Damped harmonic oscillation

[NTesla]

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.

Yes i agree. If done appropriately. One sure way of doing it appropriately would have been use of vector sign.

I very much appreciate the effort and patience of @Orodruin @haruspex @Steve4Physics and @kuruman in helping me out. Thanks to all.

 

[Orodruin]

Yes i agree. If done appropriately. One sure way of doing it appropriately would have been use of vector sign.

I very much appreciate the effort and patience of @Orodruin @haruspex @Steve4Physics and @kuruman in helping me out. Thanks to all.

You put a vector sign on one quantity and you need to put it on all quantities. If not it is certainly not doing things appropriately. There is also nothing inappropriate in looking at a component of a vector equation.

 

[NTesla]

You put a vector sign on one quantity and you need to put it on all quantities. If not it is certainly not doing things appropriately. There is also nothing inappropriate in looking at a component of a vector equation.

Yes i agree. The notation must be uniformly applied throughout. Otherwise it creates unnecessary confusion. The book uses bold letters to signify vectors everywhere else. This was the first instance when the writer meant a vector but used neither the vector sign nor the bold font. In all other places in the book, wherever vector has been meant, either the vector sign has been used(such as cap over the unit vector) or bold font has been used. But when dealing with the description of damped harmonic oscillation, neither was used. In my opinion, it is a good practice to write a vector with the vector notation when writing it in an equation.

 

[Orodruin]

This was the first instance when the writer meant a vector but used neither the vector sign nor the bold font.

Again, the component of a vector is not a vector. This is all very standard notation in one-dimensional problems.

 

[kuruman]

Yes i agree. The notation must be uniformly applied throughout. Otherwise it creates unnecessary confusion. The book uses bold letters to signify vectors everywhere else. This was the first instance when the writer meant a vector but used neither the vector sign nor the bold font. In all other places in the book, wherever vector has been meant, either the vector sign has been used(such as cap over the unit vector) or bold font has been used. But when dealing with the description of damped harmonic oscillation, neither was used. In my opinion, it is a good practice to write a vector with the vector notation when writing it in an equation.

Let's do it formally and top down. I will use boldface for vectors. The net force is always the sum of the restoring and damping force vectors, $$\begin{align}\mathbf F_{\text{net}}=\mathbf F_{\text{restoring}}+\mathbf F_{\text{damping}}~.\end{align}$$The restoring-force vector is always opposite to the displacement vector, $$\mathbf F_{\text{restoring}}=-k\mathbf x~.$$ The damping-force vector is always opposite to the velocity vector, $$\mathbf F_{\text{damping}}=-b\mathbf v~.$$Substitute in equation (1) to get the net force, $$\begin{align}\mathbf F_{\text{net}}=-k\mathbf x-b\mathbf v~.\end{align}$$ At this point, because we are in one dimension, we can change the boldface vectors to italic scalars and write $$\begin{align} F_{\text{net}}=-k x-b v~.\end{align}$$ with the understanding that the scalars are signed quantities that can be positive or negative. This means that ##x## stands for 1D displacement, not distance and ##v## stands for 1D velocity, not speed.

 

[NTesla]

Again, the component of a vector is not a vector. This is all very standard notation in one-dimensional problems.

I know that component of a vector is not a vector. But if you happen to use vector all throughout your writings and then resort to using a scalar symbol to mean that a vector is being talked about, then that is not standard.

 

[NTesla]

Let's do it formally and top down. I will use boldface for vectors. The net force is always the sum of the restoring and damping force vectors, $$\begin{align}\mathbf F_{\text{net}}=\mathbf F_{\text{restoring}}+\mathbf F_{\text{damping}}~.\end{align}$$The restoring-force vector is always opposite to the displacement vector, $$\mathbf F_{\text{restoring}}=-k\mathbf x~.$$ The damping-force vector is always opposite to the velocity vector, $$\mathbf F_{\text{damping}}=-b\mathbf v~.$$Substitute in equation (1) to get the net force, $$\begin{align}\mathbf F_{\text{net}}=-k\mathbf x-b\mathbf v~.\end{align}$$ At this point, because we are in one dimension, we can change the boldface vectors to italic scalars and write $$\begin{align} F_{\text{net}}=-k x-b v~.\end{align}$$ with the understanding that the scalars are signed quantities that can be positive or negative. This means that ##x## stands for 1D displacement, not distance and ##v## stands for 1D velocity, not speed.

