Simple Harmonic Motion solution problem

[Matuku]

I've been going through and proving to myself that the solution to the SHM equation is correct; at A level we were just told what it was but never really shown why.

I've gone through the problem as a standard 2nd Order ODE (with complex roots of course) and ended up at:

[tex] x = A_{real} cos(\omega t) - A_{im} sin(\omega t) [/tex]

But the form I want (as the one were were given) is [tex] Bsin(\omega t + \phi) [/tex]

http://mathworld.wolfram.com/SimpleHarmonicMotion.html seems to imply that you can get:
[tex] B cos(\phi) = A_{real}[/tex]
[tex] B sin(\phi) = A_{im} [/tex]

Which then allows use of a trigonometric identity to get it to cos (still not quite what I want but you can just put an extra pi/2 into the phi and make it sin). Is this always the case? Can you always find values of B and [tex]\phi[/tex] which will allow this?

 

[HallsofIvy]

I've been going through and proving to myself that the solution to the SHM equation is correct; at A level we were just told what it was but never really shown why.

I've gone through the problem as a standard 2nd Order ODE (with complex roots of course) and ended up at:

[tex] x = A_{real} cos(\omega t) - A_{im} sin(\omega t) [/tex]

But the form I want (as the one were were given) is [tex] Bsin(\omega t + \phi) [/tex]

http://mathworld.wolfram.com/SimpleHarmonicMotion.html seems to imply that you can get:
[tex] B cos(\phi) = A_{real}[/tex]
[tex] B sin(\phi) = A_{im} [/tex]

Which then allows use of a trigonometric identity to get it to cos (still not quite what I want but you can just put an extra pi/2 into the phi and make it sin). Is this always the case? Can you always find values of B and [tex]\phi[/tex] which will allow this?

[itex]sin(x+ y)= sin(x)cos(y)+ cos(x)sin(y) is the trigonometric identity you refer to. Of course, we must have [itex]sin^2(x)+ cos^2(x)= 1[/itex] but what you can do is this:

Given [itex]A_{real} cos(\omega t)+ A_{imag}sin(\omega t)[/itex], if [itex]\sqrt{A^2_{real}+ A^2_{imag}}\ne 1[/itex], factor that out:
[itex]\sqrt{A^2_{real}+ B^2_{imag}}\left(\frac{A_{real}}{\sqrt{A^2_{real}+ B^2_{imag}}}cos(\omega t)+ \frac{A_{imag}}{\sqrt{A^2_{real}+ B^2_{imag}}}sin(\omega t)\right)[/itex].

And, now, because [itex]\left(\frac{A_{real}}{\sqrt{A^2_{real}+ B^2_{imag}}}\right)^2+ \frac{A_{imag}}{\frac{A_{real}}{\sqrt{A^2_{real}+ B^2_{imag}}}}= 1[/itex], we can find [itex]\phi[/itex] such that [itex]cos(\phi)= \frac{A_{real}}{\frac{A_{real}}{\sqrt{A^2_{real}+ B^2_{imag}}}}[/itex] and [itex]sin(\phi)= \frac{A_{imag}}{\frac{A_{real}}{\sqrt{A^2_{real}+ B^2_{imag}}}}[/itex] and thus,
[tex]A_{real}cos(\omega t)+ A_{imag}sin(\omega t)=\sqrt{A^2_{real}+ A^2_{imag}}\left( sin(\phi)cos(\omega t)+ cos(\phi)sin(\omega t)\right)= \sqrt{A^2_{real}+ A^2_{imag}}sin(\omega t+ \phi)[/itex]

 

[Matuku]

Ah, that brings back memories from A level actually; let me see if I've got this right.

Let,
[tex]\]
Acos(\theta) + Bsin(\theta) = Rsin(\theta + \phi)\\
\therefore Acos(\theta) + Bsin(\theta)=Rcos(\theta)sin(\phi) + Rsin(\theta)cos(\phi)\\
\therefore A = Rsin(\phi), B= Rcos(\phi)\\
\therefore R = sqrt{A^2 + B^2}, \phi=arctan(A/B)\\
\[[/tex]