Simple harmonic motion displacement equation confusion

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SUMMARY

The discussion centers on the confusion surrounding the displacement equations for simple harmonic motion (SHM), specifically the use of the equations X = A*cos(wt + φ) and X = A*sin(wt + φ). Both equations represent the same physical phenomenon but differ based on the phase constant, φ, which is determined by the initial conditions of the motion. The choice between sine and cosine depends on the initial displacement of the particle at time t=0, which sets the phase of the motion. Understanding this relationship clarifies that both equations yield equivalent results when the phase is appropriately accounted for.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM) principles
  • Familiarity with trigonometric functions, specifically sine and cosine
  • Knowledge of angular frequency (ω) and amplitude (A)
  • Ability to interpret phase constants in wave equations
NEXT STEPS
  • Study the relationship between phase constants and initial conditions in SHM
  • Learn how to graph sine and cosine functions to visualize their equivalence
  • Explore the implications of different phase shifts in SHM equations
  • Practice solving SHM problems using both sine and cosine equations to reinforce understanding
USEFUL FOR

Students of physics, educators teaching simple harmonic motion, and anyone seeking to deepen their understanding of wave mechanics and trigonometric applications in physics.

randomgamernerd
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Okay, so I have just started with simple harmonic motion(SHM).
So the equation of displacement in my textbook is given as:
X= ACos(wt +x)
where A is the amplitude
X is displacememt from mean position at time t
w is angular frequency
x is phase constant.
Everything going good till now, then comes an illustration where suddenly the equation becomes:
X= ASin(wt +x)
where A is the amplitude
X is displacememt from mean position at time t
w is angular frequency
x is phase constant.
and then onwards in most of the sums they solved with the sin equation and not cos.
I don't get which equation to use when, how and why?
I know every Sin function can be expressed in terms of cos and vice versa, also every SHM is either in the form of Sin or Cos.
But I can't figure out which one to use and why and when.
One observation i made- usually in case of linear SHM about X axis, cos equation is used and in case of linear SHM about y axis, sin equation is used. I used the word usually beacuse that was not the case everytime. Also i don't think this can be a logic as its upto me which one will i take as X axis and which one y if its just a linear SHM.
Someone please help!
 
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randomgamernerd said:
Someone please help!

If you look at the graphs of ##\sin## and ##\cos## you'll see they are the same function displaced by ##\pi/2##. In other words:

##\cos(x) = \sin(x + \pi/2)##

Hence:

##A\cos(wt + \phi) = A\sin(wt + \phi + \pi/2)##

Therefore, any solution to any equation that can be expressed as a cosine can also be expressed as a sine and vice versa.
 
yeah, i know this fact as I have already mentioned above, but what I mean is which formula to use when we just need to plug in the given values into the formula.
I mean let's suppose we want to find the displacement of a particle undergoing SHM after 2s and its given amplitude is 2m, w= 5rad/s and x= 3 rad.
Then my answer will vary as if I use the sin formula my answer is going to be 2*sin13 which is equal to 0.84m
but if i use the cos equation my answer will be 2*cos13 which is 1.8m
so which one is the correct answer then?and why??
 
randomgamernerd said:
yeah, i know this fact as I have already mentioned above, but what I mean is which formula to use when we just need to plug in the given values into the formula.
I mean let's suppose we want to find the displacement of a particle undergoing SHM after 2s and its given amplitude is 2m, w= 5rad/s and x= 3 rad.
Then my answer will vary as if I use the sin formula my answer is going to be 2*sin13 which is equal to 0.84m
but if i use the cos equation my answer will be 2*cos13 which is 1.8m
so which one is the correct answer then?and why??

If you have a solution of ##2 \cos(13)## then if you look for a sine solution you would get ##2 \sin(13 + 90)##.

You will get the same physical answer whether you look for a sine solution or a cosine solution.
 
hows that possible, in one case it is having a displacement of 1.8m and in other case it is 0.84m?
 
randomgamernerd said:
lets suppose we want to find the displacement of a particle undergoing SHM after 2s and its given amplitude is 2m, w= 5rad/s and x= 3 rad.
That is not enough information. You need to know the displacement at time 0. That sets the phase. If you use sin you will calculate one phase, if cos you will calculate a different phase. The resulting two equations will be completely equivalent.
 
randomgamernerd said:
hows that possible, in one case it is having a displacement of 1.8m and in other case it is 0.84m?

##\cos(13) = \sin(13 + 90) = 0.97##

You don't seem to understand that:

##A\cos(wt + \phi) = A\sin(wt + \phi + \pi/2)##

These are precisely the same function; they have precisely the same graph; and, they are precisely the same solution to a SHM problem.
 
haruspex said:
That is not enough information. You need to know the displacement at time 0. That sets the phase. If you use sin you will calculate one phase, if cos you will calculate a different phase. The resulting two equations will be completely equivalent.

ok...now i got it
 
thnx everyone
 

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