# Homework Help: EM wave described using a sine function.

1. Jan 23, 2014

### Darren93

I'm going through my notes and I don't understand how they have included position in an equation to describe an EM wave. The equation is of the form http://upload.wikimedia.org/math/f/6/3/f6386c1751b91ec23c7123b15a11b52f.png [Asin(kx-ωt)]. This equation is just stated in my notes and there is no description of where this came from. Surely if you input both position and time in this manor they both cancel each other out and you always get sin(0). Like at x=λ/4, k=2π/λ, ω=2π/T, t=T/4. Then kx-ωt=π/2-π/2 = 0. It's the same for every value of t and x, always 0? I mean I could understand using either Asin(kx) or Asin(ωt) but does both in the same equation not cancel each other out? What am I missing?

2. Jan 23, 2014

### bp_psy

Why would kx-ωt be always 0? In this case x would be the position where you measure the field, which can be any point and t is the time when you measure it, which can be any time.You will of course have some points where the amplitude is 0 at any given time but not all.

3. Jan 23, 2014

### Darren93

Well all the values I use I end up getting kx-ωt=0. That's the problem it shouldn't always be zero. It should be zero for x=0,λ,2λ and ect. However x=λ/4 should give kx-ωt= π/2. When I tried calculating this for x=λ/4 I got:

x=λ/4, k=2π/λ, ω=2π/T, t=T/4
kx-ωt=(2π/λ)*(λ/4)-(2π/T)*(T/4)
kx-ωt=(π/2)-(π/2)
kx-ωt=0

So I'm doing something wrong but what? I suspect its the values of k,ω or t I'm using but I really don't know.

4. Jan 23, 2014

### bp_psy

Then you are using very few values.

x=λ/4, k=2π/λ, ω=2π/T, t=T/4
kx-ωt=(2π/λ)*(λ/4)-(2π/T)*(T/4)
kx-ωt=(π/2)-(π/2)
kx-ωt=0

but after T/4
x=λ/4, k=2π/λ, ω=2π/T, t=T/2

kx-ωt=(2π/λ)*(λ/4)-(2π/T)*(T/2)
kx-ωt=(π/2)-(π)
kx-ωt=-3n/2

or at another position at the same time

x=λ, k=2π/λ, ω=2π/T, t=T/4
kx-ωt=(2π/λ)*(λ)-(2π/T)*(T/4)
kx-ωt=(2π)-(π/2)
kx-ωt=-3n/2

or any other time δt
x=λ/4, k=2π/λ, ω=2π/T, t=δt
kx-ωt=(2π/λ)*(λ/4)-(2π/T)*δt
kx-ωt=(π/2)-(2πδt/T)

The field at any point x will oscillate between -A,0,A in time. You can always find a time when the field is -A,A at a given point but that does not mean that it is constant in time.

5. Jan 23, 2014