1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spring mass spring mass spring - 2 balls oscillations

  1. Jan 14, 2010 #1
    http://fatcat.ftj.agh.edu.pl/~i7matras/hej.jpg

    Both masses are in point.

    I need to count displacement x(t) but I don't know how to write derivative equations? Could someone help? Or at least give me a tip?
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Jan 14, 2010 #2

    HallsofIvy

    User Avatar
    Science Advisor

    Are we to include gravity or can we think of this as horizontal? Can we assume that the two masses are equal? Can we assume that the two springs have the same spring constant and natural length?

    Assuming all of that, let [itex]x_1[/itex] be the distance from the lower edge to the center of mass of the first ball, [itex]x_2[/itex] be the distance from the lower edge to the center of mass of the second ball and let m be the mass of each ball. Let L be the distance from the lower edge to the upper, let k be the spring constant of each spring and let l be their natural lengths.

    "F= ma" of course, so [itex]m d^2x_1/dt^2[/itex] is equal to the total force on the first mass. There is a force from the spring below it equal to the spring contant times its extension: [itex]-k(x_1- l)[/itex]. There is a force from the spring above it, equal to the spring constant times its extension: [itex]k(x_2- x_1- l)[/itex] (positive because if the spring is stretched, it will pull the mass upward).
    [tex]m\frac{d^2x_1}{dt^2}= -k(x_1-l)+ k(x_2- x_1- l)= -2kx_1+ kx_2[/tex]

    [itex]md^2x_2/dt^2[/itex] is equal to the total force on the second mass. There is a force from the spring below it of [itex]-k(x_2-x_1- l)[/itex], equal and opposite to the force of that spring on the lower mass. There is a force from the spring above it: [itex]k(L- x_2- l[/itex].
    [tex]m\frac{d^2 x_2}{dt^2}= -k(x_2-x_1-l)+ k(L- x_2- l)= -2kx_2+ kx_1+ kL[/itex].

    Those are your two equations. When you solve them, you should find that you have two "natural frequencies", [itex]\sqrt{k/m}[/itex] and [itex]\sqrt{3k/m}[/itex].-
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook