- #1
patric44
- 303
- 39
- Homework Statement
- derive the general formula of the superposition of two waves which are orthogonal ?
- Relevant Equations
- x^2-2xy*cos(δ)+y^2 = A^2 sin(δ)^2
hi guys
i was trying to derive the general formula of two orthogonal waves
$$x^{2}-2xycos(δ)+y^{2} = A^{2} sin(δ)^{2}$$
where the two waves are given by :
$$x = Acos(ωt)$$
$$y = Acos(ωt+δ)$$
where ##δ## is the different in phase , i know it seems trivial but i am stuck on where should i begin ?
my trial :
$$x^2 = A^2 cos(ωt)^2$$
$$y = A[cos(ωt)cos(δ)-sin(ωt)sin(δ)]$$
squaring ##y## :
$$y^2 = A^2[cos(ωt)cos(δ)-sin(ωt)sin(δ)]^2⇒ y^2 = A^2 [cos(ωt)^2cos(δ)^2+sin(ωt)^2sin(δ)^2-2cos(ωt)cos(δ)sin(ωt)sin(δ)] $$
it seems that it got more complicated
where to continue from here , or is there another method to get :
$$x^{2}-2xycos(δ)+y^{2} = A^{2} sin(δ)^{2}$$
i was trying to derive the general formula of two orthogonal waves
$$x^{2}-2xycos(δ)+y^{2} = A^{2} sin(δ)^{2}$$
where the two waves are given by :
$$x = Acos(ωt)$$
$$y = Acos(ωt+δ)$$
where ##δ## is the different in phase , i know it seems trivial but i am stuck on where should i begin ?
my trial :
$$x^2 = A^2 cos(ωt)^2$$
$$y = A[cos(ωt)cos(δ)-sin(ωt)sin(δ)]$$
squaring ##y## :
$$y^2 = A^2[cos(ωt)cos(δ)-sin(ωt)sin(δ)]^2⇒ y^2 = A^2 [cos(ωt)^2cos(δ)^2+sin(ωt)^2sin(δ)^2-2cos(ωt)cos(δ)sin(ωt)sin(δ)] $$
it seems that it got more complicated
![Thinking face :thinking: 🤔](https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f914.png)
$$x^{2}-2xycos(δ)+y^{2} = A^{2} sin(δ)^{2}$$