation becomes: ##\vec F## = ##-kx\hat i## + ##bv\hat i##.
Would you agree to this equation..?

No! See above. There is only one equation, ##\vec F## = ##-kx\hat i - bv\hat i##, and it applies in all 4 cases.

[Minor edits made.]

 

[NTesla]

Yes, but note that the restoring force is ##-kx\hat i## in all 4 cases.


No. The damping force is always ##-bv\hat i##.

In case 2 the mass is moving left, so ##v## is negative. The damping force is:

##F_{damping} = -bv\hat i##

##= - \text {(a positive quantity} \times \text {(a negative quantity)} ~\hat i##

##= \text {(a positive quantity)} ~\hat i##

So in case 2, ##F_{damping}## acts to the right, as required,


No! See above. There is only one equation, ##\vec F## = ##-kx\hat i## - ##bv\hat i##, and it applies in all 4 cases.

Kindly see my calculation below:

 

[Orodruin]

This is the way I've defined velocity: ##v## is only the magnitude of the velocity.

This is a very bad idea. Now you alternaningly have ##\dot x = v## and ##\dot x = -v## depending on the direction of motion.

Since the mass is going in the left direction so the ##\vec {velocity} ## = -##v\hat i##.

Yes, and you also have ##\dot x = -v##, which resolves your sign issue in the differential equation.

 

[Orodruin]

Kindly see my calculation below:
View attachment 345403

This calculation is even missing the v in the force …

 

[NTesla]

This calculation is even missing the v in the force …

The calculation is not missing v. Actually, the value 2 in the calculation is the magnitude of velocity. It's considered in the calculation.

 

[Steve4Physics]

Kindly see my calculation below:
View attachment 345403

I assume your attachment is referring to ‘case 2’, with the mass on the right of the equilibrium point.

Your eqation ##m \ddot x = -kx\hat i+ 2b \hat i## is only ‘a snapshot’. The (variable) velocity ##\dot x## has disappeared because you have replaced it by a specific value (-2m/s).

(By the way, if you need to use '##\hat i##' you have forgotton to include it on the left side of the above equation.)

Your ensuing differential equation therefore does not apply to the overall motion of the mass.

Sorry, I don't think I can contribute further.

 

[Orodruin]

The calculation is not missing v. Actually, the value 2 in the calculation is the magnitude of velocity. It's considered in the calculation.

Still doesn’t change the fact that ##\dot x## in your case is then ##-v##. And as stated in the previous post, by taking a particular velocity your equation is no longer generally applicable.

 

[haruspex]

Seems to me your whole confusion comes from taking v to be a magnitude whereas in the equation you quote in post #1 it is a signed scalar.

 

[NTesla]

Still doesn’t change the fact that ##\dot x## in your case is then ##-v##. And as stated in the previous post, by taking a particular velocity your equation is no longer generally applicable.

Yes, I understand that the calculation shown in post 17 is not generally applicable to the overall motion of the mass. That calculation I showed to let you know how I'm using the speed in my calculation. In my calculation in post 13, I've defined velocity vector as -##v\hat i## since it is going in the negative x direction. However, now i think that in your descriptions, whenever you wrote ##v## you meant it not as magnitude but as vector. Is that right ?

 

[Orodruin]

I mean, the bottom line is that if you want the dissipative force to be in the opposite direction of velocity, then you must use ##F = - b\dot x##, not ##F = -b|\dot x|##.

The former changes sign when ##\dot x## does. The latter does not.

 

[NTesla]

I mean, the bottom line is that if you want the dissipative force to be in the opposite direction of velocity, then you must use ##F = - b\dot x##, not ##F = -b|\dot x|##.

The former changes sign when ##\dot x## does. The latter does not.

Yes, that I understand now.

However, in the calculation in post 17, I wrote the equation: $$m\ddot x \hat i= - kx\hat i + 2b\hat i $$. Just for that particular moment of the motion, is that equation right ?

If it's right for that particular moment, then for that moment gamma is negative. Is that right..?

 

[Orodruin]

Yes, that I understand now.

However, in the calculation in post 17, I wrote the equation: $$m\ddot x \hat i= - kx\hat i + 2b\hat i $$. Just for that particular moment of the motion, is that equation right ?

Yes.

If it's right for that particular moment, then for that moment gamma is negative. Is that right..?

No. You are still missing the point that ##\gamma## is the factor in front of ##\dot x## in the differential equation when the term is moved to the LHS. Since ##\dot x = -2##, ##\gamma## is still ##b/m > 0##.

 

[NTesla]

Seems to me your whole confusion comes from taking v to be a magnitude whereas in the equation you quote in post #1 it is a signed scalar.

