In Physics, the Simple Harmonic Oscillator is represented by the equation d2x/dt2=−ω2x. By using the characteristic polynomial, you get solutions of the form x(t)=Aeiωt+Be−iωt, I get that you Euler's formula eiθ=cos⁡θ+isin⁡θ, but I can't seem to find my way all the way to the ''traditional form'' of Dcos⁡ωt+Csin⁡ωt.

Name: In Physics, the Simple Harmonic Oscillator is represented by the equation d2x/dt2=−ω2x. By using the characteristic polynomial, you get solutions of the form x(t)=Aeiωt+Be−iωt, I get that you Euler&#039;s formula eiθ=cos⁡θ+isin⁡θ, but I can&#039;t seem to find my way all the way to the &#039;&#039;traditional form&#039;&#039; of Dcos⁡ωt+Csin⁡ωt.
Uploaded: 2022-02-23T17:22:57+00:00
Description: In Physics, the Simple Harmonic Oscillator is represented by the equation d2x/dt2=−ω2x. By using the characteristic polynomial, you get solutions of the form x(t)=Aeiωt+Be−iωt, I get that you Euler&#039;s formula eiθ=cos⁡θ+isin⁡θ, but I can&#039;t seem to find my way all the way to the &#039;&#039;traditional form&#039;&#039; of Dcos⁡ωt+Csin⁡ωt.

Question

In Physics, the Simple Harmonic Oscillator is represented by the equation d2x/dt2=−ω2x. By using the characteristic polynomial, you get solutions of the form x(t)=Aeiωt+Be−iωt, I get that you Euler&#039;s formula eiθ=cos⁡θ+isin⁡θ, but I can&#039;t seem to find my way all the way to the &#039;&#039;traditional form&#039;&#039; of Dcos⁡ωt+Csin⁡ωt.

kalupunangh · Accepted Answer

However, if you assume the function x(t) is real, then they are related as A―=B. This can be seen since x(t)―=x(t) which gives us A―=B (since eiωt and e−iωt are orthonormal) Rewriting the complex exponentials as sine and cosine we get x(t)=A(cos⁡(ωt)+isin⁡(ωt))+B(cos⁡(ωt)−isin⁡(ωt))=(A+B)cos⁡(ωt)+i(A−B)sin⁡(ωt) Now as proved before A―=B and hence (A+B) is real and (A−B) is imaginary and hence we can write (A+B)=C and i(A−B)=D where C, D are real number. And the solution becomes x(t)=Ccos⁡(ωt)+Dsin⁡(ωt) where C,D∈R

Robert Pina · Accepted Answer

Remember that x(t)=Aeiωt+Be−iωt+Be−iωt is a solution for any values of A and B which means that choosing A=B=12 result in the solution x1(t)=eiωt+e−iωt Euler&#039;s fomula, along with the properties cos⁡(−ωt)=cos⁡(ωt) and sin⁡(−ωt)=−sin⁡(ωt), simplifies this solution to x1=cos⁡(ωt) Similarly, choosing A=1/(2i) and B=−1/(2i) gives the solution x2(t)=sin⁡(ωt) Finally, by linearity you can conclude that the linear combination Cx1+Dx2 is also a solution.

alenahelenash · Accepted Answer

the answers given are absolutely fine as they show the way from e±iωt to sin and cos. It might be even shorter if you use that sin⁡(x)=12i(eix−eix) and cos⁡(x)=12(eix+e−ix) This can be interpreted as a definition for the functions or seen from its series definitions.

In Physics, the Simple Harmonic Oscillator is represented by the equation d^{2}x/dt^{2}=-\omage^{

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