That's what I'm talking about. So precise and clear. Appreciate it.

 

[TSny]

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

But this is not the meaning of ##v## as used in the textbook for the oscillator.
In Example 11.1 of the book we find (for the undamped oscillator):

Clearly, here, ##v## is used to denote ##\dot x##. So, ##v## is the rate of change of ##x##.
##v## is positive for motion toward the right and negative for motion toward the left.

Also,

So, in the section on the damped oscillator we again see ##v = \dot x##:

In other parts of the book, when dealing with motion in 2 or 3 dimensions, the symbol ##v## is used for the speed (magnitude of the velocity vector). So, this might be a source of confusion for the student.

 

[haruspex]

Again, the component of a vector is not a vector. This is all very standard notation in one-dimensional problems.

That does not appear to be completely standard.
It certainly seems a more reasonable use of the word "component" to say that a vector equals the sum of its components, which would mean those are vectors too. That is the view at https://www.toppr.com/ask/en-au/question/component-of-a-vector-can-be-scalar-state-true-or-false/ and https://en.wikipedia.org/wiki/Vector_projection.
The scalar multiplier of a basis vector can be called the coefficient.

 

[Orodruin]

That does not appear to be completely standard.
It certainly seems a more reasonable use of the word "component" to say that a vector equals the sum of its components, which would mean those are vectors too. That is the view at https://www.toppr.com/ask/en-au/question/component-of-a-vector-can-be-scalar-state-true-or-false/ and https://en.wikipedia.org/wiki/Vector_projection.
The scalar multiplier of a basis vector can be called the coefficient.

Now consider what happens to those when you change your coordinates?

Edit: See also eg https://www.grc.nasa.gov/www/k-12/rocket/vectpart.html

 

[haruspex]

Now consider what happens to those when you change your coordinates?

I see no problem. All we are discussing is terminology. If ##\vec v=x_1\hat e_1+x_2\hat e_2## should we call ##x_1\hat e_1, x_2\hat e_2## components and ##x_1, x_2## coefficients or call and ##x_1, x_2## components and have no word to refer to ##x_1\hat e_1, x_2\hat e_2##?

Edit: See also eg https://www.grc.nasa.gov/www/k-12/rocket/vectpart.html

Sure, there are plenty of places where you can find that usage. My point is that it is not universal even amongst sites which you would think are fairly authoritative.

 

[kuruman]

It certainly seems a more reasonable use of the word "component" to say that a vector equals the sum of its components, which would mean those are vectors too.

It is reasonable and appropriate to say that. We say "a vector is the sum of its components". To translate this statement from English into mathematese, we write the equation$$\mathbf F=F_x~\mathbf {\hat x}+F_y~\mathbf {\hat y}.$$ However, when we ask "What is the x-component of vector ##\mathbf F##?", we teach (as we were taught) that the answer is ##F_x.##

 

[haruspex]

However, when we ask "What is the x-component of vector ##\mathbf F##?", we teach (as we were taught) that the answer is ##F_x.##

We all, naturally, teach as we were taught (or as we understood at the time, which is not always the same). But as the links I posted illustrate, we were not all taught the same, it seems.

This one is interesting: https://www.cuemath.com/geometry/components-of-a-vector/
"The values ##a, b, c ## are called the scalar components of vector A, and ##a\hat i, b\hat j, c\hat k## are called the vector components."

