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Homework Help: Finding the Time in a SHM Problem

  1. Jan 15, 2010 #1
    1. The problem statement, all variables and given/known data

    Bungee Man is a superhero who does super deeds with the help of Super Bungee cords. The Super Bungee cords act like ideal springs no matter how much they are stretched. One day, Bungee Man stopped a school bus that had lost its brakes by hooking one end of a Super Bungee to the rear of the bus as it passed him, planting his feet, and holding on to the other end of the Super Bungee until the bus came to a halt. (Of course, he then had to quickly release the Super Bungee before the bus came flying back at him.) The mass of the bus, including passengers, was 1.20×104 kg, and its speed was 21.2 m/s. The bus came to a stop in 50.0 m.
    B) How much time after the Super Bungee was attached did it take the bus to stop?

    x = 50 m
    m = 1.20x104 kg
    v = 21.2
    k = 2160 N/m (calculated in Part A of the question. I used the conservation of energy to figure it out)

    2. Relevant equations

    T = 2pi x sqrt(k/m)
    v = -Awsin(wt)
    x = Acos((2pi/T) x t)

    3. The attempt at a solution

    I solved this problem by finding the period T, and dividing this by 2, since the point at which the bus stops would be equal to half of the period in SHM. Now I'm starting to wonder whether the initial velocity of the bus would cause the bus to actually take longer than T/2 to reach that point where it stops. I even tried using simple kinematics to solve this problem, but that didn't work either. For T, I got 14.8, and divided it by 2 to get 7.40, which is wrong...does anyone know what I'm doing wrong?
  2. jcsd
  3. Jan 15, 2010 #2
    Once the spring has reached a standstill and commences to retract, the SHM eqns should work. It makes no difference how that distance from equilibrium was reached.
    From your comments, I believe you may have computed the swing back time, but neglected the time required to slow the bus?

    Edit: on second thought, 1/2 T is right--I'll check the K computation.
  4. Jan 15, 2010 #3


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    In SHM the velocity reduces from maximum to zero in time T/4 s.
  5. Jan 15, 2010 #4
    Right, I guess i misread the question as well, but the T seems too large:

    I get .43(2*pi) for a complete period?
  6. Jan 15, 2010 #5
    I double checked my calculations and I'm getting that too. So now I just divide this value by 4?
  7. Jan 15, 2010 #6
    I divided it by 4 and got 0.67, which is incorrect...
  8. Jan 15, 2010 #7
    I have a feeling I only have to divide it by 2. Can someone verify this?
  9. Jan 15, 2010 #8
    There is an error in the formula for the period in the first post.
    It is
    [tex] $ T= 2\pi\sqrt{\frac{m}{k}} $ [/tex]
    (and not k/m)
    It gives T= 14.8 s with k=2157 N/m
    The time from the position where the spring is not stretched to the maximum extension is T/4.
  10. Jan 16, 2010 #9
    AHH woops...thanks so much for noticing that! :)
  11. Jan 13, 2011 #10

    Ive been working on this problem and i used T=2pi * root k/m and got 14.81..im not sure where to go from here to get the time after though. i tried using the v=-Asinwt equation and got 5.77 which is wrong. Any help is greatly appriciated.

    Thanks in advance!
  12. Jan 13, 2011 #11
    We're looking for the period of time starting when the bus is first snagged by the bungee cord, and ending when the bus is at its maximum distance. At the former instant, the bus is at "position zero." It is at the same position as the bungee cord's anchors, and therefore feels no net force due to the cord. In other words, it is at the equilibrium position. The latter instant corresponds to the position of local maximum for the bus. We can interpret these points using the graph of the position function:

    [tex] x(t) = A cos( \frac{2 \pi }{T} t ) [/tex]

    The time interval we want is the one between an equilibrium position (i.e. a zero) and a local maximum on this graph. You should be able to determine what fraction of the period it is from this.
  13. Jan 13, 2011 #12
    I tried using that equation, however we do not have the amplitude, leaving us with two unknowns in the equation :S
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