Why is the term w(t-x/c) used in the cosine representation of a traveling wave?

[mcheung4]

This is about wave reflection and transmission.

For an infinite string with a density change at x=0, consider an incident wave propagating to the right from x = -∞. The most general form is W = A cos(w(t-x/c)+θ), with amplitude A, angular freuqency w, time t, distance x (from origin), wave speed c and phase θ.

I do not understand the term w(t-x/c). what is x/c negative?


Update : http://www.animations.physics.unsw.edu.au/jw/travelling_sine_wave.htm

I understand this derivation apart from x' = x-vt. Shouldnt it be x' = x-vt since x' is moving relative to x?

 

[Simon Bridge]

c is the speed of the wave.

If you distort a string so that the height forms a wave of arbitrary shape y(x,t=0)=f(x), and the wave subsequently travels to the right (+x direction) without changing shape, then for t>0, y(x,t)=f(x-ct), where c is the speed of the wave.

If $f(x)=A\cos( kx + \theta)$

What is $y(x,t)=$?

Ignore the derivation you linked to for now: it is unnecessarily convoluted.

 

[dauto]

The question is not clear to me. Are you trying to understand why there is a negative sign in the factor w(t-x/c)? Or are you trying to understand the physical meaning of those variables?

 

[Simon Bridge]

or why it is the "x/c" that is negative as opposed to the "t"
i.e. why w(t-x/c) and not w[(x/c)-t] ... which is what you get off the standard derivation.

... or myriad other possibilities - which is why I wanted to see the derivation done first.

 

[sophiecentaur]

I do not understand the term w(t-x/c). what is x/c negative?

This is confusing at first. However, if you think that, as you get further away from the source (increasing x) the phase of the wave is 'earlier' (because the 'later' bit hasn't got there yet). So increasing x has to decrease the phase, if you are using the conventional signs for everything else.

 

[Simon Bridge]

When you do the derivation for the traveling wave, though, the sine ends up on the time rather than the space contribution to the overall phase. A waveform f(x) traveling to the right with speed c changes as f(x-ct) ... see what I mean?

But I suppose of x is the observer position, the the observer is looking at an earlier part of the wave.
... I just don't think that's the context. Really need feedback.
i.e. the second question makes no sense.