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S.H.M and calculating depth of water

  1. Oct 20, 2009 #1
    Greetings,

    Kindly, may I have my answers checked, please.


    1. The problem statement, all variables and given/known data
    d. A 40kg child sits on a swing of length 8.0m and is swinging with a maximum displacement of 2.0m from the equilibrium position. If the motion can be considered to be SHM, calculate the:

    i. Period of the swing.
    ii. Maximum vertical distance the child rises above the equilibrium position
    iii. Maximum velocity of the child during the motion
    iv. The total energy produced due to this motion



    2. Relevant equations
    i. T = 2π√l/g
    ii. x = rsinwt
    iii. V = coswt
    iv. (Kinetic Energy) Ek = 1/2mv²

    3. The attempt at a solution
    i. From T = 2π√l/g = 2πw, we see that w = √g/l = √9.8/1 = 9.8, assuming g = 9.8 m/s ²
    T = √ 2π/9.8
    = 0.8 s

    ii. x = rsinwt
    w = 2πf
    f = 1/0.8
    = 0.25 Hz
    w = 2π x 1.25
    = 7.85 1/rads
    x = 2m x sin7.85 1/rads x 0.8
    = 1.6

    iii. V = wrcoswt
    = 0.8 1/rads x 2π x cos7.851/rads x 0.8
    = 0.02 m/s

    iv. Ek = 1/2mv²
    = 40 x (0.02)²/2
    = 0.008 J

    QUESTION 2

    In a harbour, the equation for the depth h of water is h = 5.0 + 3.0sin(2π/45600)
    where h is given in metres and t is the time in s. (The angle 2πt/45600 is in radians)
    For this harbour, calculate:
    i. the maximum depth of water

    5.0 + 3.0 = 8.0 m

    ii. the minimum depth of water

    = 5.0 m

    iii. the time interval between high and low-water
    iv. two values of t at which the water is 5.0m deep
    v. the length of time for each tide during which the depth of water is more than 7.0 m.

    I don't understand iii, iv v. May I have some hints on how to approach them please.
     
  2. jcsd
  3. Oct 20, 2009 #2

    Delphi51

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    T = 2π√(l/g) = 2π√(8/9.81) = 5.67 s.

    I don't follow your work in ii.
    You say w = 2πf and f = 0.25 but then you put "w = 2π x 1.25"
    You have
    x = rsinwt = 2m x sin7.85 1/rads x 0.8
    but r is known to be 8 and the time is unknown.
    Perhaps knowing the displacement is 2 m, you can find the angle the swing is at and use x = r*sin(θ)
     
  4. Oct 20, 2009 #3
    Thanks for the correction.
    Revised calculations

    ii. x = rsinwt
    w = 2πf
    f = 1/5.67
    = 0.18 Hz
    t = 5.67s
    w = 2π x 0.18
    = 1.13 1/rads
    x = 2m x sin1.13 1/rads x 5.67
    = 10.3

    iii. V = wrcoswt
    = 1.13 1/rads x 2π x cos1.13 1/rads x 5.67
    = 17.18 m/s

    iv. Ek = 1/2mv²
    = 40 x (17.18)²/2
    = 5903 J
     
  5. Oct 20, 2009 #4

    Delphi51

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    Why are you replacing r with 2 m? Shouldn't it be 8m ?
    Why are you using a time of 5.67 seconds? That is the period, the time for one complete swing to one side and back and to the other side and back. The time to swing out 2 m will be much less than 5.67 seconds.
     
  6. Oct 20, 2009 #5
    So, switching the 2m to 8m would make it more proportional?
    5.67s was used because it is the only time available from the problem. Also noting that the calculation of time period preceded this question.

    x = 8 m x sin1.13 1/rads x 5.67?
     
  7. Oct 20, 2009 #6

    Delphi51

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    Oops, sorry! I am mistaken and you are right. Maximum displacement of 2 m so the r = 2. I was thinking of the swing going around in a circle.

    According to x = rsinwt at time zero the swing is vertical, so it would take time 5.67/4 (a quarter of the period) to swing out to the maximum displacement. Yes, I think that works out to x = 2 m so it all hangs together.

    What was the question? Oh yes, the maximum VERTICAL distance, so nothing to do with the frequency or the x = rsinwt. Consider that the swing moves in a circle with radius 8. When it has moved 2 m along that circle, what angle is it at? Hint: it has moved 2 meters out of the full circumference, so it is a fraction of 360 degrees. Once you have that angle you can use trig to get the height.
     
