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Coupled pendulums problem please help

  1. Nov 13, 2008 #1
    1. The problem statement, all variables and given/known data
    two coupled pendulums are used at positions x,1 and x,2

    Newton’s equation for the forces leads to the two equations:

    m,1 * (second derivative of x,1 with respect to t) = -k,1x,1 + k(x,2 - x,1)

    and m,2 * (second derivative of x,2 with respect to t) = -k,2x,2 - k(x,2 - x,1)

    This leads to the two solutions:

    x,1(t) = A,1*sin(ω,1*t + α,1) + A,2*sin(ω,2*t + α,2) (equation 3)

    and

    x,2(t) = A,1*sin(ω,1*t + α,1) - A,2*sin(ω,2*t + α,2) (equation 4)

    where

    A,1 = A,2 and α,1 = α,2

    rewrite equations (3) and (4) in the very interesting form:

    x,1(t) = 2A,1*cos(((ω,1 - ω,2)/2)*t)sin(((ω,1 + ω,2)/2) (equation 3a)

    and

    x,2(t) = 2A,1*sin(((ω,1 - ω,2)/2)*t)cos(((ω,1 + ω,2)/2) (equation 4a)



    Basically i have to derive (3a) and (4a) from equations 3 and 4 using A,1 = A,2 and α,1 = α,2

    2. Relevant equations

    above

    3. The attempt at a solution

    all ive managed to do is expand out the brackets and thats where i get stuck, is there anyone that can help get me in the right direction as i have no idea where to go from here or how it changes from sin to a cos,

    thanks
     
  2. jcsd
  3. Nov 13, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi 8614smith! Welcome to PF! :smile:

    You need to learn the four equations for sinA ± sinB and cosA ± cosB.

    In this case, use sinA + sin B = 2.sin((A+B)/2).cos((A-B)/2) :wink:
     
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