 

[kuruman]

We all, naturally, teach as we were taught (or as we understood at the time, which is not always the same). But as the links I posted illustrate, we were not all taught the same, it seems.

This one is interesting: https://www.cuemath.com/geometry/components-of-a-vector/
"The values ##a, b, c ## are called the scalar components of vector A, and ##a\hat i, b\hat j, c\hat k## are called the vector components."

Yes it's interesting and as it should be. And then there is this (see below) from the people who landed people on the Moon almost 55 years ago. Resistance is futile.

 

[haruspex]

Yes it's interesting and as it should be. And then there is this (see below) from the people who landed people on the Moon almost 55 years ago.

I repeat, my point is that the usage is not as standard as claimed in post #39. Perfectly respectable sources express conflicting views.

 

[kuruman]

I repeat, my point is that the usage is not as standard as claimed in post #39. Perfectly respectable sources express conflicting views.

You don't have to repeat, I am with you 100%. I was expressing my doubts (in counterpoint to my "as it should be") about the respectability of NASA whose $125 million satellite to Mars crashed in 1999 because someone apparently used numbers without paying close attention to the units.

 

[Orodruin]

I see no problem. All we are discussing is terminology. If ##\vec v=x_1\hat e_1+x_2\hat e_2## should we call ##x_1\hat e_1, x_2\hat e_2## components and ##x_1, x_2## coefficients or call and ##x_1, x_2## components and have no word to refer to ##x_1\hat e_1, x_2\hat e_2##?

So you do not see a problem with an alleged vector changing magnitude under rotations etc?

 

[haruspex]

So you do not see a problem with an alleged vector changing magnitude under rotations etc?

##x_1\hat e_1, x_2\hat e_2## are vectors, no allegation required.
But please clarify the concern you have.

 

[docnet]

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

I would personally consider Orodruin's advice with more care, because there's probably a good reason. He's also a legit retired university professor who has authored a textbook in physics

 

[Orodruin]

##x_1\hat e_1, x_2\hat e_2## are vectors, no allegation required.
But please clarify the concern you have.

Yes, those are vectors, but they are the (coordinate) components of the resultant in the given coordinate system only.

 

[Orodruin]

He's also a legit retired university professor who has authored a textbook in physics

Wait, I don’t have to come to work on Monday?!? How do I tell my students? I am supposed to give a lecture at 10 am.

 

[docnet]

Wait, I don’t have to come to work on Monday?!? How do I tell my students? I am supposed to give a lecture at 10 am.

Apologies, I somehow mistook you for a retired professor instead of a current one

 

[kuruman]

Yes, those are vectors, but they are the (coordinate) components of the resultant in the given coordinate system only.

Say I have vector ##\mathbf C## in a given 2D coordinate system specified by unit vectors ##\mathbf {\hat e}_1## and ##\mathbf {\hat e}_2.## In that system I write $$\mathbf C=C_1~\mathbf {\hat e}_1+C_2~\mathbf {\hat e}_2.$$In the above equation, ##C_1~\mathbf {\hat e}_1## and ##C_2~\mathbf {\hat e}_2## are the vector components of ##\mathbf C## whilst ##C_1## and ##C_2## are the scalar components of ##\mathbf C##. If I consider a rotation in the standard form that takes the unprimed vectors to primed vectors$$
\tilde R =\begin{pmatrix}
\cos\varphi & -\sin\varphi \\
\sin\varphi & \cos\varphi \\
\end{pmatrix},
$$ I get $$\mathbf C'=\tilde R~ \mathbf C=\tilde R (C_1~\mathbf {\hat e}_1)+\tilde R( C_2~\mathbf {\hat e}_2)
=C_1\tilde R~(\mathbf {\hat e}_1 )+C_2\tilde R~(\mathbf {\hat e}_2) =C_1\mathbf {\hat e}'_1+C_2\mathbf {\hat e}'_2.$$ The vector components of the rotated vector ##\mathbf C'## are the rotated vector components of the initial vector ##\mathbf C## whilst the scalar components are invariant. It's what one would expect.

