1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Interpreting Simple Harmonic Motion Graphs

  1. Oct 14, 2014 #1
    1. The problem statement, all variables and given/known data
    Explain the shape of the velocity-displacement and acceleration-displacement graphs for an object undergoing simple harmonic motion. The graph is attached to this thread

    2. Relevant equations
    v = wsqrt(A^2-x^2) where w = angular frequency, A = amplitude and x = displacement

    a = - w^2x

    3. The attempt at a solution
    I get the acceleration-displacement graph. The acceleration is proportional to the displacement which means you will get a straight line (with a negative slope because of the minus sign). I can't figure out why the velocity-displacement graph is circular?
     

    Attached Files:

  2. jcsd
  3. Oct 14, 2014 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    What is the velocity at the points of maximal displacement? What is the displacement at the point of maximal velocity (positive and negative)? How do they change in between?
     
  4. Oct 14, 2014 #3
    Velocity at maximum displacement is zero and at minimum displacement is maximum. I see that on the graph but why the curved shape between these two points and why does the whole thing trace out a circle?
     
  5. Oct 14, 2014 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    If you know the formula for displacement over time, you can calculate its first derivative (or look it up) to get the velocity.
    If you square both and add them, you get an ellipse - or a circle as special case if you choose the axes in the right way.

    This is not by accident, it is related to energy conservation: both velocity squared and distance squared correspond to components of the total energy, so points of equal distance to the origin correspond to equal energy.
     
  6. Oct 14, 2014 #5

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Do you have an expression for x as a function of t ?
    What does that yield when substituted in v = wsqrt(A^2-x^2) ?

    Personally, I like the animation in this Wikipedia lemma.
     
  7. Oct 14, 2014 #6
    So, x = A cos(wt) so you would get v = wsqrt(A^2-A^2cos^2w^2t^2). How can I use this to explain the shape of the velocity curves?
     
  8. Oct 14, 2014 #7
    I have been searching online for some answers and may have found one. Is the equation given the equation for a half circle and since the velocity can have two values (positive and negative) you get a full circle, is that right? See example 4 in this link (http://www.analyzemath.com/Graphing/graphing_square_root_func.html) or have I got it completely wrong. We haven't done any circle maths in class so I didn't think it would be this complicated....
     
  9. Oct 15, 2014 #8
    Does anyone have any answers please?
     
  10. Oct 15, 2014 #9

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I see, no circle math. But then I would like to know where your v = wsqrt(A^2-x^2) came from (because that seems to be the hangup now). It is also wrong, so we need to clarify and fix that.

    v has one single value at any time, just like x and a. But its value range is from -##\omega##A to +##\omega##A.
     
  11. Oct 15, 2014 #10

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes. So I have to correct my silent approval o:) of
    Because that expression yields only non-negative v, and it is clear that the motion involves negative v half the time!

    There is the relationship ##v^2 = \omega^2 (A^2 - x^2)## which is true at all times and does help to draw the graph. It just doesn't help finding v at a given x, because for every x there are two v that satisfy the relationship: same |v| but opposite sign.

    Again, I really like the wikipedia picture (which is your picture turned 1/4 turn).

    Simple_Harmonic_Motion_Orbit.gif
     
  12. Oct 15, 2014 #11
    This
    This link has the equation in it before it has been square rooted (http://www.mathsrevision.net/advanced-level-maths-revision/mechanics/simple-harmonic-motion). If you look under further equations it is listed there.
     
  13. Oct 15, 2014 #12
    Sorry only just seen your second reply. I still don't understand the shape of the graph though? How does that equation give rise to the graph I originally attached?
     
  14. Oct 15, 2014 #13
    Please could someone kindly explain this graph, I still don't get it! The equation and animation posted by BvU shows a circular shape. What does that equation say is the relationship between displacement and velocity? Is it that the velocity is proportional to the square root of some function of x squared? If you plot any graph using an equation you should be able to predict the outcome using the equation and I don't see how that equation should give a circle/curve?

    v2=ω2(A2−x2) - this is the equation I mean which gives rise to this circular shaped graph if you plot velocity against dispalcement
     
  15. Oct 15, 2014 #14

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Yes, but that description is very broad. "Some function" is a specific function.
    If you did not have functions for circles yet, this is hard to understand. Did you calculate distances on a plane before? Or the length of a side in a rectangular triangle? Those formulas are useful here.
     
  16. Oct 15, 2014 #15
    It's not a circular shape but an ellipse. So you need to know the equation of the ellipse in Cartesian coordinates. Then you can show that V and x satisfy the equation of the ellipse in the phase space. It's quite simple.
     
  17. Oct 15, 2014 #16
    Thanks. We haven't done anything on functions of circles. So does that equation describe the equation of a circle and hence why the graph is circular? I have attached a small extract from my textbook. At the end of the first main paragraph it says that this shape for the velocity displacement graph is suprising and gives some vague guidance on how to show why the shape of the graph is like that. Does the guidance it give (dealing with cos^2theta and sine^2theta) make any sense to anyone?
     

    Attached Files:

    • SHM.jpg
      SHM.jpg
      File size:
      53.5 KB
      Views:
      173
  18. Oct 15, 2014 #17
    Is it simple for a pre-college physics student? Could you point me in the direction of source that links this to the equation of an ellipse? I always thought SHM is all to do with circular motion and didn't involve ellipses?
     
  19. Oct 15, 2014 #18
    Sorry this should be clearer, think the last attachment was too small to see
     

    Attached Files:

  20. Oct 15, 2014 #19
    Well, how can I guess your level? It does not seem you mentioned it.:)
    Here you can find the equation of an ellipse.
    http://en.wikipedia.org/wiki/Ellipse (see under "Equations")
    The one you need here is
    (x/a)^+(y/b)^2=1
    Here a and b are the semi-axes of the ellipse.

    You show an elliptical shape in the problem so if you want to solve it you have to deal with ellipses, don't you? Or give up until you learn about ellipses. You cannot pretend is a circle just because you did not cover this yet. :)

    Hint. In your case, try to calculate (x/A)^2+(v/wA)^2. Which is what they suggest at the end of the paragraph in your picture.
     
  21. Oct 15, 2014 #20
    Thanks I will try and digest all of this. So when you get taught that simple harmonic is modeled using circular motion this isn't necessarily true because if you are dealing with velocity and displacement you are technically dealing with ellipses? We were shown a ping pong ball going around in a circle which casts a 1 dimensional shadow which undergoes simple harmonic motion and we applied the equations of circular math to SHM.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted