How Does Equation (1) Translate to Equation (2) in Simple Harmonic Motion?

[Behroz]

I'm supposed to derive x as a function of time for a simple
harmonic oscillator (ie, a spring). According to my textbook
this is done by using Newton's second law and hooke's law
as this: ma=-kx and one gets a differential equation in
the second order. I can follow the calculations until this
happens: (see attached picture)

(where omega is the frequency)

I do get the equation (1) when I solve the differential
equation myself but I don't understand how equation (1) translates
to (2)?
I assume this must be done by using some trigonometric law?
if so then which one and how??
Thanks

 

[Mindscrape]

Spring mass systems often use omega to represent sqrt(k/m). It isn't a trigonometric law, though if your textbook eventually (I can't see the picture so I don't know) represents the motion as x(t) = Acos(wt + ø) then you will need to use trig.

 

[Behroz]

Spring mass systems often use omega to represent sqrt(k/m). It isn't a trigonometric law, though if your textbook eventually (I can't see the picture so I don't know) represents the motion as x(t) = Acos(wt + ø) then you will need to use trig.

That's right.. but exactly which trig law do I use and how do I use it to go from equation (1) above in the attached picture to x(t) = Asin(wt + ø).

Or in other words HOW do I go FROM x(t)=x0cos(wt)+(v0/w)sin(wt) ---- (w being = sqrt(k/m) TO x(t) = Asin(wt + ø)
how? HOW? HOW?!??!? HOW?!?

 

[Mindscrape]

That is for you to find out. :p

Try working backwards, it might be a little easier.