Just to check $$\begin{align} & \tilde R (C_1~\mathbf {\hat e}_1)=
\begin{pmatrix}
\cos\varphi & -\sin\varphi \\
\sin\varphi & \cos\varphi \\
\end{pmatrix}
\begin{pmatrix} C_1 \\ 0 \end{pmatrix}=C_1\begin{pmatrix} \cos\varphi \\ \sin\varphi \end{pmatrix}\implies
|\tilde R (C_1~\mathbf {\hat e}_1)|=C_1
\nonumber \\
& \tilde R (C_2~\mathbf {\hat e}_2)=
\begin{pmatrix}
\cos\varphi & -\sin\varphi \\
\sin\varphi & \cos\varphi \\
\end{pmatrix}
\begin{pmatrix} 0 \\ C_2 \end{pmatrix}=C_2\begin{pmatrix} -\sin\varphi \\ \cos\varphi \end{pmatrix}
\implies
|\tilde R (C_2~\mathbf {\hat e}_2)|=C_2
.
\nonumber\end{align}$$ Yup, the magnitudes of the vector components of ##\mathbf C## are invariant under rotations.

I see no problem with all this. Did I miss your point?

 

[Orodruin]

Say I have vector ##\mathbf C## in a given 2D coordinate system specified by unit vectors ##\mathbf {\hat e}_1## and ##\mathbf {\hat e}_2.## In that system I write $$\mathbf C=C_1~\mathbf {\hat e}_1+C_2~\mathbf {\hat e}_2.$$In the above equation, ##C_1~\mathbf {\hat e}_1## and ##C_2~\mathbf {\hat e}_2## are the vector components of ##\mathbf C## whilst ##C_1## and ##C_2## are the scalar components of ##\mathbf C##. If I consider a rotation in the standard form that takes the unprimed vectors to primed vectors$$
\tilde R =\begin{pmatrix}
\cos\varphi & -\sin\varphi \\
\sin\varphi & \cos\varphi \\
\end{pmatrix},
$$ I get $$\mathbf C'=\tilde R~ \mathbf C=\tilde R (C_1~\mathbf {\hat e}_1)+\tilde R( C_2~\mathbf {\hat e}_2)
=C_1\tilde R~(\mathbf {\hat e}_1 )+C_2\tilde R~(\mathbf {\hat e}_2) =C_1\mathbf {\hat e}'_1+C_2\mathbf {\hat e}'_2.$$ The vector components of the rotated vector ##\mathbf C'## are the rotated vector components of the initial vector ##\mathbf C## whilst the scalar components are invariant. It's what one would expect.

Just to check $$\begin{align} & \tilde R (C_1~\mathbf {\hat e}_1)=
\begin{pmatrix}
\cos\varphi & -\sin\varphi \\
\sin\varphi & \cos\varphi \\
\end{pmatrix}
\begin{pmatrix} C_1 \\ 0 \end{pmatrix}=C_1\begin{pmatrix} \cos\varphi \\ \sin\varphi \end{pmatrix}\implies
|\tilde R (C_1~\mathbf {\hat e}_1)|=C_1
\nonumber \\
& \tilde R (C_2~\mathbf {\hat e}_2)=
\begin{pmatrix}
\cos\varphi & -\sin\varphi \\
\sin\varphi & \cos\varphi \\
\end{pmatrix}
\begin{pmatrix} 0 \\ C_2 \end{pmatrix}=C_2\begin{pmatrix} -\sin\varphi \\ \cos\varphi \end{pmatrix}
\implies
|\tilde R (C_2~\mathbf {\hat e}_2)|=C_2
.
\nonumber\end{align}$$ Yup, the magnitudes of the vector components of ##\mathbf C## are invariant under rotations.

I see no problem with all this. Did I miss your point?

