Simple harmonic motion displacement equation confusion

  • #1
Member warned to use the homework template for posts in the homework sections of PF.
Okay, so I have just started with simple harmonic motion(SHM).
So the equation of displacement in my text book 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 dont 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 cant 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 dont 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!!!!!!!
 

Answers and Replies

  • #2
PeroK
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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.
 
  • #3
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 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.
Then my answer will vary as if I use the sin formula my answer is gonna 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??
 
  • #4
PeroK
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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 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.
Then my answer will vary as if I use the sin formula my answer is gonna 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.
 
  • #5
hows that possible, in one case it is having a displacement of 1.8m and in other case it is 0.84m?
 
  • #6
haruspex
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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.
 
  • #7
PeroK
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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.
 
  • #8
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
 
  • #9
thnx everyone
 
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