Optics: Finding the wave equation given position and amplitude information

AI Thread Summary
To find the wave equation for a harmonic wave traveling in the +x-direction, the initial displacements at t = 0 are given as 13 units at x = 0 and -7.5 units at x = 3λ/4. The wave equation is expressed as r = asin(kx - vt + θ), where parameters include amplitude (a), wave number (k), and initial phase angle (θ). By setting up equations for both positions, the relationship between amplitude and phase angle can be derived, leading to the conclusion that the amplitude is 15. The final wave equation, confirmed by the book, is 15sin(kx + π/3).
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A harmonic wave traveling in +x-direction has, at t = 0, a displacement of 13 units at x = 0 and a displacement of -7.5 units at x = 3λ/4. Write the equation for the wave at t = 0.



Homework Equations



The equation for a harmonic wave is

r = asin(kx-vt+θ)

a being the amplitude
k being the wave number k=2π/λ
v being the velocity of the wave
θ being the initial phase angle

The Attempt at a Solution



I set up the wave equations at both positions because we have two unknowns so we need two equations

13 = asin(θ) & -7.5= asin((2π/λ)(3λ/4) + θ)
13/sin(θ ) = a -7.5=asin(3π/2 + θ)

Now I plugged in 13/sin(θ ) for a in the other equation and I ended up with

-7.5=(13/sin(θ ))sin(3π/2 + θ)
-7.5sin(θ ) = 13sin(3π/2 + θ)

This is where I got stuck. Am I on the right track? I imagine there is a trig identity that will help me solve for θ and then I can easily solve for the amplitude. The answer according to the book is:

15sin(kx+π/3)
 
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I solved it, but thanks to those who may have read it. For those that are curious you use the trig identity sin(u+v) = sin(u)cos(v) + sin(v)cos(u) and from there it is simple algebra
 
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