My point is that the x-component in one system is not the same as the x-component in another system. Taking the x-component of a vector in system A does not give you the same thing as taking the x-component in system B. In fact, they generally have both different magnitude and obviously correspond to different directions. As such, ”the x-component” is not coordinate invariant.

 

[Steve4Physics]

My point is that the x-component in one system is not the same as the x-component in another system. Taking the x-component of a vector in system A does not give you the same thing as taking the x-component in system B. In fact, they generally have both different magnitude and obviously correspond to different directions. As such, ”the x-component” is not coordinate invariant.

If I may interject…

I think that in Post #59, @kuruman is considering the case where the vector C and the axes are all rotated through the same angle.

But you are considering the case where C is fixed and the axes are rotated (i.e rotation of coordinate system only).

In the former case, the ‘scalar components’ stay the same, as @kuruman says.

In the latter case, the ‘scalar components’ change, as you say.

 

[haruspex]

My point is that the x-component in one system is not the same as the x-component in another system. Taking the x-component of a vector in system A does not give you the same thing as taking the x-component in system B. In fact, they generally have both different magnitude and obviously correspond to different directions. As such, ”the x-component” is not coordinate invariant.

I cannot see what that has to do with my post #44. The question I raised there was merely one of terminology: having resolved a vector into components in the ##\hat x## and ##\hat y## directions, as ##\vec v=x\hat x+y\hat y## say, what are the "components"? Are they ##x## and ##y## or ##x\hat x## and ##y\hat y##?

If anyone wants to continue this sidebar, I think we should do it by PM.

 

[Orodruin]

I cannot see what that has to do with my post #44. The question I raised there was merely one of terminology: having resolved a vector into components in the ##\hat x## and ##\hat y## directions, as ##\vec v=x\hat x+y\hat y## say, what are the "components"? Are they ##x## and ##y## or ##x\hat x## and ##y\hat y##?

If anyone wants to continue this sidebar, I think we should do it by PM.

I mean, ultimately it comes down to if you want everyone to necessarily write vector arrows on everything that would need it in three dimensions also for one-dimensional problems. I don’t. I expect anyone that argues for this to also start writing double arrows on top of any variable representing string tension as it is actually a rank two tensor. What you actually call things later is not as important.

 

[NTesla]

I mean, ultimately it comes down to if you want everyone to necessarily write vector arrows on everything that would need it in three dimensions also for one-dimensional problems. I don’t. I expect anyone that argues for this to also start writing double arrows on top of any variable representing string tension as it is actually a rank two tensor. What you actually call things later is not as important.

A good physicist ought to use proper symbols/notations/units everywhere, especially when writing a book, or when working on a project or when explaining things to others. If one is going to not use such symbols/notations/units for something, then justification and reminder for doing it like that, must be mentioned at appropriate places. If you are studying/doing some physics on your own for your own sake, then maybe not. This makes things precise & clear for everyone, otherwise it may lead to such disasters as mentioned in post#51.

 

[Orodruin]

A good physicist ought to use proper symbols/notations/units everywhere, especially when writing a book, or when working on a project or when explaining things to others. If one is going to not use such symbols/notations/units for something, then justification and reminder for doing it like that, must be mentioned at appropriate places. If you are studying/doing some physics on your own for your own sake, then maybe not. This makes things precise & clear for everyone, otherwise it may lead to such disasters as mentioned in post#51.

You are missing the point completely. Writing out the projected equation is absolutely proper. There is not a single thing inappropriate about it. Do you also want to use double arrows on string tensions? If you want to be consistent with this point of view you must. I do not know any single source that does this, you are free to find counter examples if you can.

 

[NTesla]

You are missing the point completely. Writing out the projected equation is absolutely proper. There is not a single thing inappropriate about it. Do you also want to use double arrows on string tensions? If you want to be consistent with this point of view you must. I do not know any single source that does this, you are free to find counter examples if you can.

You are completely missing the point that I made regarding writing justifications and reminders if one is not going to do that.