Circle approximation of a wave pulse on a string or spring

[Brian_D]

Homework Statement

In general, this concerns problems involving maximum speed of a point on a wave pulse on a string or spring. For example, consider the following: "A transverse wave with 3.0 cm amplitude and 75 cm wavelength is propagating on a stretched spring whose mass per unit length is 170 g/m. If the wave speed is 6.7 m/s, find (a) the spring tension and (b) the maximum speed of any point on the spring."

The answer key says (a) 7.6 N; (b) 1.7 m/s

Relevant Equations

V=\sqrt{\frac{F}{\mu}}\mathit{where}V\mathit{is}\mathit{wave}\mathit{speed},F\mathit{is}\mathit{the}\mathit{tension}\mathit{force},
\\
\boldsymbol{\land}\mu\mathit{is}\mathit{the}\mathit{mass}\mathit{per}\mathit{unit}\mathit{length}\mathit{of}\mathit{the}\mathit{string}.

\mathit{circumference}=\pi \mathit{diameter}

I tried using LaTeX and Preview, but when I click Preview I didn't see the preview. The tool bar turns red, but the LaTeX code does not appear as typeset text. The relevant equations are (1) V= sqrt(F/mu) where V is the wave speed, F is the tension force, and mu is the mass per unit length of the string/spring; and (2) circumference = pi * diameter.

My physics textbook gives a circle approximation of a wave pulse on a string, where the amplitude is the radius and the top of the pulse is an arc of the circle. I don't understand how this can work. In this kind of problem, the amplitude should be independent of the wavelength. But in the circle approximation, the amplitude is the radius and the wavelength is twice the radius (where the arc is a semicircle representing half of the wave cycle). In the above-referenced example, the amplitude and the wavelength are given as separate parameters, and I can't make sense of that using the circle approximation. Can anyone help with this?

 

[BvU]

Hi,

I tried using LaTeX and Preview, but when I click Preview I didn't see the preview.

Can happen when MathJax is unable to interpret the stuff. Sometimes it helps to refresh the page (F5)
But here I suspect the delimiters are missing.

Check out the $\LaTeX$ guide (button on lower left of edit window):

You must put special delimiters at the beginning and end of your LaTeX code, in order for MathJax to recognize it and process it. If you want the equation to appear inline with the surrounding text, put ## before and after it. If you want the equation to stand alone in a separate "paragraph", centered horizontally on the screen as in most textbooks, put $$ before and after it.

$$V=\sqrt{\frac{F}{\mu}}$$ where ##V## is wave speed, ##F## is the tension force},
and ##\mu## is the mass per unit length of the string.

## {\text {circumference }}\ =\pi\   {\text {diameter}}##

yields
$$V=\sqrt{\frac{F}{\mu}}$$ where $V$ is wave speed, $F$ is the tension force, and $\mu$ is the mass per unit length of the string.

${\text {circumference }}\ =\pi\ {\text {diameter}}$

- - - - - - -

Your relevant equation helps solve part a) of the exercise.

For part b) you need to differentiate the expression for the transverse motion of a point. So you need an equation for that.

I can't make sense of

My physics textbook gives a circle approximation of a wave pulse on a string, where the amplitude is the radius and the top of the pulse is an arc of the circle. I don't understand how this can work. In this kind of problem, the amplitude should be independent of the wavelength. But in the circle approximation, the amplitude is the radius and the wavelength is twice the radius (where the arc is a semicircle representing half of the wave cycle). In the above-referenced example, the amplitude and the wavelength are given as separate parameters, and I can't make sense of that using the circle approximation. Can anyone help with this?

The amplitude is in the transverse direction, the wavelength is in the longitudinal direction. I don't believe 'the wavelength is twice the radius' is a correct interpretation...

$\$

 

[haruspex]

My physics textbook gives a circle approximation of a wave pulse on a string, where the amplitude is the radius and the top of the pulse is an arc of the circle.

That sounds bizarre. Forgive me, but I have to wonder if you are misinterpreting it. Can you post the actual text and diagram as an image?

 

[renormalize]

That sounds bizarre. Forgive me, but I have to wonder if you are misinterpreting it. Can you post the actual text and diagram as an image?

He's probably referring to a diagram similar to this:

(https://xmphysics.com/2023/01/11/appendix-a-angular-frequency-vs-angular-velocity/)

 

[haruspex]

He's probably referring to a diagram similar to this:
View attachment 357319
(https://xmphysics.com/2023/01/11/appendix-a-angular-frequency-vs-angular-velocity/)

Yes, that is what I was suspecting.

 

[Brian_D]

Thank you, all, for these helpful comments. I created a two-page PDF with the relevant section of the textbook, including a diagram. Can someone tell me how to attach the PDF or otherwise insert the pages? I tried the "insert image" command, but it would not let me insert a PDF file.

The entire discussion in the book and the diagram are apparently intended to derive the formula $v=\sqrt{\frac{F}{\mu}}$. This formula can be used to solve part (a) of the problem, as BvU noted. BvU said "For part b) you need to differentiate the expression for the transverse motion of a point. So you need an equation for that." I need to give that more thought.

BvU also said, "I don't believe 'the wavelength is twice the radius' is a correct interpretation..." Yes, I understand that now. I was basing that statement on a half-circle arc, but I think the book is assuming that the approximation only holds for a small part of the arc near the top of the pulse. Also, I was incorrectly thinking that the point moves along the arc, while in fact it only moves up and down.

 

[TSny]

For a harmonic wave (sinusoidal wave), each element of the spring moves in simple harmonic motion (SHM). The maximum speed of a particle moving in SHM can be found from the amplitude and frequency of the SHM. You might need to review this from your study of SHM.

 

[renormalize]

Can someone tell me how to attach the PDF or otherwise insert the pages?

 

[haruspex]

I created a two-page PDF with the relevant section of the textbook, including a diagram. Can someone tell me how to attach the PDF or otherwise insert the pages? I tried the "insert image" command, but it would not let me insert a PDF file.

You can attach pdf, but a lot of viewers won’t bother to download it to view it. Better to attach an image.
On my iPad I can click the image button, just left of the smiley button, then tap the "drop here" instruction. It gives me the option of taking a photo.

BvU said "For part b) you need to differentiate the expression for the transverse motion of a point. So you need an equation for that."

The motions of the physical material are transverse for a string but longitudinal for a spring. Your problem statement says it could be either. But the equation (the derivation of which for a string does involve an approximation for small oscillations) is the same.

 

[Brian_D]

For a harmonic wave (sinusoidal wave), each element of the spring moves in simple harmonic motion (SHM). The maximum speed of a particle moving in SHM can be found from the amplitude and frequency of the SHM. You might need to review this from your study of SHM.

View attachment 357335

View attachment 357335

Thank you, renormalize, I don't know how I overlooked that. :-) Thank you, TSny and Haruspex, more things that I need to think about further. For now, I'm just going to attach the PDF. These two pages are the section from the textbook is entitled "Waves on a String." It includes a diagram (Figure16-13) showing the circular approximation of wave pulse.

 

[haruspex]

Thank you, renormalize, I don't know how I overlooked that. :-) Thank you, TSny and Haruspex, more things that I need to think about further. For now, I'm just going to attach the PDF. These two pages are the section from the textbook is entitled "Waves on a String." It includes a diagram (Figure16-13) showing the circular approximation of wave pulse.

Ok.
The book is only claiming that an arbitrarily short section of the string at the peak of the displacement can be approximated by an arc of a circle. For this purpose, the wave as a whole can be any reasonably smooth shape.

 

[BvU]

Found your exercise (14.2 no 27 in an older edition). Given the transverse wave amplitude and wavelength, you are expected to use $(16.3)$: $y(x,t)=A\cos(kx-\omega t)$ (sinusoidal wave) as the equation for part b).

