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Why is Sin the convention for the harmonic oscillator? 
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#1
Apr1214, 09:10 PM

P: 504

In the course of solving the simple harmonic oscillator, one reaches a fork in the road.
x(t) = A1Sin(wt) + A2Cos(wt) At this point, you exploit a trig identity and arrive at one of two solutions x(t) = B1Sin(wt+phi1) or x(t) = B2Cos(wt+phi2) Both of these are correct solutions and either one can be used to suit the particular problem. However, convention usually has us using Sin instead of Cos. Is there any particular reason for this? Is it to exploit the small angle / taylor series approximation? Sin(x) ≈ x  (x^3)/6 Cos(x) ≈ 1  (x^2)/2 


#2
Apr1214, 11:36 PM

P: 1,969

There is no convention about using sin rather than cosine.
The initial phase can be adjusted to use whichever of the two functions you like. 


#3
Apr1214, 11:57 PM

Mentor
P: 16,982

I thought the convention was to use cos since the cos terms are the real ones in the Fourier transform.
I guess that just goes to show that the convention isn't very strong at all. 


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