I Standing Waves by Reflection: Losses of the Reflected Wave

[greypilgrim]

TL;DR Summary

The reflected wave should have lost quite some energy, why does this not seem to matter?

Hi.
I've seen quite a couple of demonstrations of standing waves that were excited at one end only, such as transverse waves on a string attached to a vibration generator and clamped at the other, or the motion of cork dust in a Kundt's tube with a speaker at one end and the other either open or closed. The justification normally is
$$A\sin\left(\omega t-kx\right)+A\sin\left(\omega t+kx\right)=2A\sin\left(\omega t\right)\cos\left( kx\right)$$
and imposing boundary conditions.

However, this seems to be heavily idealised. There is a transmitted part at the boundary, so the amplitude of the reflected wave should be reduced, and especially in the case of Kundt's tube with open end this should be significant. And there will also be damping along the chain of oscillators, which will reduce the amplitude gradually the further the waves has travelled from the generator.

However, I haven't noticed asymmetries in those experiments, so I assume even including those losses the steady-state solutions will approach above idealised solution. Is there a rigorous mathematical treatment (or maybe even a simple argument) supporting this?

 

[BvU]

Hi,

TL;DR Summary: The reflected wave should have lost quite some energy, why does this not seem to matter?

maybe even a simple argument

Perhaps:

excited at one end

I suppose the energy balance must hold: the source of excitation adds the energy dissipated along the string.

An argument to underpin this may be found in the startup situation: as long as the dissipation doesn't equal the excitation energy, the amplitude increases.

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[greypilgrim]

Right. But how does this explain that in those very asymmetric situations all anti-nodes seem to have the exact same amplitude?

 

[BvU]

I don't know which situations you are referring to. Startup ? Can you be more specific ?
References, examples ?

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[greypilgrim]

Kundt's tube and standing waves on a string. For example in this picture:

The wave is only driven from the left, but it looks like all anti-nodes have perfectly the same amplitude. Why? Is damping along the string just too low such that no effects are noticeable?

Let's assume a very simple model where the amplitude of the wave is damped exponentially with distance from the generator, and there is no loss at the boundary. Let the generator be at $x=0$ and reflection happen at $x=\lambda$, and assume anti-nodes at both ends. Then the wave travelling to the right is given by
$$f\left(t,x\right)=A\cdot e^{-\kappa x}\cdot\sin\left(2\pi\left[\frac{t}{T}-\frac{x}{\lambda}\right]\right)$$
with some damping parameter $\kappa$.
At position $x$, the reflected wave has travelled by a total distance from the generator $\lambda+\left(\lambda-x\right)=2\lambda-x$, hence can be written as
$$g\left(t,x\right)=A\cdot e^{-\kappa\cdot\left(2\lambda-x\right)}\cdot\sin\left(2\pi\left[\frac{t}{T}+\frac{x}{\lambda}\right]\right)\enspace.$$
I ran a simulation and don't get a standing wave for the superposition $f\left(t,x\right)+g\left(t,x\right)$ for larger $\kappa$.

Also, if I assume no damping $\kappa=0$ but that only part of the wave is reflected, i.e.
$$f\left(t,x\right)+\mu\cdot g\left(t,x\right)$$
for $0<\mu<1$, I don't get a standing wave either.

 

[BvU]

don't get a stable solution

What is your energy source ?



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[greypilgrim]

The idealised case does not take into account energy either (so it assumes energy input and losses to be equal in the stationary case), so why should this be relevant now?

 

[sophiecentaur]

I suppose the energy balance must hold: the source of excitation adds the energy dissipated along the string

Another loss can be in the reflector. A closed pipe/ opened ended pipe / solid fixing of a string etc. will reduce the reflected wave amplitude. So the 'nodes' , which would ideally have zero amplitude, are filled in by an amount which is the non-zero difference between the two wave amplitudes. As you say, the energy supplied by the source will cause the energy in the waves to increase up to a limiting value where the energy supplied is the energy dissipated. Depending on the Q factor, this can take many cycles before conditions settle down.

What is your energy source ?

This is often not considered . In order to supply energy to the system, the energy source (power source) may have an appreciable source impedance. In any simulation, you need to include values of loss or , as the OP has found. With unsuitable values and with an arbitrary length of tube / string, you will initially get a progressive wave going back and forth, interfering with its own multiple reflections. Eventually, you will get a standing wave with a phase difference between the driving force and the resulting displacement at the end. This is very familiar for RF engineers with lengths of Feeder and real impedances at transmitter and antenna. In that situation, they want low reflection and ideally a Standing Wave Ratio as near to unity as possible (over as wide a frequency range as possible.

I don't get a standing wave either.

This may take many cycles if the tube length is not an integral number of half waves long.

 

[greypilgrim]

This may take many cycles if the tube length is not an integral number of half waves long.

Wait, are you saying in the lossy case there are always stationary solutions in the long time limit, even if the tube length is not an integer multiple of half the wavelength?

In the idealised case with no losses, to my knowledge only standing waves that satisfy this condition exactly are allowed.

