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Some questions about S=S0*cos(ωf-kz+φ)

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Member level 3
Nov 26, 2006
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Hi! I have some problem with S=S0*cos(ωf-kz+φ)

1.why "wave" can express as S=S0*cos(ωf-kz+φ)?

2.If I can take"sin"instead of cos as S=S0*sin(ωf-kz+φ)?

3.What is "ωf" means in the expression?

4.Whas is "-kz" and why not "+kz"?

5.what is "φ" and why not "-φ"?


1. A wave is something that moves with time in space and this can be described with any function that satisfies f(z-vt) (shifted to the right by vt) or f(z+vt) (shifted to the left). So for example the function sin(z - vt) will propagate to the right with speed v as time t elapses. To see this plot it for some times t (choose v = 1 for example). Why did we take harmonic functions like cos(z) and sin(z) or exp(j*z)? This is because any other function can be built up from a superposition of harmonic functions. Now the most general cosine function with a linear argument would be cos[k(z-vt)+φ] and this still satifies our critera f(z-vt). Multiplying by a constant k and adding a phase constant φ makes this function general. It then remains to to understand what k is and we see that when z advances by 2*pi/k then this wave have moved one cycle (2*pi). But one cycle is one wavelength λ so we have the relation k = 2*pi/λ. We also know that f = v/λ or in terms of angular frequency ω = 2*pi*f instead (it is more convenient for harmonic functions) we get the relation ω = kv. So finally we have cos(kz - ωt + φ).

2. Yes, this still satisfies f(z-vt). We most often use the cosine function because it is the real part of exp[i(kz -ωt)]. Sine would be the imaginary part and for some reason the real part is "nicer".

3. I think you mean ω and t. ω is the angular frequency and t is time.

4. Changing to cos(kz+ωt) changes the direction of propagation from +z to -z.

5. φ this is only a constant parameter, called the phase constant. It sets the starting point of the function at time t = 0. φ can be +23 or -1.02 or whatever so the sign doesn't really matter but plus signes are nicer than minuses.

You'll find more on this in any introductory text on physics.

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