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Basic Waves help!

Screen Shot 2017-07-21 at 14.36.06.png

So, you see how there's two axis on this diagram: Horizontal is distance which isn't a vector and vertically displacement (x) which is a vector.

My question is:
Can a wave only have distance in the direction it is travelling because it can only travel in one direction? And is that why displacement is 90 degrees from the direction the wave is travelling because it is oscillating?

Also if that is incorrect, why is the definition of amplitude written as "Maximum magnitude of displacement".

Does displacement always have a fixed definition when referring to waves (as how far a point on the wave has moved from its undisturbed position)?
Ok, so the amplitude is the maximum displacement of a wave from equilibrium (if the wave was at 'rest'). In other words, it is the displacement from the peak/trough to equilibrium. There's no why this occurs, it's a fundamental property of transverse waves.

I'm not quite understanding what you mean by "have distance"? Elaborate?

And what do you mean "does displacement always have a fixed position?"
Original post by BTAnonymous
Ok, so the amplitude is the maximum displacement of a wave from equilibrium (if the wave was at 'rest':wink:. In other words, it is the displacement from the peak/trough to equilibrium. There's no why this occurs, it's a fundamental property of transverse waves.

I'm not quite understanding what you mean by "have distance"? Elaborate?

And what do you mean "does displacement always have a fixed position?"


Thanks;

What I meant by distance is that - distance is a scalar so it generally indicates one direction right? So would that mean the direction of travel can only be in one direction?

Also, does the meaning of displacement always mean one thing in reference to waves ( i.e. the one shown in the text book being "how far a point on the wave has moved from its undisturbed position") ?

Actually the last question is quite a stupid one i just realised lol.
Original post by CorpusLuteum
Thanks;

What I meant by distance is that - distance is a scalar so it generally indicates one direction right? So would that mean the direction of travel can only be in one direction?

Also, does the meaning of displacement always mean one thing in reference to waves ( i.e. the one shown in the text book being "how far a point on the wave has moved from its undisturbed position":wink: ?

Actually the last question is quite a stupid one i just realised lol.


I feel like you're over thinking this way too much.

The distance on the model is just indicating how the motion of transverse waves behave as they oscillate. Waves can be travelling in one direction then be refracted or reflected to then start 'travelling' in a different direction but they are still 'travelling' a distance.

I think you may misunderstand displacement and distance? Just to clear it up, displacement is the distance from the point of origin. So if I start on a circle and complete one cycle of the circle, my displacement is 0 because I arrive at my origin. However, the distance I travel is the circumference of the circle and as you can tell, the numbers will be very different.

So in this model of a transverse wave, the displacement represents how far the wave oscillates from the point of origin (the correct wording in waves is equilibrium, not origin). How as you know, transverse waves oscillate perpendicular to the direction of travel. To get an important grasp of transverse waves, you need to know that waves transfer energy only. There is no movement of matter. Take my example here:

I have a string and I have a ball threaded through the string as it is secured in place so it doesn't move so it cannot slide through the string, it is fixed. You hold the other end of the string. Now if I move my hand vertically upwards and downwards, what you will see is the ball only oscillating (moving) upwards and downards; it will not change its position on the string. However, you will see the direction of energy transfer (it's motion) to be be forward or perpendicular to the motion of the ball if that makes sense? You will see the peaks and troughs of the waves to appear moving forward. When really, the energy is being transferred to make the particles in the string move upwards and downwards.

The amplitude of this wave is the maximum displacement of the ball (so when I move the string vertically, there will be a point where the ball cannot displace itself any higher - this is the peak) from rest position (when me and you are holding the string).

Displacement is an important concept to understand in physics and maths so do not get mixed up with distance.

Btw are you starting in September? It's good you're starting revision now if you are :smile:

Ask me any more questions if I have not made myself clear. I'll be happy to help
Original post by CorpusLuteum
Screen Shot 2017-07-21 at 14.36.06.png

So, you see how there's two axis on this diagram: Horizontal is distance which isn't a vector and vertically displacement (x) which is a vector.

My question is:
Can a wave only have distance in the direction it is travelling because it can only travel in one direction? And is that why displacement is 90 degrees from the direction the wave is travelling because it is oscillating?

Also if that is incorrect, why is the definition of amplitude written as "Maximum magnitude of displacement".

Does displacement always have a fixed definition when referring to waves (as how far a point on the wave has moved from its undisturbed position)?


The set-up is wave packet moving to the right only, so we only note distance travelled. Oscillations within the packet have a vert displacement at each point relative to the equilibrium line. This is a vector e.g) + A or - A.

Amplitude is a scalar so you take the magnitude of either to give you A. This is the vert distance from peak/trough to the equilibrium line. Oscillations displace perpendicular to direction of travel in transverse waves, giving a sinusoidal form.
(edited 6 years ago)
Original post by BTAnonymous
Ok, so the amplitude is the maximum displacement of a wave from equilibrium ...

Just to clear it up, displacement is the distance from the point of origin ...


Displacement is a vector e.g) + A or - A. Amplitude is a scalar so it's only positive, so you take the magnitude in either case (A). It represents vert distance from peak/trough to the equilibrium line.

The example set-up has a wave packet move 1-way (right) only, so you only need to care about distance travelled. If it reflects back they may introduce horiz displacement too.
(edited 6 years ago)
Original post by CorpusLuteum
My question is:
Can a wave only have distance in the direction it is travelling because it can only travel in one direction? And is that why displacement is 90 degrees from the direction the wave is travelling because it is oscillating?
The answers to your questions will probably over-complicate the subject for you at this stage, so I wouldn't worry too much about it if they don't immediately make sense.

Waves can be reflected, at which point you no longer see the individual components, but the net effect, e.g. standing waves. Think about what happens to ripples in a pond when they hit the exterior.

Displacement for a wave is not always at 90 degrees to the direction of travel, although that's common. Sound waves have the oscillation along the direction of travel (compression / extension of the air).
Original post by Physics Enemy
Displacement is a vector e.g) +5 or - 5. Amplitude is a scalar so it's only positive, so you take the magnitude in either case (5). It represents the vertical distance from peak/trough to the equilibrium line.

The example was set-up to have a wave packet move 1-way (right) only, so you only need to care about distance travelled. If it reflected back, they should introduce horizontal displacement too.


Oh right ok, I think I have misunderstood and interpreted the diagram a lot.
@Physics Enemy


@RogerOxon


@BTAnonymous



Thanks so much guys this clears everything up! :smile:
Original post by CorpusLuteum
@Physics Enemy


@RogerOxon


@BTAnonymous



Thanks so much guys this clears everything up! :smile:


Np.

I thank the others for correcting me as well.

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