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Plotting graph for acceleration against displacement

  1. Aug 6, 2017 #1
    A mass of 0.3 kg is suspended from a spring of stiffness 200 N m–1. If the mass is displaced by 10 mm from its equilibrium position and released, for the resulting vibration:

    Plot a graph of acceleration against displacement (x) (for values of x from x = –10 mm to x = +10 mm)

    Hugely struggling to plot and draw this graph. Have been reading about a formula to give values for the acceleration but this formula is giving me numbers which seem way off.

    Formula: -(w^2 times displacement)

    So, -25.81^2 x 0.1= -66.61561???

    Am I on the right track or miles off?? First post and any help is greatly appreciated!
  2. jcsd
  3. Aug 6, 2017 #2


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    First: I'm not sure what formula you are talking about: what = what? And what does w represent?
    Second: I'm not sure where the 25.81 came from?

    To help: What are the sources of acceleration for the mass?
  4. Aug 6, 2017 #3
    My Apologies, this is actually the second part of the question. The first part I have already answered.

    So firstly I calculated the frequency at 25.81 rad s^-1

    Then the maximum velocity (Vmax) at 0.26 m/s^-1

    And the maximum acceleration (a max) at 6.66 m/s^-2

    So the 'w' is the frequency which I know sometime is shown using 'f'

    I read somewhere that that formula could help in plotting points for the graph but it seems a little off to me.

    Struggling to know where to start in plotting and drawing the graph.... :confused:
  5. Aug 6, 2017 #4


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    Ok thanks for clarification.

    To plot graph, you need to know a bit more than accn at one point. What determines the acceleration of the mass ?
  6. Aug 6, 2017 #5
    I don't know to be honest, the question doesn't say.

    My understanding was that there is a way of determining the acceleration at + and -10mm during the displacement and plotting a graph that way.

    People have talked of a formula on different forums but none seem to work
  7. Aug 6, 2017 #6


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    Ok. I'm not trying to suggest a formula, because I think the answer is simple, when you understand it.
    It is a mass on the end of a spring. When it just hangs there at its equilibrium position, it is not accelerating. When you move it away from there and release it from rest, it accelerates and starts moving.

    What does Newton tell us is needed to make a mass accelerate?
  8. Aug 6, 2017 #7
    That it requires a force exerted on it ?
  9. Aug 6, 2017 #8


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    What provides that force?
  10. Aug 6, 2017 #9
    the spring ? but how does that translate into the graph?
  11. Aug 6, 2017 #10


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    There's a force due to gravity (weight) and a force due to stretching of the spring. Since the string is stretched just enough at equilibrium to balance gravity, we can just measure stretching away from equilibrium and ignore the weight and just think about the extra force of the spring caused by displacement from equilibrium.
    Ok. You know the connection between force and acceleration, so if you know how the force varies, you know how acceleration varies.

    How is the force of the spring related to displacement? Remember, "a spring of stiffness 200 N m–1."
  12. Aug 6, 2017 #11
    Because it's under tension, when it's stretched it wants to pull and when compressed it wants to push.

    So is the acceleration not a product of the string stiffness, the weight and gravity?

    Also, this is another of the formulas is was talking about...

    a = -Aw²cos(wt)
    x = Acos(wt),

    a = -w²*x

    So, -frequency^2 times displacement value
  13. Aug 6, 2017 #12


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    Yes. I'm sorry. I see what you're doing now. It's just not how I thought to approach the question, because it diid not originally need the frequecny calculating. (and it seemed to have given you a wrong value.)
    So you have your answer and just need to plot the graph!. a = -w²*x

    My approach did not get into SHM, just spring & mass

    So F = - kx (- because the force always opposes the displacement -)

    This tells you two useful things:
    First you can work out the force for any displacement. (And hence the acceleration for any displacement.)
    Secondly, proportionality tells you what sort of graph it is. (This is the key to the simplicity of the final answer.)
  14. Aug 6, 2017 #13
    Ahh okay, no worries.

    To be honest, in my head, the graph looks a bit like a sin wave, due to the oscillating element of the question. Is that what I should expect?

    So, -25.81^2 x 0.1=-66.62
    & -25.81^2 x -0.1= 66.62

    Does that look right to you?

    Apologies, those figures are wrong, they should be

    So, -25.81^2 x 0.01=-6.66
    & -25.81^2 x -0.01= 6.66
    Last edited by a moderator: Aug 6, 2017
  15. Aug 6, 2017 #14


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    Yes. So in fact your second calculation is correct, not the first. (I'd read it as 10cm displacement, like you did in the first calculation.)
    So you know tha accn at +/- 10mm displacement.

    Using either a = -w²*x or F=-kx , so a ∝ x tells you what sort of graph it is - not a sine.
    Remember, ω is now a constant for this system. It does not vary with displacement.

    The sine function is how things are varying with time rather than with displacement. Displacement itself is a sin function of time.
    Last edited: Aug 6, 2017
  16. Aug 6, 2017 #15

    My graph looks along the lines of this if you imagine the line going through point (0,0). Displacement on the x axis and acceleration on the y axis.

    Does that seem okay?

    Many thanks for all your help so far.
  17. Aug 6, 2017 #16


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    Well you have the essential point - it's linear.
    You're right it goes through ( 0,0 ).
    You already calculated accn as ±6.67m/sec2[/SUP at ±0.01m displacement.
    So it's just a matter of sketching that and marking in the values at the ends.
  18. Aug 6, 2017 #17
    Many thanks for all your help Merlin, it's been much appreciated!!
  19. Aug 6, 2017 #18


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    UR welcome.
  20. Aug 6, 2017 #19
    That exactly how mine turned out too. Couldn't figure out how to upload the picture on here!

    Again, many thanks!
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