X = Acos(ωt) + Bsin(ωt) derivation

[sparkle123]

How do you derive x = Acos(ωt) + Bsin(ωt) from F = -mω²x and what is the former used for?

Thank you!

 

[I like Serena]

Hey sparkle!

If this is homework, you should show some effort before we're allowed to help you (PF regulations I'm afraid).
What's it for?

 

[sparkle123]

This was on a list of things you should know for physics contests. :)

 

[I like Serena]

Well, what can you make of it?

 

[sparkle123]

Well, F = -mω2x is hooke's law
and SHM for a spring is like x = Asin(ωt)

 

[I like Serena]

So what's your question?

Actually, F=ma and "a" is the second derivative of "x" with respect to time.
So you have mx''=-mω²x.
The general solution to this differential equation is x = Acos(ωt) + Bsin(ωt).

 

[sparkle123]

so would you get from Acos(ωt) + Bsin(ωt) to Asin(ωt)?
thanks!

 

[I like Serena]

so would you get from Acos(ωt) + Bsin(ωt) to Asin(ωt)?
thanks!

The relation is:
$$A\cos(ωt) + B\sin(ωt) = \sqrt{A^2+B^2}\sin(ωt+θ_0)$$

 

[sparkle123]

thank you! :)