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Harmonic motion of a bungee jumper

  1. Oct 25, 2011 #1
    1. The problem statement, all variables and given/known data

    A bungee jumper undergoes simple harmonic motion with amplitude 6.9 m and frequency 0.115 Hz. Assume the bungee jumper follows the simple harmonic motion equation x=A\cos (\omega t).

    Find the velocity of the jumper at 0.25s

    2. Relevant equations

    x=Acos(\omega*t)

    3. The attempt at a solution

    Found the angular velocity by multiplying the frequency by 2pi,
    then plugged my values into the harmonic motion equation to find the x value, and divide by the time interval to find the velocity, but the homework says thats the wrong answer.
     
  2. jcsd
  3. Oct 25, 2011 #2
    Please explain the angular velocity aspect of this problem. Secondly, the question asks for the velocity at 0.25 seconds. They do not ask for average velocity.
     
  4. Oct 25, 2011 #3
    Would the angular velocity be the bungee sway as the jump goes on? I don't really understand it to be honest... Are you saying we don't use the given equation in this question?
     
  5. Oct 25, 2011 #4
    No, I was questioning your terminology. I do not think they are referring to the swaying motion of the bungee. In my opinion the solution supplied was for the up and down motion. But nevertheless, let's get on with the problem.

    They supply you with a function for position. It has the variable time in it. They are looking for the velocity. So how does one get velocity from a position function?
     
  6. Oct 25, 2011 #5
    By dividing the position by time
     
  7. Oct 25, 2011 #6
    No, that is not correct. If you divide the CHANGE in position by time you get AVERAGE velocity over that period of time. But the problem is not asking for that. The problem is asking for the velocity when the time is 0.25 seconds. The velocity is not constant in this problem.

    So what do you do?
     
  8. Oct 25, 2011 #7
    I can't think of how to obtain velocity from the position function.
     
  9. Oct 25, 2011 #8
    Have you had a course in calculus?
     
  10. Oct 25, 2011 #9
    I have, I just can't recall how to obtain the value needed here, which is why I'm coming for help.
     
  11. Oct 25, 2011 #10
    OK, is the function written correctly?

    x=Acos(\omega*t)

    What is the back slash? Typo?

    If you have a relationship that gives the position as a function of time, the derivative of it gives the velocity. So take dx/dt.
     
  12. Oct 25, 2011 #11
    It's x=Acos(ωt), I should've clarified.

    We've never had to take any derivatives for this class, but I'll try.
     
  13. Oct 25, 2011 #12
    Are you sure there's not a simpler way to solve this? I don't know how to derive this down...
     
  14. Oct 25, 2011 #13
    Not for the instantaneous velocity. The derivative of the cosine function is the negative sine function.

    V = d (A*cos(omega*t))/dt = -omega * A * sin(omega * t)

    So now all you have to do is plug in the values.
     
  15. Oct 25, 2011 #14
    Okay, so for the instantaneous velocity you need to derive the function with a position function. I got the correct answer, thank you.

    My calculus teacher was really sub-par, I barely passed the class, had to do a ton of tutoring. That was a year ago, need to scrub up I guess.
     
  16. Oct 25, 2011 #15
    You're welcome. Calculus is very important in technical fields such as engineering, physics, etc. What are you studying?
     
  17. Oct 25, 2011 #16
    I'm a biology major, physics is a major requirement for me to graduate. I'm good with the other sciences, not so much in mathematics.
     
  18. Oct 25, 2011 #17
    Good luck with your studies and have a successful career. Signing off now.
     
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