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Springs Assignment

  1. Feb 17, 2015 #1
    1. The problem statement, all variables and given/known data
    When a 100 cm long thing aluminium rod is lodged into the ground, so that it is vertical, it requires 25.0 N of force to deflect it 20.0 cm from the rest position. The rod is then lodged into the side of a wall, and a 100 g of mass is placed at its end and it is then pulled down a distance of 20.0 cm. As the rod vibrates, it loses 50% of its initial energy every second. Draw a graph of the mass's position with respect to time for the first 2 seconds it vibrates.

    2. Relevant equations
    F = -kx
    T = 2*pi*sqrt(mass/k)
    http://www.regentsprep.org/regents/math/algtrig/att7/sinuso44.gif

    3. The attempt at a solution
    From the given information, i found the spring constant and the period to be 125 N/kg and 0.057*pi Hz. And i know that the amplitude is (1/2)^t since it loses 50% per second. So the equation is currently y = (1/2)^t*sin(35.35x) - 20. My question is does this take into account the vertical acceleration since the rod is sideways on a wall.

    i got the 35.35 from 0.057*pi = 2pi/B, B = 35.35.
     
  2. jcsd
  3. Feb 17, 2015 #2

    lightgrav

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    The spring's stiffness is 125 Newton per meter (not N/kg ... that is the Sun's gravity field).
    How far down will the spring sag, when the 100g is attached to its end? ... that becomes the center of the oscillation.
    your 35.36 is correct for ω , but you haven't written its units ... this allows you to err in using it:
    you have written that y (vertical location) depends sinusoidally on the value of x (? a horizontal location? )
    You should see where the mass is at time zero, and at time 1 cycle ... is it a sine, or a cosine ... + or - ?
     
  4. Feb 17, 2015 #3
    Right my bad on the units part, so 125 N/m
    The spring will sag: From F = -kx,
    0.1*-9.8 = -125*Δx
    Δx = 0.00784, it will sag 0.784 cm
    35.36 is the period which is measured in seconds
    Also my bad, it should be y = something in terms of t, y being distance and t being time
    I think it should be at -20.784 cm at time 0, not sure how to calculate at time 1 cycle, i think it should be sin with initial vertical displacement down 20.784cm, and it should be positive.
    Revised equation: y=(1/2)^t*sin(0.1776t) -20.784?
     
  5. Feb 17, 2015 #4
    Scratch the equation, doesn't look right at all with the (1/2)^t, looks like simple harmonic without it, so now the problem i think is how to incorporate the losing 50% energy every second so that it becomes damped harmonic motion?
     
  6. Feb 17, 2015 #5

    lightgrav

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    right, 8mm sag. So yes, yequilibrium = -.028 m.
    but 35.36 is NOT the period in seconds/cycle ... it is angular frequency in radian/second ... verify how to get this! (the 2π is radians/cycle)
    phase: at time t=0, what is sin(0)?
    amplitude: about how high will it be, ½ cycle later?
    if it loses ½ its Energy by 1 second:
    a) what does that say about its stretch?
    b) how many oscillations is that?
     
  7. Feb 17, 2015 #6
    80ce1126e4146b562fd459501dcf6221.png from wiki, makes sense now since its a sinusoidal graph the units should be radians per second
    finally figured it out now: angular frequency: ω = 35.36, period: 0.178s, ordinary frequency: 5.628 Hz
    at time 0, it is at equilibrium so -.028m, sin(0) is 0 and since it completes 1 cycle in 0.178s, 1/2 a cycle should be at time equals (0.178/2) = 0.089.
    From, y = sin(1.118*t) - 0.028 (didn't add in the losing 50% energy part) you would get 0.099-0.028 = 0.071m (calculator should be in radians right?)
    a) at time 1 s, its position should be multiplied by 0.5?
    b) an oscillation is 1/2 a cycle on a sine graph, since it completes 1 cycle in 0.178s, at time 1s, it would have completed 5.61 cycles, times 2 11.23 oscillations?
     
