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## Main Question or Discussion Point

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?

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?