Optics, Superposition of waves

[heycoa]

Homework Statement

Using expansion of sin and cos functions, show that the resultant of adding the following two waves:
a) E₁ = E₀₁*sin(wt-k(x+Δx))
b) E₂ = E₀₁*sin(wt-kx)

Gives: E = 2E₀₁*cos((1/2)kΔx)*sin[wt-k(x+Δx/2)]

Homework Equations

N/A

The Attempt at a Solution

I don't know how to sum the two waves using superposition. I don't understand how to get to the final solution. Please help.

 

[TSny]

To sum the waves using superposition, you just add the mathematical expressions for E1 and E2. See if one of these trig identities will help: http://en.wikipedia.org/wiki/List_o...#Product-to-sum_and_sum-to-product_identities

 

[heycoa]

yes that worked! thank you very much

 

[TSny]

Good work!

 

[Nickie]

I can understand your confusion and will try my best to explain the solution in a clear and concise manner.

First, let's define some variables for the two waves:

E1 = E01*sin(wt-k(x+Δx))
E2 = E02*sin(wt-kx)

Now, we can use the trigonometric identity for the sum of two sine functions:

sin(a+b) = sin(a)cos(b) + cos(a)sin(b)

Applying this identity to the two waves, we get:

E = E01*sin(wt-k(x+Δx)) + E02*sin(wt-kx)
= E01*sin(wt)*cos(k(x+Δx)) + E01*cos(wt)*sin(k(x+Δx)) + E02*sin(wt)*cos(kx) + E02*cos(wt)*sin(kx)

Next, we can use the double angle identity for sine:

sin(2x) = 2sin(x)cos(x)

Applying this to the above equation, we get:

E = E01*sin(wt)*[2*cos((1/2)k(x+Δx))*sin((1/2)k(x+Δx))] + E02*sin(wt)*[2*cos((1/2)kx)*sin((1/2)kx)]

Now, we can see that the terms within the square brackets are just the expansion of the sine and cosine functions. So, we can rewrite the equation as:

E = 2E01*cos((1/2)k(x+Δx))*sin[wt-k(x+Δx/2)] + 2E02*cos((1/2)kx)*sin[wt-kx/2]

Finally, we can combine the two terms with the same sine function and get:

E = 2E01*cos((1/2)kΔx)*sin[wt-k(x+Δx/2)]

Which is the same as the given solution.

I hope this explanation helps you to understand how superposition is used to sum two waves. If you have any further questions, please do not hesitate to ask. Keep up the good work!