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Homework Help: Simple harmonic motion of small planet

  1. Apr 4, 2007 #1
    1. The problem statement, all variables and given/known data
    The residents of a small planet have bored a hole straight through its center as part of a communications system. The hole has been filled with a tube and the air has been pumped out of the tube to virtually eliminate friction. Messages are passed back and forth by dropping packets through the tube. The planet has a density of 4120kg/m^3, and it has a radius of R= 5.67E+6 m. What is the speed of the message packet as it passes a point a distance of .420R from the center of the planet?
    How long does it take for a message to pass from one side of the planet to the other?
    2. Relevant equations

    1.density=mass/volume= mass/[(4/3)pi*R^3]
    2. x=Acos(wt)
    3. v=-wAsin(wt)

    3. The attempt at a solution

    I first found the mass in equation one. Then I thought .420R was x, the radius is A and, that is as far as I got with that.

    Am I wrong so far?
  2. jcsd
  3. Apr 4, 2007 #2
    Since it's simple harmonic, you can find (acceleration) a=w^2*x first. (express the gravitational force exerting on the packet as a linear function of r(the distance between the packet and the center of the planet)
  4. Apr 5, 2007 #3


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    DonĀ“t forget that the mass of the planet to be considered in the calculation of the gravitational force is the contained in the sphere of radius r.
  5. Apr 5, 2007 #4


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    The only mass exerting a nett force on the packet is that contained inside the current radius because the shell of matter outside the radius has no effect ( as pointed out above). So the force at point x is

    [tex]F(x) = -\frac{4}{3}\pi Gm\rho x^3/x^2[/tex]

    which gives

    [tex]F(x) = -Kx[/tex]

    which is the equation for a simple harmonic oscillator.
    Last edited: Apr 5, 2007
  6. Apr 5, 2007 #5
    Mentz114, so m in that equation is the mass of the packet? Because in this problem we are not given the mass of the packet, and that has been confusing me. (well a lot of this problem period is confusing me). I figured that since the mass of the packet wasn't mentioned then it somehow gets canceled out in the problem.

    Andy_ToK, I don't understand how to find the acceleration. Using the equation Mentz114 suggests, how do I find the acceleration?

    I tried at first to set ma=GMm/R^3 and then the m of packet cancels out.

    Our class just started doing simple harmonic motion yesterday so I am still trying to understand the concepts. This problem is just so confusing.

    So Mentz114, am I to assume that x in F= -kx, is .420R in my problem? and if so then the k is -\frac{4}{3}\pi m
  7. Apr 5, 2007 #6
    ok the last part I meant k = -(4/3) pi*m*p
  8. Apr 5, 2007 #7


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    Hi Fruit,

    I made an error and left out a 'G' in the force formula,

    [tex]F(x) = -\frac{4}{3}\pi Gm\rho x^3/x^2[/tex]

    You'll find from your SHO studies that the frequency of oscillation depends only on K and m so if you have K you can work out the time part.

    By the way, acceleration is force/mass ( Newton).
    Last edited: Apr 5, 2007
  9. Apr 5, 2007 #8
    Wouldn't it be easier to solve this problem using energy methods once k is determined?
  10. Apr 5, 2007 #9
    Shouldn't that be

    [tex]F(x) = -\frac{4}{3}\pi Gm\rho R^3/x^2[/tex]

    although, if it were, the motion of the packet would not be simple harmonic.
  11. Apr 5, 2007 #10


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    Only the matter inside the current position ( ie with r < x) exerts a nett force. The remaining matter forms a hollow shell which, as you know exerts no field inside.

    See post #3 and #4.

    Using energy we can get the maximum PE of the oscillator and equate to the maximum grav potential, so

    [tex] KR^2 = \frac{GMm}{R}[/tex]


    [tex] K = \frac{GMm}{R^3}[/tex]


    [tex]K = \frac{4}{3}\pi Gm\rho [/tex]

    which is the same as above.
    Last edited: Apr 5, 2007
  12. Apr 5, 2007 #11
    oh, ok. so am I right to assume that when I set up the equation for the period, 2pi * sqrt m/k, that the m's will cancel?
  13. Apr 5, 2007 #12


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    Yes, you got it. I think this might be the only gravitational field that can give harmonic oscillations.
    To solve the velocity part of the question you'll need the solution to

    [tex]\frac{dx}{dt} = f(K, x)[/tex]
  14. Apr 5, 2007 #13
    I keep getting the period incorrect
    I did 2*pi *sqrt m/GMm/R^3, I canceled out the m's
    leaving 2*pi *sqrt R^3/G*M = 2pi sqrt (5.67E6)^3/[(6.67E-11)*(3.145E24)]
    PERIOD = 5856 s

    Plus, a friend just told me to try a= GM/R^2, then find angular speed
    with angular speed = sqrt a/R.
    Since the period is 2pi/angular speed as well, I used the answer I got and got the same thing!

    For some reason I still cannot get the first part of the problem.

    Please tell me if my rationale on this portion is wrong: x=Acos(wt)
    so the x would be the .420R, A would be simply R and the angular speed is w.
    when I tried the k Mentz114 suggested I got the same answer that I got when I tried w=sqrt a/R
    so I think I'm doing it correctly but when I tried to find t, I think I solved it incorrectly....I'm not the greatest at math:blushing:
    I did x/A*cosw which became x/cos w since the R's cancel. I ended up getting a value of t= .4200s However, when I put this value in the velocity equation, v= -wAsin(wt), I got a small negative velocity that was wrong.

    and to find the mass of planet, it is density times volume and the forumla for the volume is 4/3 pi R^3...I just don't understand what I am doing wrong?
  15. Apr 5, 2007 #14
    Hmm... I understand that dx/dt is the dirst derivative of the position equation, but I do not understand what f(K,x) means
  16. Apr 5, 2007 #15


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  17. Apr 5, 2007 #16
    Mentz114, I guess I didn't do a good job at explaining what I thought you meant. Isn't the velocity equation the first derivative of the position equation? That was all I meant.

    I am looking at the equations, but I don't know what t is, and I don't know how to solve for t

    Could I actually use energy conservation in this problem?
  18. Apr 5, 2007 #17
    I just tried to do 1/2 mv^2= 1/2 kA^2
    v^2 = (GM/R^3)*A^2
    v= sqrt (GM/R^3) *A
    and got v=6.083 *10^3 m/s and that was wrong.
  19. Apr 5, 2007 #18


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    Use the position equation to solve for t, then use t in the velocity equation.

    0.42R= Rsin(wt)
    sin(wt) = 0.42
    wt = sin^-1(0.42)
    t = sin^-1(0.42)/w

    w = 1/T
  20. Apr 5, 2007 #19
    my calculator should be in radians, correct?
  21. Apr 5, 2007 #20
    Mentz114, thank you, I was able to get the correct answer for part one.

    Yet, I still cannot get part two correct
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