Were you referring to this v? (kindly see the screenshot): i understand what a scalar quantity is, but did you mean something else when you wrote a "signed scalar".

I now suppose in the screenshot, v has been taken as a vector, even when there's no vector sign on it, & neither is it bold font, which is the norm in that book.

 

[haruspex]

Were you referring to this v? (kindly see the screenshot): i understand what a scalar quantity is, but did you mean something else when you wrote a "signed scalar".

No, just emphasising that, being a scalar, it can be positive or negative.

I now suppose in the screenshot, v has been taken as a vector, even when there's no vector sign on it, & neither is it bold font, which is the norm in that book.

Why do you suppose it is a vector? The equation here is for motion in one dimension, so a scalar is as good as a vector.

 

[NTesla]

No, just emphasising that, being a scalar, it can be positive or negative.

Why do you suppose it is a vector? The equation here is for motion in one dimension, so a scalar is as good as a vector.

In that expression, ##f_{fric}## has been taken as ##-bv##. Since, damped force will always be opposite to the velocity vector, therefore, a negative sign has been used, so v should more aptly be written with a vector sign.

 

[Orodruin]

In that expression, ##f_{fric}## has been taken as ##-bv##. Since, damped force will always be opposite to the velocity vector, therefore, a negative sign has been used, so v should more aptly be written with a vector sign.

It is unclear why you keep insisting on this. As already stated, this is not really an issue in one dimension. There is only one direction so you do not need to write things out with basis vectors and everything. If you really want, you can call it the x-component of a vector equation with ##v## being the x-component of the velocity. Components are not vectors and should not come with vector arrows or boldface.

 

[NTesla]

It is unclear why you keep insisting on this. As already stated, this is not really an issue in one dimension. There is only one direction so you do not need to write things out with basis vectors and everything. If you really want, you can call it the x-component of a vector equation with ##v## being the x-component of the velocity. Components are not vectors and should not come with vector arrows or boldface.

I understand what you've written. I had some inclination from the very beginning that the writer of that equation in the book meant a vector but has choosen to write only the components in the equation. I also thought that maybe I'm wrong in assuming that. That's why I started to think about specific cases.

However, in order to make it ambiguity free and mathematically rigorous, it would have been for the benefit of all, if the book had written the force equation using the vector sign, and then the writer could have said: since in this example, only one dimension has been considered, we may choose to write the equation as follows: as it has been written in the book presently.

For someone who is studying damped oscillation for the first time, it would have been much clear if the vector sign had been used.

 

[Orodruin]

I understand what you've written. I had some inclination from the very beginning that the writer of that equation in the book meant a vector but has choosen to write only the components in the equation. I also thought that maybe I'm wrong in assuming that. That's why I started to think about specific cases.

However, in order to make it ambiguity free and mathematically rigorous, it would have been for the benefit of all, if the book had written the force equation using the vector sign, and then the writer could have said: since in this example, only one dimension has been considered, we may choose to write the equation as follows: as it has been written in the book presently.

For someone who is studying damped oscillation for the first time, it would have been much clear if the vector sign had been used.

I don’t have access to the book, but I would presume that all of the definitions are in it. In particular in section 3.6, which is referenced here. If done appropriately, then there is no source of possible confusion.

 

[kuruman]

I think part of the problem with writing Newton's second law in this case is that the velocity, which is not normally part of an FBD, is crucial here for determining the direction of the damping force. The way to sort this out is to draw the FBD at the moment when the velocity is in the whatever direction is positive, e.g. to the right. Then the damping force, written as ##F_{damping}=-bv## will always be opposite to the velocity, i.e. to the left when ##\dot x=v>0## and to the right when ##\dot x=v<0##.

Understanding this is entirely analogous to understanding the restoring force as ##F_{restoring}=-kx.##
That said, it is important to note that when writing the second law from a FBD, the direction of the forces must come out right.

In the FBD of the stretched spring on the right both forces are to the left and ##x## and ##v## are positive. If one writes $$F_{net}=-kx-bv$$ both forces are negative, which is as it should be.

In the FBD of the compressed spring on the right, the restoring force is positive whilst the damping force is negative. However, ##x## has changed direction but ##v## has not. If one writes $$F_{net}=-kx-bv$$the restoring force is positive and the damping force is negative, which is as it should be.

 

[kuruman]

It occurred to me that I should complete the picture by adding FBDs of the two cases above when the velocity of the mass is reversed (see below). The reader can readily verify that the equation $$F_{net}=-kx-bv$$ gives the correct direction of the restoring and damping forces.