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Oscillations in strings

  1. Jul 5, 2008 #1
    Astronauts in space cannot weigh themselves by standing on a bathroom scale. Instead, they determine their mass by oscillating on a large spring. Suppose an astronaut attaches one end of a large spring to her belt and the other end to a hook on the wall of the space capsule. A fellow astronaut then pulls her away from the wall and releases her. The spring's length as a function of time is shown in the figure

    What is her speed when the spring's length is 1.2 ?

    Given (seen from a length vs. time graph):
    Spring's constant: 240N/m
    Max length of string: 1.4m
    Min length of string: 0.6m
    Equilibrium position: 1.0m (if taking the middle)
    Mass of person attached to string: 54.7kg
    Frequency: 2 oscillations per 6 seconds

    Associated formulas:
    w=root (k/m)
    w=2(pi)f
    1/2mv^2 + 1/2kx^2 = 1/2kA^2 = 1/2 m(Vmax)^2
    might be more that I am missing.
    I have tried the question using the conservation of energy and yielded 2.37m/s, which I think is not possible since the entire string is less than that.
     
  2. jcsd
  3. Jul 6, 2008 #2
    Conservation of energy will give you the correct result

    You are given k and m, A is the maximum deflection from the equilibrium position (also given) and x is the deflection from equilibrium (also given) at the desired velocity. Solve for V.
     
  4. Jul 6, 2008 #3
    But how can a velocity of -2.37m make sense if the minimum distance from the equilibrium is 0.6m and that 1 oscillation takes less than a second?
     
  5. Jul 6, 2008 #4

    alphysicist

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    I haven't checked your numbers, but is it the idea of the velocity being 2.37 m/s when the spring is not 2.37 m long that you think doesn't make sense? You can't compare this speed and distance directly like that. Having a speed of 2.37 m/s only means that if it kept that same speed for a second it would travel 2.37 meters; but here the speed is always changing, so the 2.37 m/s is a speed it only has for one instant of time.

    Also, instead of 2.37 m/s, you could say it 2.37 millimeters/millisecond, for example, and it would be the same.

    (Baseball pitchers can pitch a baseball over 100 miles/hour, but it's definitely not 100 miles from the pitcher to home plate. Or more to the point, if you look down at the speedometer of you car and notice you are travelling 50 miles/hour, that by itself tells you nothing at all about the length of your trip--you might be going 1 mile or 500 miles.)
     
  6. Jul 6, 2008 #5
    Hmm right. I thought it was constant for some reason. I know it's not. Okay, I'm not getting the right answer though...The diagram is shaped like a W with the highest point at 1.4m and lowest point at 0.6m. Lasts for 6 seconds...2 oscillations...
     
  7. Jul 6, 2008 #6

    alphysicist

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    Can you show the details of what you did (especially what numbers you used)?

    For example, a common mistake to make here would be to assume the amplitude is 1.4 meters, but without knowing what you did , it's hard to tell what might be going wrong.
     
  8. Jul 6, 2008 #7
    I used
    (1/2)mv^2 + 1/2kx^2 = 1/2k(A)^2 since at the length 1.2m, the total energy in the system should equal the potential energy at the amplitudes.
    So, I just plugged in 54.7kg as m, 240 as k and 0.4m as amplitude and 1.2m as x.
    The amplitude concerns me because if I use
    x(0) = Acos(theta), and the graph shows that at x=0, the string's length is 1.4m, it will not work. ie) 1.4m = 0.4m cos (theta)...theta cannot be defined.
    I wonder what is the amplitude? Some one mentioned something about differentiation, which I have no idea how to do in this case.
     
  9. Jul 6, 2008 #8

    alphysicist

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    This amplitude looks right; however, the amplitude is the maximum stretch of the spring, so x can't be 1.2m; x is how much the spring is stretched when the length equals 1.2 meters. What do you get for x?
     
  10. Jul 6, 2008 #9
    Last edited: Jul 6, 2008
  11. Jul 6, 2008 #10

    alphysicist

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    I don't think that's right. The variable x is how much the spring has stretched. Isn't the length of the spring 1 meter when it is not stretched at all? Then if the length of the spring is 1.2 m, what is x?

    (Remember that 1.4m is the length when it is at maximum stretch, and 0.6m is the length when it is at maximum compression.)
     
  12. Jul 6, 2008 #11

    alphysicist

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    That is the right x value, but the way to think about it is that x is the displacement from the equilibrium value. So the way to think about it is:

    x=1.2-1.0=0.2 m

    So for example, if they had asked for the velocity when then length was 1.3m, the x value would be 0.3m.



    The spring is normally at a length of 1 meter when it is not oscillating. When it is undergoing this oscillation, it goes back and forth between having a length of 1.4m and 0.6 m.
     
  13. Jul 6, 2008 #12
    Thank you very much! Also, how do you determine the sign of the velocity? It doesn't say whether it is going forward or backward...
     
  14. Jul 6, 2008 #13

    alphysicist

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    At different times, it is going forward and backward at that spot, so depending on the time the velocity could be either positive or negative. But the problem just asks for the speed, right?
     
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