 

[sophiecentaur]

Wait, are you saying in the lossy case there are always stationary solutions

Yes. I think so because the impedance of the power source will 'provide' the line with a phase shift to give the equivalent of line length appropriate for a stationary wave - the line will behave as if a bit shorter or longer. For an 'infinite or exactly zero source impedance the line will never be resonant. This is the equivalent function of making a simple resonator (vibrating reed / mass on spring) respond off the natural frequency

 

[BvU]

The idealised case does not take into account energy either (so it assumes energy input and losses to be equal in the stationary case), so why should this be relevant now?

In your simulation I don't see reflection modelled. No way for $A$ to grow.

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[greypilgrim]

I played around with the following applet:

https://phet.colorado.edu/sims/html/wave-on-a-string/latest/wave-on-a-string_all.html

However, they unfortunately do not say how they model the damping, but it looks exponential to me:

Then the wave travelling to the right is given by
$$f\left(t,x\right)=A\cdot e^{-\kappa x}\cdot\sin\left(2\pi\left[\frac{t}{T}-\frac{x}{\lambda}\right]\right)$$
with some damping parameter $\kappa$.

Then I placed a loose end at $x=\lambda$ and assumed continuity there for the reflected part:

$$g\left(t,x\right)=A\cdot e^{-\kappa\cdot\left(2\lambda-x\right)}\cdot\sin\left(2\pi\left[\frac{t}{T}+\frac{x}{\lambda}\right]\right)$$

So I think what I assume is that both those waves have reached a state where for each oscillator the amplitude $A\cdot e^{-\kappa x}$ or $A\cdot e^{-\kappa\cdot\left(2\lambda-x\right)}$ has reached an equilibrium between energy input and loss. But isn't the idealised case in #1 (which is my model for $\kappa\rightarrow 0$) doing the same? They also just assume the amplitudes of both waves to be constant.

The applet also allows placing a fixed or loose end on the right, but it's quite hard to find the frequencies for standing waves, and there is not enough information provided to calculate them...

 

[hutchphd]

Also, if I assume no damping κ=0 but that only part of the wave is reflected, i.e.
f(t,x)+μ⋅g(t,x)
for 0<μ<1, I don't get a standing wave either.

If you wish to solve this you should learn to use complex exponentials for waves. The solution you write down is incorrect because the partial reflection in this case will require a phase shift for reflected wave (i.e μ will be complex). Similarly the case with extended attenuation is much much easier to understand and calculate with complex wavenumbers.
Anyhow the supposition above is incomplete, and an exact solution is not that difficult if you wish to do it. Any introductory quantum mechanics text should treat this as 1D scattering from a barrier.

 

[BvU]

The applet also allows placing a fixed or loose end on the right, but it's quite hard to find the frequencies for standing waves, and there is not enough information provided to calculate them...

Yes, it does take some patience. I wasted time playing with it (instead of thinking ) and approached a resonance with 0.25 Hz (medium tension, no damping)

Takes quite some time to build up

but after 400 seconds the amplitude is about 2 cm. Unfortunately, the smallest increment in damping saps all energy in something like four periods. In two periods the amplitude drops from 2 cm to 2 mm, so $\kappa$ is huge and $Q$ is abysmal:

meaning you can't get a decent resonance. You need to increase the driving amplitude (e.g. to 2 mm), and thus also go to a lower frequency (0.23 Hz), but the amplitude of the wave doesn't go above 4 mm.

Perhaps instead of diving into quantum mechanics and complex numbers, a suitable next step is to study the classical damped harmonic coscillator

In the mean time we could ask the app chap to allow the slider to set a lot less less damping ....

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[Herman Trivilino]

TL;DR Summary: The reflected wave should have lost quite some energy, why does this not seem to matter?

There is a transmitted part at the boundary, so the amplitude of the reflected wave should be reduced,

Ok. So it's reduced. Why should that cause each wave peak to have a different height? Wouldn't they all be affected the same?

 

[A.T.]

Kundt's tube and standing waves on a string. For example in this picture:
View attachment 356297
The wave is only driven from the left, but it looks like all anti-nodes have perfectly the same amplitude. Why?

Which anti-nodes do you expect to have higher amplitudes, and why?

 

[tech99]

The vibrating system is an energy store, so that when we first switch on, the generator is mostly occupied in filling the store. When equilibrium is reached, the energy of the generator now supplies only the losses in the system, such as movement of the fixings and sound radiation. During this state of continuous oscillation, there is a large, perfect standing wave and a small travelling wave. The former is the energy store and the latter conveys energy from the generator to the various loads (or sources of dissipation). When the generator is switched off, the energy store then supplies the losses and slowly dies away.
Notice that the standing wave is not altered by the losses in the system, whereas the travelling wave carries only the energy being supplied to the losses. We can find that the phase of the standing wave is everywhere the same except for a sudden 180 degree reversal at nodes. On the other hand, the phase of the travelling wave progresses uniformly along the line. Additionally, if we observe, say, the pressure along a sound pipe, then if it has losses we see that the node is not quite zero, and we see the travelling wave alone at that point, so we actually observe a small component of pressure at that point and find that it is at +/-90 degrees, or in quadrature, to the standing wave.