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Harmonic motion

  1. Sep 13, 2006 #1
    Hello there!
    I started a new chapter in my script and have done some problems about harmonic motion. After searching in this forum for a similiar problem for ages but without any succees, I need to ask you guys to help me out.

    Here it is:

    A toy car of mass 1kg is attached with a spring to a wall. The spring's constant is 20 N/m. The car is beeing pulled 1 m (horizontal) away from the wall and then let go. It is going back towards the wall up to a distance of 0.4 m.
    What is the system's energy?

    Every way I took led me into a death end. The most logical way to me is to find first the equilibrium point of the spring which is supposed to be at the same spot where the velocity is the greatest:

    [tex] V_{max}= A\times \omega [/tex]

    I use 0.6 m for A and I compute omega with

    [tex]\omega=\sqrt{\frac{k}{m}}= 0.6m\sqrt{20}= 2.68 m/s[/tex]

    The equilibrium point is where KE and KP are equal:

    [tex] \frac{1}{2}mv^2=\frac{1}{2}kx^2[/tex]

    that gives me the x= 0.366 m.

    My guess is that the equilibrium point is 1m - 0.366m.

    From here I do not have any clue how to get the over all energy of the mass-spring system.

    I would be very happ about any kind of help !
    Last edited: Sep 13, 2006
  2. jcsd
  3. Sep 13, 2006 #2
    Sometimes it is easier than it seems!
    I just figured that A is half of the distance between the two spots (1m and 0.4 m). The equation for the energy of a mass-spring distance is:

    [tex] K=\frac{1}{2}kA^2=0.9 Joules[/tex] and that matches with the answer in my script!
    Time to go to sleep now ....
  4. Sep 13, 2006 #3


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    We all love that eureka moment :smile:
  5. Sep 16, 2006 #4
    we certainly do ;)
    Coudl someone give me a hint how to find the distances to the wall where the car and the spring have the same energy?

    I know that if I draw the graph for the spring's energy (1/2*k*s^2) and the one for the mass's energy (1/2*m*(A*w*sin(wt))^2) into a diagramm, they probably meet and where they intersect is where the energy is equal, right? Unfortunately, I cannot the a way to do that with math :(
  6. Sep 16, 2006 #5
    and here was that eureka moment again!
    the one graph is actually a cos function and the other one is a sin function. In the end, they intersect at pi/4!
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