Damped harmonic oscillation

[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.

 

[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.