Perhaps also check out $(12.1.3)$ under 'Harmonic Waves' here

$\$

 

[Brian_D]

View attachment 357337

 

[Brian_D]

Thank you, BvU, very helpful. This is straightforward, except for two things. First, I don't know how to find k. The formula is $k=m \omega^{2}$, but isn't the "m" in the formula the mass of a bob or ball on a spring? For the current problem, we have the mass per unit length of the whole spring, but there is no bob or ball, so how do we find k?

Second, the maximum speed of a simple harmonic system would occur when displacement is zero. But the equation you suggested is a function of two variables, so how do we solve for x and t when displacement is zero?

 

[haruspex]

Thank you, BvU, very helpful. This is straightforward, except for two things. First, I don't know how to find k. The formula is $k=m \omega^{2}$, but isn't the "m" in the formula the mass of a bob or ball on a spring? For the current problem, we have the mass per unit length of the whole spring, but there is no bob or ball, so how do we find k?

Second, the maximum speed of a simple harmonic system would occur when displacement is zero. But the equation you suggested is a function of two variables, so how do we solve for x and t when displacement is zero?

Two different meanings of k. Compare the dimensions.
In $y(x,t)=A\cos(kx-\omega t)$, $kx$ must be an angle (in radians). Increasing x by one wavelength must increase that angle by $2\pi$, so $k=\frac{2\pi}{\lambda}$.
In $k=m \omega^{2}$, $k$ has dimension $MT^{-2}$.

 

[BvU]

Thank you, BvU, very helpful. This is straightforward, except for two things. First, I don't know how to find k. The formula is $k=m \omega^{2}$, but isn't the "m" in the formula the mass of a bob or ball on a spring? For the current problem, we have the mass per unit length of the whole spring, but there is no bob or ball, so how do we find k?

I grant you it's confusing . Matter of context, and (as @haruspex clarifies) often you can focus using dimensions.

With a spring, the force is $F = kx$, with $k$ in Newton/m.

In the formula ($(16.3)$ if I am not mistaken) that describes a travelling transverse wave $$y(x,t)=A\cos(kx-\omega t)$$ the angular wave number is $\displaystyle k ={ 2\pi\over\lambda}$ , with $k$ in radians/m.

Same issue with force $F$ (Newton, kg m/s²) and frequency $f$ (Hz, 1/s)

Second, the maximum speed of a simple harmonic system would occur when displacement is zero. But the equation you suggested is a function of two variables, so how do we solve for x and t when displacement is zero?

That is the subject of 'Wave Math' (16.2 I expect).

That the wave travels to the right you can see by looking at the argument $kx - \omega t$: a crest of the moving wave satisfies $kx - \omega t=n\lambda$ with $n$ an integer. Follow the crest: for a given constant $n$, the value of $x$ has to increase with $t$ to the tune of $kx = n\lambda + \omega t$. in other words $\displaystyle {dx\over dt } = {\omega\over k}$ .
Or -- in other words $v = 2\pi f / (2\pi/\lambda) = \lambda f$.

(And sure enough, with a plus sign instead of a minus sign the wave travels in the negative x direction.)

A point on the spring has a fixed position on the $x$-axis. It moves up and down only, so it has longitudinal speed zero (!).
Pick a point, for example with $x = \lambda/4$ so that $kx = \pi/2$. Or $x = (n+{1\over 4})\,\lambda$. Its transverse deviation is $y = A\cos(\pi/2-\omega t)= A\sin(\omega t)$. And its transverse speed is ${dy\over dt } =\omega A\cos(\omega t)$.

And indeed, when t = 0 it has deviation 0 and maximum transverse speed $A\omega$.

Tip:

You can learn a lot by drawing a sine wave on a transparency, Draw a coordinate system on a piece of paper; place the transparency on top and move it along the x-axis.
Makes these pictures come to life:

$\$

 

[haruspex]

With a spring, the force is $F = kx$, with $k$ in Newton/m.

No, $k$ is in force per unit length. The units can be anything of that dimension; dynes per light year would be valid.

