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Simple harmonic motion questions

  1. Nov 9, 2006 #1
    A u-tube manometer is half filled with water. The liquid levels are displaced so that one side is higher than the other. The water is then left free to oscillate from side to side. The tube has a cross sectional area of 1.cm^2 and the inital displacement is 0.1m from the rest position, which is 0.5m above the middle of the bottom of the tube.

    The total length of tube filled with water is 1.1m.

    g = 10 N/kg

    1) Using the relationship of pressure and density and depth, calculate the pressure at the bottom of the tube due to each of the unbalanced columns of water and hence the resultatn force acting..

    Please someone explain/help.
     
  2. jcsd
  3. Nov 9, 2006 #2
    the answers are meant to be "4000Pa, 6000Pa, 0.2N"
     
  4. Nov 9, 2006 #3
    Never Mind I Got It..
     
  5. Nov 9, 2006 #4
    Well I got the pressures. How can I work out the force?
     
  6. Nov 9, 2006 #5

    rsk

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    don't you just use p = F/area?
     
  7. Nov 9, 2006 #6
    yeah that's it thanks.

    also:

    a bungee jumper jumps from a high platform. he has a mass of 60kg and the rope, which is tied to the platform is light, 20m long and perfectly elastic with a stiness of 30 N/m.

    g = 10N/kg

    How far will the bungee have stretched when the jumper first comes to rest?
     
  8. Nov 9, 2006 #7

    Hootenanny

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    HINT: Conservation of energy.
     
  9. Nov 9, 2006 #8
    thanks i will have a look
     
  10. Nov 9, 2006 #9
    thanks for the help.

    how do I find out the tension in the bungee when he's at rest? i'd have thought it's T = mg = 600N, but the answer says it's 1600N.
     
  11. Nov 10, 2006 #10

    Hootenanny

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    No problem. What is the extension in the bungee at this point (lowest point)?
     
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