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Graphical analysis of a spring mass system over a function of time

  1. May 4, 2013 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations


    3. The attempt at a solution

    part a) frequency of oscillation can be found on the graph, to find frequency I need to find the period (T) first, the period is when the object is back to it's initial conditions. Knowing this I looked at the points from 0s - 0.9s because it took 0.9 seconds for the object to displace and return to it's original position.

    $$ freq = 1 / T $$

    part b) since I've found the frequency I can use the equation

    $$ freq = 1/2pi * sqrt(k/m) $$
    $$ k = (2*pi*freq)^2 * m $$

    part c) I'm having trouble figuring this out, I know linear momentum is p = mv, I'm thinking I am able to find V by using the equation V = R[itex]\omega[/itex], but I don't know what the R would be? would be just the amplitude? or is there another way to go about this question?

    part d) maximum force would equal to F = -kx , since I have found out the k in part b, I just plug that value in and use the amplitude for x, which is 0.12m.

    part e) use the kinetic energy equation for springs?

    with x = 0.06m

    [tex] 1/2kx2 [/tex]
  2. jcsd
  3. May 4, 2013 #2


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    a) 0.9 s is not the period. Think about it: the period is the time for one full cycle or repetition, which means that what the curve does during one period should be identical to what the curve does during the next period. Does the first 0.9 s on the graph look exactly the same as the next 0.9 s? No, it doesn't. Another hint: at the beginning (t=0) the mass is at 0 m and moving forwards. After 0.9 s, is it at 0 m and moving forwards? No. When is the next time that this is true?

    b)sounds okay to me. By the way, your LaTeX got messed up because you were missing important backslashes. Try \pi and \sqrt{k/m} to produce ##\pi## and ##\sqrt{k/m}##

    c) v = R*omega is totally the wrong equation to use, because it pertains to circular motion. "R" means radius in that formula, and there is no radius here, because there is nothing moving in a circle. You know that the max speed occurs at x = 0 m, and you can figure out what this max speed v is from the conservation of energy. Hint: how much of the energy is kinetic at x = 0.12 m? How about at x = 0 m?

    d) sounds good

    e) Again, use the conservation of energy. At any point in the oscillation, some of the energy is kinetic, and some is elastic potential energy. The relative amounts of these two vary, but their total is always the same, because energy is conserved.
  4. May 4, 2013 #3
    a) Oh I see, the period would be 0 to 1.8s,

    c) to find momentum, I need final velocity, and since energy is conserved I can apply the equation Eo = Ef which gives me,,

    PEspring(initial) = KEfinal
    1/2kx2 = 1/2mvf2
    vf = sqrt(kx2/m)

    and then use the velocity in the equation P = mv

    e) since it is conservation of energy, wouldn't my solution remain the same?

    change in kinetic = change in potential
    so it would just be = 1/2kx^2 with x evaluated at 0.06?
  5. May 5, 2013 #4


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    For e), not it wouldn't just be 1/2kx^2. The tricky part is that the total energy is constant, but HOW MUCH of it is kinetic vs. HOW MUCH is potential varies with position x. The two extremes are x = x_max = +/- 0.12 m and v = 0 (in which case all of the energy is potential) and the other extreme is x = 0 and v = v_max , where all the energy is kinetic. Since energy is conserved, it must be true that:

    E_total = (1/2)m(v_max)^2 = 1/2k*(x_max)^2

    Now, at *intermediate* points (any value of x) between these two extremes, SOME of the energy is kinetic and some is potential. However, the total is still the same by conservation of energy. So, at ANY position, it is true that:

    E_total = (1/2)mv^2 + (1/2)kx^2

    You need to compute what the first (kinetic) term is. You can do this because you know E_total, and you know what the second (potential) term is, because you know x.
  6. May 5, 2013 #5
    Oh I see,

    just to see if I got the answer right,

    since the total energy is constant, I know that

    E_total = 1/2k(0.12m)^2

    then I get that value and set it equal to

    E_total = 1/2mvf^2 + 1/2kx^2, then I solve for 1/2mvf^2?

    1/2k(0.12m)^2 - 1/2k(0.06m)^2 = KE
  7. May 5, 2013 #6


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    Yeah, that sounds about right. If you look at your last equation, it makes total sense. The difference between the max possible PE and the PE at 6 cm is equal to the KE, since that's where that energy went.
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