What does the phase angle phi mean in the harmonic oscillation function?

AI Thread Summary
The phase angle phi in the harmonic oscillation function Acos(ωT + φ) determines the initial position of the oscillating object. While omega (ω) indicates the oscillation frequency, phi shifts the graph horizontally, affecting the starting point of the oscillation. The discussion clarifies that changing phi does not convert the function from sine to cosine but rather adjusts the phase of the cosine wave. The relationship between sine and cosine is highlighted, noting that a phase shift of π/2 converts cosine to sine. Understanding phi is crucial for accurately representing the initial conditions of simple harmonic motion.
CrazyNeutrino
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The function for simple harmonic oscillation is:
Acos(ωT)+\phi
Why is there an angle phi added to the function acos(ωT)?
 
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it's Acos(ωT+ϕ) omega stands for how fast it' s oscilating, but phi determines the initial position( wold be Acosϕ)
 
I was reading a book on wave and found that when they derive the equation of shm from the equation force varies with negetive displacement , they had taken a propotionality constant to make the force and displacement equal and they had taken frequency of the shm as the constant . So my question is , is there any derivation which can show that the constant is the frequency of the shm .
 
If phi had a value, what would the shm graph look like? Or how would it change from Acos(wT)?
 
CrazyNeutrino said:
If phi had a value, what would the shm graph look like? Or how would it change from Acos(wT)?
The graph would change from sine to cosine, (remember, that sin(x+π)=cosx)
 
isnt the function already Acos(wT)+phi?
 
so how could it change from sine if it doesn't start at sine?
 
CrazyNeutrino said:
so how could it change from sine if it doesn't start at sine?
It never started as cosine, it's allways either sine or cosine or somethin in between. It's like you move the cosine a bit to the right and if you move it by π/2 you get sine! SO all phi determines is the initial position!
 
Ok... Thanks! I almost understand.
 
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