  8. Feb 17, 2015 #7

    lightgrav

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    good: Nature wants us to use radians ... equations are simpler in radians : ω = √k/m ... y = A sin(ωt + φ) .
    A: What is the amplitude of a sine curve? how can you make your curve have location units?
    φ: is it knocked downward at t=0 to get it moving, or is it pushed upward before being released at t=0?
    decay: how much Energy does it have at .200m from equilibrium? ... how much of that would it have at .100m ?
     
  9. Feb 17, 2015 #8
    Amplitude is half the distance between the max and min, but since it is damped, the max and min will both be "shrinking" with time. Are location units like coordinates? so wouldn't i plug in values?, but the problem is how do i find max/min without amplitude, or do i just find amplitude and then maxs and mins.
    φ: It is knocked downwards at time 0 (pulled down distance of 20 cm), question: wouldnt the -.028 be the vertical shift outside the brackets? Ex. y = sin(1.118*t) - 0.028
    decay: forgot to use energy formula? PE = 1/2kx^2, at .200, PE = 2.5J, .100m, PE = 0.625J, but this isn't taking into account it losing 50%? Do i find their position at time 1 and 2 seconds and then find energy divide that by 2, and then solve for position taking into account losing 50% energy? well at 2 s it would be back to equilibrium again since it has lost all of its energy.
     
  10. Feb 17, 2015 #9

    lightgrav

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    eq: the center of the sine function sags how far from where the spring was attached?
    A: what is the maximum value for a sine curve? what is the minimum value for a sine curve? can you now tell me the Amplitude for a sine curve?
    . . . notice that the sine function is unitless ... ALL functions are themselves unitless, with unitless arguments (the variables inside them).
    what are the units of "y" supposed to be? what quantity is going to carry those units there? how far does this mass need to go? (which direction)
    φ: Being pulled down is a very different thing from being knocked down :
    . . . pulled down has a distance or location (or Force to hold it there) - knocked downward has a speed or velocity (or impulse)
    decay: is .625J equal to half of 2.5J ? (no) ... what fraction is it?
    . . . option a) what spring stretch WOULD yield ½ ×2.5J ? . . . option b) how many seconds "τ" should occur before max stretch IS 0.100m ? (½)^t/τ
    ...it is supposed to lose ½ its Energy the 1st second, then ½ of the rest in the 2nd second, ½ the remainder in the 3rd ...
    at equilibrium, all its Energy is Kinetic
     
  11. Feb 17, 2015 #10
    max for sine is 1, min is -1, amplitude is 1.
    units of y are m or i was thinking of doing them in cm, the mass will go upwards when it is released, to i dont know how to calculate the max height.
    When released at time 0, it will be "pushed" upwards since it will be trying to restore to its equilibrium after being pulled down 20 cm
    decay: .625 is a quarter of 2.5 so 1/4
    option a) the spring would have to stretch 0.141m to yield 1/2*2.5J

    Updated: y = A sin(ωt + φ), still dont know A, because isn't it constantly changing as in the first max height it reaches will be the absolute max height and the others would only be local maxs? (the amplitude is also changing with time) φ: Ok i know that this is the phase shift in rads so, Is it the fraction? so 1/4? and where does decay come into this? if at all?

    (If this is wrong, i guess i just need help on how to find the amplitude and the phase shift)
     
    Last edited: Feb 17, 2015
  12. Feb 17, 2015 #11
    Found out how to get A and φ, but now i would need v0 , more relevant equations i guess:
    v=Aw cos(wt+Ø)
    a=Aw² sin(wt+Ø)
    A = (y)^2 +(v/w)^2
    Ø = tan-1 (y*w/v)
     
  13. Feb 18, 2015 #12

    lightgrav

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    At the beginning, A=20cm. but after 1 second, it has changed smoothly to be only 14.1cm; at 2s it is 10cm.
    I would write it as (20cm) [½^(t/2s)] .
    the phase offset φ is ¼ of a cycle, not ¼ of a radian.
     
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