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Thread: Plotting graph acceleration against displacement

  1. #1
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    Plotting graph acceleration against displacement

    A mass of 0.3 kg is suspended from a spring of stiffness 200 N m1. Ifthe mass is displaced by 10 mm from its equilibrium position andreleased, for the resulting vibration:


    Plot a graph of acceleration against displacement (x) (for values of xfrom 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!
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  2. #2
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    Re: Plotting graph acceleration against displacement

    Quote Originally Posted by james1 View Post
    A mass of 0.3 kg is suspended from a spring of stiffness 200 N/m. 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 for values of x from x = 10 mm to x = +10 mm
    $a = -\omega^2 \cdot x = -\dfrac{k}{m} \cdot x = -\dfrac{2000}{3} \cdot x$

    note ...

    $-0.01 \, m \le x \le 0.01 \, m$

    $a$ vs $x$ shown in first attached graph

    $a$ vs $t$ shown in second attached graph
    Attached Thumbnails Attached Thumbnails Plotting graph acceleration against displacement-a_vs_x.jpg   Plotting graph acceleration against displacement-a_vs_t.jpg  
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  3. #3
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    Re: Plotting graph acceleration against displacement

    What does "w" mean? And where did you get "-25.81"?

    The way I would do this problem is, using "force= mass times acceleration", if the x m from its equilibrium position then the force is -200x = 0.3 x'' (negative since the force is opposite x- if x is downward, the force is upward and vice-versa), where x'' is the second derivative of position- the acceleration. That is the "differential equation" x''+ (200/0.3)x= x''+ (266.67)x= 0 which has the general solution x= Acos(\sqrt{266.67}t)+ Bsin(\sqrt{266.67}t)= A cos(25.81t)+ Bsin(25.81t). We can then calculate that v= dx/dt= -25.81A sin(25.81t)+ 25.81B cos(26.81t).

    We are told that the original displacement is 10mm= 1 cm= 0.01 cm (not 0.1) with initial velocity 0 (since it is "released" from that position). That is, we have x(0)= A cos(0)+ B sin(0)= A= -.01 and v(0)= -25.81A sin(0)+ 25.81B cos(0)= 25.81B= 0.

    So position x at time t is given by x(t)= -.01 cos(25.81 t). The velocity, as above, is the derivative, 0.2581 sin(25.81 t), and the acceleration is the derivative of that, (25.81^2)(0.01) cos(25.81t)= 6.66 cos(25.81 t).

    So the acceleration is \frac{6.66}{-0.01}= -666= -(25.81)^2 times x. The graph of that is a straight line, passing through the origin with slope -(25.81)^2= -666.

    I don't know how much of that makes sense because you don't say what you do understand and what you can do. I suspect that you were given some formulas but you don't say what. You will get a lot better (and simpler) answers next time if you give us all that information to start with.
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    Re: Plotting graph acceleration against displacement

    Hi there, thanks for replying!
    I managed to figure out how the graph is plotted and drawn. Exactly like the first attached picture
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