In the formula ($(16.3)$ if I am not mistaken) that describes a travelling transverse wave

It would apply to longitudinal waves too, no? The only difference is whether $y$ is a lateral or longitudinal displacement.
The question in post #1 is a bit strange in that it describes a wave along a spring as transverse. It could be, I suppose, but rather unusual.

A point on the spring has a fixed position on the $x$-axis. It moves up and down only, so it has longitudinal speed zero (!).

Again, that would be for a string, but unusual for a spring.

 

[BvU]

No, $k$ is in force per unit length. The units can be anything of that dimension; dynes per light year would be valid.

True. I didn't want to make it too complicated. Still have nightmares about the furlongs per fortnight my colleagues in the developing states used instead of m/s like on my speedometer ... (oops!)

Again, that would be for a string, but unusual for a spring.

Exercise composers....

apparently part (b) was added in the later edition

And the pdf fragment in #10 also shows a transverse pulse in a slinky

$\$

 

[Orodruin]

No, k is in force per unit length. The units can be anything of that dimension; dynes per light year would be valid.

This is actually a pet peeve of mine. You see quite often that people use units for doing dimensional analysis. It will of course give the correct answer because the base units themselves of course have dimension, but it somewhat hides the fundamental idea of dimensionality and can be confusing if different units are used.

Ideally of course everyone would use \mathsf for typesetting dimensions …
$\mathsf M / \mathsf T^2$

 

[haruspex]

This is actually a pet peeve of mine. You see quite often that people use units for doing dimensional analysis. It will of course give the correct answer because the base units themselves of course have dimension, but it somewhat hides the fundamental idea of dimensionality and can be confusing if different units are used.

Ideally of course everyone would use \mathsf for typesetting dimensions …
$\mathsf M / \mathsf T^2$

Thanks for the tip. And I see it works using {} too: $\mathsf {MT^{-2}}$

 

[haruspex]

Exercise composers....

View attachment 357541

apparently part (b) was added in the later edition

And the pdf fragment in #10 also shows a transverse pulse in a slinky

$\$

Yes, I wrote unusual, not wrong. And intuitively I am uncomfortable with it, even though an elastic string is, in principle, just a spring with a zero compression constant. I feel lateral oscillations would induce longitudinal ones too, making the whole a bit messy.

 

[BvU]

Ideally of course everyone would use \mathsf for typesetting dimensions …
$\mathsf M / \mathsf T^2$

I seem to remember square brackets were desirable too... $\left[\mathsf M/T^2\right]$

 

[Orodruin]

Yes, I wrote unusual, not wrong. And intuitively I am uncomfortable with it, even though an elastic string is, in principle, just a spring with a zero compression constant. I feel lateral oscillations would induce longitudinal ones too, making the whole a bit messy.

They do indeed. This is true for a regular taut string as well, but is usually ignored to first approximation. They can generally have different wave velocities. Not only that, but the slinky also has a torsion mode with a different wave velocity. You can actually see it in the short clip I made of a slinky drop a while back:

Towards the end of the video - if you pay close attention - you can see the the end start wobbling a bit and a reflected pulse propagating up through the slinky. While the longitudinal wave speed is not sufficient to outrun the shock front (hence, shock front ...), the torsional mode is.

I seem to remember square brackets were desirable too... $\left[\mathsf M/T^2\right]$

Square brackets are typically used to indicate the physical dimension of a quantity. For example:
$$
[k] = \mathsf{M T^{-2}}
$$
See, e.g., my Insight on the basics of dimensional analysis (dear lord, it is already 6 years old ...).

 

[BvU]

Of course, thanks! A welcome refresher (after 53 years )

$\$

 

[Brian_D]

I grant you it's confusing . Matter of context, and (as @haruspex clarifies) often you can focus using dimensions.

With a spring, the force is $F = kx$, with $k$ in Newton/m.

In the formula ($(16.3)$ if I am not mistaken) that describes a travelling transverse wave $$y(x,t)=A\cos(kx-\omega t)$$ the angular wave number is $\displaystyle k ={ 2\pi\over\lambda}$ , with $k$ in radians/m.

Same issue with force $F$ (Newton, kg m/s²) and frequency $f$ (Hz, 1/s)




That is the subject of 'Wave Math' (16.2 I expect).

That the wave travels to the right you can see by looking at the argument $kx - \omega t$: a crest of the moving wave satisfies $kx - \omega t=n\lambda$ with $n$ an integer. Follow the crest: for a given constant $n$, the value of $x$ has to increase with $t$ to the tune of $kx = n\lambda + \omega t$. in other words $\displaystyle {dx\over dt } = {\omega\over k}$ .
Or -- in other words $v = 2\pi f / (2\pi/\lambda) = \lambda f$.

(And sure enough, with a plus sign instead of a minus sign the wave travels in the negative x direction.)

A point on the spring has a fixed position on the $x$-axis. It moves up and down only, so it has longitudinal speed zero (!).
Pick a point, for example with $x = \lambda/4$ so that $kx = \pi/2$. Or $x = (n+{1\over 4})\,\lambda$. Its transverse deviation is $y = A\cos(\pi/2-\omega t)= A\sin(\omega t)$. And its transverse speed is ${dy\over dt } =\omega A\cos(\omega t)$.

And indeed, when t = 0 it has deviation 0 and maximum transverse speed $A\omega$.

Tip:

You can learn a lot by drawing a sine wave on a transparency, Draw a coordinate system on a piece of paper; place the transparency on top and move it along the x-axis.
Makes these pictures come to life:

View attachment 357508

View attachment 357509​

$\$

There is a lot of discussion in this thread that is more advanced than I need for this problem, so I am going to focus on BvU's comments here, which seem to adopt the simplifying assumptions made by the authors of the textbook. For purposes of finding the maximum speed of a point on the spring (adopting the simplifying assumption that it is moving only vertically), let us look at the system when x=0. Then the (vertical) displacement would be $A \cos\! \left(\omega t\right)$ and the velocity would be the derivative of that, namely $-A \omega \sin\! \left(\omega t\right)$.

Next, let's find omega. Since the wave speed is 6.7 m/s (given) and the wavelength is 75 cm (given), the period (T) must be .112 meters per second, and omega is therefore $\frac{2 \pi}{T}$ or 56 radians per second. Next, the speed will be maximum where the displacement is zero, so setting the first equation to zero and solving($3 \cos\! \left(56 t\right)=0$), maximum speed occurs at $t=\frac{\pi}{112}$. Finally, putting this value of t into the second equation, the maximum speed is -168 m/s, which doesn't make sense. (The textbook answer is 1.7 m/s). Where am I going wrong?

 

[renormalize]

Next, let's find omega. Since the wave speed is 6.7 m/s (given) and the wavelength is 75 cm (given), the period (T) must be .112 meters per second, and omega is therefore $\frac{2 \pi}{T}$ or 56 radians per second. Next, the speed will be maximum where the displacement is zero, so setting the first equation to zero and solving($3 \cos\! \left(56 t\right)=0$), maximum speed occurs at $t=\frac{\pi}{112}$. Finally, putting this value of t into the second equation, the maximum speed is -168 m/s, which doesn't make sense. (The textbook answer is 1.7 m/s). Where am I going wrong?

• Period $T$ has units of $s$ (no meters involved).
• Working in SI units, the amplitude $A$ has the value $0.03\text{ m}$.
• Speed is the magnitude of the instantaneous velocity-vector and is always positive or zero.

 

[Brian_D]

Fabulous, thank you renormalize. With these corrections, I get the textbook answer. It is all clear to me now.

 

[haruspex]

• Period $T$ has units of $s^{-1}$ (no meters involved).

You mean $s$: $(0.75 m/wave)/(6.7 m/s) = 0.112 s/wave$.

 

[renormalize]

You mean $s$: $(0.75 m/wave)/(6.7 m/s) = 0.112 s/wave$.

Yes, $T$ has units of $s$. My post has been corrected. Thank's for catching my error!

 

[Brian_D]

Thank you, haruspex, renormalize and all.