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Mass swinging on a string

  1. Sep 28, 2014 #1
    1. The problem statement, all variables and given/known data
    A mass,m, hangs from a string and swings with a frequency of 0.8 Hz with
    a maximum
    displacement of 0.1 rad. The equation of motion is given by
    x=Acos(ωt).

    A) What is the length of the string?
    B) What is the maximum displacement of the mass in meters?
    C) What is the velocity of the mass as a function of time? Leave the answer as a function of m,g,L, and θmax
    (Hint: Take the derivative of the equation of motion).
    D)
    What is the restoring force acting on the mass as a function of time? Leave the answer as a function of m,g,L, and θmax
    (Hint: Find the acceleration)

    2. Relevant equations
    ω=sqrt(g/l)
    ω=2pif

    3. The attempt at a solution
    for part a) ω = 2pif and ω = sqrt(g/l) (since this is a simple pendulum)
    so
    2pif=sqrt(g/l)
    or
    l=g/(4pi^2f^2) = 9.8/(4pi^2(.8)^2) = .39m

    for part b i cant figure out how to get .1 rads into meters. this is my attempt so far
    (.1 rad) (cycle/2pi rad) (second/.8cycle) = .02 seconds

    for part c take the derivative


    dx/dt = d/dt(Acos(sqrt(g/l)t))
    v = -Asin(sqrt(g/l)t)(sqrt(g/l)
    im guessing the theta max that the question wants v defined in terms of will be part of A


    part d
    a = dv/dt = d/dt(-Asin(sqrt(g/l)t)(sqrt(g/l))
    = -Acos(sqrt(g/l)t)(g/l)

    then use F = ma = -kx

    m * -Acos(sqrt(g/l)t)(g/l)

    any feedback?
     
  2. jcsd
  3. Sep 28, 2014 #2
    wait i think i got it.
    for part b
    x= l*theta = .39m * .01
    or is it .39 * .01 * 2pi? (radians confuse me)
     
  4. Sep 28, 2014 #3

    haruspex

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    You mean .39m * .1, right? Yes, that's near enough. (The whole point of using radians as the unit of angle is to make this relationship easy: radius x angle = arc length.)
    But note that it really gives you the arc length displacement, not the horizontal displacement. If you want the horizontal displacement you need to use sin(), but for such a small angle there's very little difference. Besides, in order to solve (a) you already had to assume the angle is so small that the two are near enough the same.
     
  5. Sep 28, 2014 #4
    oops yes i meant .1 And yea I assuming sin(theta) is approximately equal to theta. I probably should have mentioned that. The rest of it looks ok though? particularly part d? I wasnt entirely sure what to do once I found the acceleration so i just set ma= -kx and plugged in a
     
  6. Sep 28, 2014 #5

    haruspex

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    Yes, but you need to replace A by some function involving thetamax.
     
  7. Sep 28, 2014 #6
    couldnt i replace the A with l * thetamax?
     
  8. Sep 28, 2014 #7

    haruspex

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    That'll do it.
     
  9. Sep 29, 2014 #8
    thanks for your help
     
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