  8. Oct 20, 2009 #7
    l = 2 m
    = 2/2π
    = 0.32
     
  9. Oct 20, 2009 #8

    Delphi51

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    2 meters out of a circumference of 2*pi*8 = angle out of 360 degrees
     
  10. Oct 20, 2009 #9
    2π x 8 = 50.3
    50.3 - 2 = 48.3?
     
  11. Oct 20, 2009 #10

    Delphi51

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    2 meters out of a circumference of 2*pi*8 = angle out of 360 degrees
    2/(2*pi*8) = A/360
    Solve for A in degrees.
     
  12. Oct 21, 2009 #11
    2/(2π x 8) = A/360
    720/(2π x 8) = A
    A= 720/(2π x 8)
    = 14.32

    I'm not sure of how to calculate the height, considering that theta = 14.32, r = 8
     
    Last edited: Oct 21, 2009
  13. Oct 22, 2009 #12
    Question 2 should read

    In a harbour, the equation for the depth h of water is h = 5.0 + 3.0 sin (2π t/45600)
    where h is given in metres and t is the time in s. (The angle 2πt/45600 is in radians)
    For this harbour, calculate:
     
  14. Oct 22, 2009 #13

    Delphi51

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    swing.jpg
    If you can find the side "a", then you can get "h" easily.

    #2: Better check the minimum. Suggest you graph the sine function on your calculator so you can see how it goes. You can get a good estimate of the time interval between high and low that way, too. If you calculate the period of the sinusoidal, you can easily get the exact value for that interval. In sin(kt) the period is 2π/k.
     
  15. Oct 22, 2009 #14
    sin14.32 = x/8
    x = 8sin14.32
    = 1.98 m

    (8)² = a² + (1.98)²
    64 - 3.92 = a²
    60.08 = a²
    a = √60.08
    = 7.75 m

    x = 1.98 m; 2 m =hyp; h =?
    2² = 1.98² + (hyp)²
    4 - 3.92 = hyp²
    0.08 = (hyp)²
    hyp = √0.08
    = 0.28 m; Hence h = 0.28 m

    I'm not understanding this.

    t = 0 is the minimum depth of water

    2πt/45600 = π/2
    t = 45600π/ 4π
    = 11 400 s (low water)

    2πt/45600 = π
    t = 45600/2π
    = 22 800 s (high water)

    Time interval between high and low water: 22 800 - 11 400
    = 11 400 s

    depth of water (l) = 8.0 m
    g = 9.8 m/s ²
    T= ?
    Eqn. for T = 2π√(l/g)
    = 2π√8/9.8
    = 1.81 s
     
    Last edited: Oct 22, 2009
  16. Oct 23, 2009 #15

    Delphi51

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    You have side a = 7.75 correct. To find the height, just subtract
    h = 8 - 7.75 = 0.25.

    If you graph #2, you will see:
    sinegraph.jpg
    In sin(kt) the period is 2π/k. Your k = 2π/45600 so you have
    period = 2π/(2π/45600) = 45600. The time from one crest to the next is the period, 45600 seconds. Think about where the minimum is compared to the two crests. Check your answer by looking at the graph.
     
  17. Oct 25, 2009 #16
    I don't get 45600 when I subtract a 1000 (initial) from 35000 (minimum point) - the difference between the two crests. Explain please.
    In sin(kt) the period is 2π/k. Your k = 2π/45600 so you have
    period = 2π/(2π/45600) = 45600.
    I don't understand the equations you put forth?!
     
  18. Oct 25, 2009 #17

    Delphi51

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    The first 2 crests are at about 11000 and 57000.
    A crest is a maximum, the top of the wave.
     
  19. Oct 25, 2009 #18
    Is the minimum = 35000?
    So, how do I go about obtaining the time interval between high and low-water?
     
  20. Oct 25, 2009 #19

    Delphi51

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    Yes, roughly 35000. 34000 would be better.
    So, you've got a low at 34000 and a high at 11000.
    Just subtract to get the time interval between them - approximately.
    To get an accurate answer you must look again at the wave formula and calculate the period. Use that T = 2π/k trick I mentioned earlier.
     
  21. Oct 25, 2009 #20
    I'm not sure what you mean by accurate here. I get answers at different ends of the spectrum:)

    T = 2π/(2π/45600) = 45 600
    Time interval = 34 000 - 11 000 = 23 000

    iv. two values of t at which the water is 5.0m deep

    (0,0) seeing that these are the coordinates

    v. the length of time for each tide during which the depth of water is more than 7.0 m.

    Readings from the graph.
    Tide 1 app. 7 000s Tide 2 is app. 50 000
     
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