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Child oscillating on playground swing

  1. Aug 1, 2005 #1
    Here is a homework problem that I believe is solved right.

    A child oscillates freely on a playground swing. It takes 2.5 seconds for the child to complete one full swing and return to the person that is pushing her. How long is the swing? This is how I solved for the answer.

    l = (980)(2.5)^2/4n^2 = 15112.83174 (the n stands for pi)
    then I took the square root of 15112.83174 to get 122.9342578 or 122.93 cm

    The next part of the problem I don't know what to do.
    A child twice as heavy sits on the same swing. What is his period of oscillation? How do you Know?
    I'm not real sure where I'm supposed to start with this problem.
  2. jcsd
  3. Aug 1, 2005 #2


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    In the first part you're pretty much right, but it should say [itex]l^2=[/itex] instead of l=, and the first term should be [itex]\sqrt{g}[/tex] intead of g. For the second part, a hint: does the mass appear in the equation for the period? What does that tell you?
  4. Aug 1, 2005 #3
    Hmmm not totally sure about that hint.
    The period is the time so 2.5^2 is the period of time, but I don't see where the mass fits in. Do you divide how long the swing is by the time.
    122.93/2.5 = 49.172 to get the mass and then double that to get 99.42?
  5. Aug 1, 2005 #4
    the hint is, mass doesnt make a difference...
  6. Aug 1, 2005 #5
    So, that means that mass doesn't make a difference in this particular problem only.
    Wouldn't a heavier child make the length of the swing change, or is this question asking for the length of the swing not the length of the oscillation?
  7. Aug 1, 2005 #6


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    The mass of the kid doesn't affect the period of oscilation. That means that if they gave you the period of oscilation and asked for the length of the swing, the kid's mass would not be a factor. This is the case here because gravity is the force and the force of gravity is proportional to the mass. This causes a cancelation in the differential equation from which the formula for the period is derived. If we were talking about a block on a spring lying on the table, then the mass would change the period.
  8. Aug 2, 2005 #7


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    Since gravitational force is proportional to m and F= ma, acceleration, and therefore movement, due to gravity, in general, does not depend on the mass of the accelerating body. For a pendulum, which is essentially what you have here, period depends only on the square root of the length of the pendulum.
  9. Aug 2, 2005 #8
    I understand this problem, but I tried to work another problem in the same manner and my professer said I did it the wrong way. Here is that problem and the correct way to work it.
    How long should the lenth of a pendulum be to have a period of exactly 1.000 seconds on a "grandfather" clock?
    l=[(980)(1.000)^2]/(4pi^2) = 24.82368999 or 24.82cm
    My question is what is the basic difference between these two questions and why wouldn't I take the square root of the answer in this question?
  10. Aug 2, 2005 #9


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    Do you understand the difference between "square" and "square root"? What I said was that the period is proportional to the square root of the length. That is,[tex]T= k\sqrt{l}[/tex] (and the k is, as you gave, [tex]\frac{2\pi}{\sqrt{g}}[/tex]) so [tex]T^2= k^2l[/tex] and [tex]l= \frac{T^2}{k^2}[/tex], exactly the formula you give. The length is proportional to the square of the period.
    Last edited: Aug 2, 2005
  11. Aug 2, 2005 #10
    Square is when I square the number like 2^2 and square root is when I take the square root of the answer. I'm in intro to Physics and doing the lab part and I understand some of the basics but tend to have problems of knowing what to do and when to do it. I'm trying hard to understand all of it but the equations give me problems, especially since every problem has a new formula to use.
  12. Aug 2, 2005 #11

    Doc Al

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    The two problems are essentially identical. This one is done correctly. The one you did in your first post was done incorrectly: (1) You made an arithmetic mistake, (2) for some reason, you took the square root.

    The period of a simple pendulum is:
    [tex]T = 2 \pi \sqrt{\frac{L}{g}}[/tex]

    Solving for the length:
    [tex]L = \frac{g T^2}{4 \pi^2}[/tex]
  13. Aug 2, 2005 #12
    I have trouble remembering when to take the square root of something. You all are very helpful but algebra and physics are hard for me to understand. I do understand the basics but trust me I will be glad when I'm through with these courses. It's like a foreign language to me that I have to force myself to concentrate on.
    I did rework the first problem l=((980)(2.5)^2)/(4pi^2)) = 155.15cm
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