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Simple Harmonic Motion problem

  • Thread starter Behroz
  • Start date
  • #1
5
0
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
 

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Answers and Replies

  • #2
1,860
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Spring mass systems often use omega to represent sqrt(k/m). It isn't a trigonometric law, though if your text book 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.
 
  • #3
5
0
Spring mass systems often use omega to represent sqrt(k/m). It isn't a trigonometric law, though if your text book 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?!?!?!?
 
  • #4
1,860
0
That is for you to find out. :p

Try working backwards, it might be a little easier.
 

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