# A Question about transverse waves...

• B
Okay....I have a question that, is it possible for a transverse waves to only consists of crests and not troughs (or vice versa)??..Like is it possible for the particles of the medium to only displace upwards from mean position , and not downwards???Any help is appreciated...

sophiecentaur
Gold Member
2020 Award
How would you define "mean position"?

How would you define "mean position"?

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How would you define "mean position"?
The position at which the particles of the medium initially were before displacement (or disturbance)...OR , where no net force acts on particles...

davenn
Gold Member
Okay....I have a question that, is it possible for a transverse waves to only consists of crests and not troughs (or vice versa)??..Like is it possible for the particles of the medium to only displace upwards from mean position , and not downwards???Any help is appreciated...

I cannot see how ?
if the initial displacement is upwards from your starting point, that cannot continue upwards for ever.
At some point it is going to cycle back down to the starting value before going back up again
so that low value starting point IS your trough.
Or it will continue down past that starting point before rising again
either way you still have peaks and troughs

I don't really approve of your "mean" definition

Dave

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Kaneki123
jtbell
Mentor
How would you define "mean position"?
...OR , where no net force acts on particles...
If the net force were zero at the bottom of a "trough", what would make the oscillating particles turn around and start going upwards again?

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sophiecentaur
Gold Member
2020 Award
If you had a rope, laying on the floor, you could lift the end and return it to the floor. The resulting wave would only have positive displacement. Is that good enough?PS The mean displacement would not be zero.

I cannot see how ?
if the initial displacement is upwards from your starting point, that cannot continue upwards for ever.
At some point it is going to cycle back down to the starting value before going back up again
so that low value starting point IS your trough.
Or it will continue down past that starting point before rising again
either way you still have peaks and troughs

I don't really approve of your "mean" definition

Dave
Okay...From your answer. I get that there can be no transverse wave without crests and troughs (the highest and lowest points from mean)...Two things I want to ask you, The mean position I indicated in the diagram is not exactly THE ''mean position'', right???...Second, if mean position is the position where no net force acts, then where is such position in a transverse wave???

davenn
Gold Member
Okay...From your answer. I get that there can be no transverse wave without crests and troughs (the highest and lowest points from mean)...Two things I want to ask you, The mean position I indicated in the diagram is not exactly THE ''mean position'', right???...Second, if mean position is the position where no net force acts, then where is such position in a transverse wave???

OK before we go further, I just want to make sure you understand what a traverse wave is ....

can you see in this traverse wave animation how there is a peak and trough about a mean point

Second, if mean position is the position where no net force acts, then where is such position in a transverse wave???

it would be at the "zero" crossing point of the wave, that is, where there is neither upwards or downwards motion ( or left / right sideways motion if the motion is that style)

Dave

olivermsun
two ways to view one possible answer (among many here)...
mathematical: "mean" must be higher than the lowest points if there are also crests (higher points)
continuity: in something like a water wave, if "material" goes to form the crests, there has to be a deficit of material in the troughs.

sophiecentaur
sophiecentaur
Gold Member
2020 Award
Okay...From your answer. I get that there can be no transverse wave without crests and troughs (the highest and lowest points from mean)...Two things I want to ask you, The mean position I indicated in the diagram is not exactly THE ''mean position'', right???...Second, if mean position is the position where no net force acts, then where is such position in a transverse wave???
How much reading around did you do before asking your question? I googled "Basic Wave Theory" and there were, of course, more hits than we would ever have time to read. Perhaps you should try that approach at this stage????
Your choice of the word "mean" has caused a problem. If you read around, you will find the phrase Equilibrium Position is used for what you are probably referring to. That is the position that sections of the string / water etc will take up when undisturbed by the wave. In the long term, the mean position will be the same as the equilibrium position. This would happen if the string (for instance) is returned to its original position. If not, there is a permanent offset (a 'DC' component, in EE terms)
If you read around, as I have suggested, all this will become apparent to you and you may be able to see for yourself exactly what you you actually meant by your question. From the answers, I can see that PF is a bit at loss to decide what you actually mean.
You may have been confused by the fact that some descriptions of waves stick to sinusoidal waves and others involve a single pulse (as when a string is jerked up and down once). Both are OK, of course but the descriptions of one do not fit the other . . . . Not all waves are 'symmetrical'.

nasu
Gold Member
Solitons may look like what you have in mind. In a soliton pulse the particles move one way form the equilibrium position and return to equilibrium without going past the equilibrium position.

sophiecentaur
Gold Member
2020 Award
I don't think there's anything particularly magic about solitons. It's just that the duty cycle is so low that the 'trough' is very near the equilibrium level to keep the overall volume of water unchanged.

nasu
Gold Member
The solution of the KdV equation (one of the equations with soliton solutions) is strictly positive.
I did not say that solitons are special, just that some of them may illustrate what the OP has in mind.
I suppose that if the medium has enough damping you can have other single pulse regimes.

sophiecentaur
Gold Member
2020 Award
The solution of the KdV equation (one of the equations with soliton solutions) is strictly positive.
I did not say that solitons are special, just that some of them may illustrate what the OP has in mind.
I suppose that if the medium has enough damping you can have other single pulse regimes.
I see that. But the wave has to start somewhere with a pile of water that must come from somewhere. If a train of solitons (another oxymoron?) were launched with a nice big space between them, it would involve perhaps a measurable lowering of the level. I think the solution to the equation would have to ignoring that. And it makes no difference I guess.
PS it was a good example. Even a blip on a string can be very lopsided.

it would be at the "zero" crossing point of the wave, that is, where there is neither upwards or downwards motion ( or left / right sideways motion if the motion is that style)
In case of a transverse wave in a rope, in which there is only upwards displacement from starting position, there is no such point where net force would be zero (if there is, please point out)...Hence,can we conclude that ,in such types of waves (only upwards or downwards waves), no equilibrium position exists???
P.S If mean position is taken to be the "mean between two extremes'', then that kind of demands the statement that the ''mean position is not always THE equilibrium position"???Please point out anything wrong....

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sophiecentaur