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Homework Help: Simple harmonic motion problem

  1. Sep 29, 2008 #1
    1. The problem statement, all variables and given/known data
    A certain oscillator satisfies the equation

    x''+4x=0

    x=[tex]\sqrt{3}[/tex]
    |x'|= 2
    show that , in subsequent motion,

    x=[tex]\sqrt{3}[/tex] cos(2t)-sin(2t)

    Deduce the amplitude of the oscillations. How long does it take for the particle reach its first origin?

    2. Relevant equations

    General equation: x(t)=A*cos([tex]\Omega[/tex]*t) + B*sin([tex]\Omega[/tex]*t)
    x(t)=C*cos([tex]\Omega[/tex]*t-[tex]\gamma[/tex])
    x''+[tex]\Omega[/tex]2*x=0
    3. The attempt at a solution
    x(t)=A*cos([tex]\Omega[/tex]*t) + B*sin([tex]\Omega[/tex]*t)
    x'(t)= A[tex]\Omega[/tex]*sin([tex]\Omega[/tex]*t) + B*[tex]\Omega[/tex]*cos([tex]\Omega[/tex]*t)

    x''(t)=A[tex]\Omega[/tex]2*cos([tex]\Omega[/tex]*t)-B*[tex]\Omega[/tex]2*cos([tex]\Omega[/tex]*t)

    t=0 at x=[tex]\sqrt{3}[/tex} , therefore , A=[tex]\sqrt{3}[/tex]
    [tex]\Omega[/tex]=2


    x' = [tex]\sqrt{3}[/tex]*2*sin(2t)-2*B*cos(2t)
    x''=[tex]\sqrt{3}[/tex]*4*cos(2t)+4*B*sin(2t)

    x''= -4[tex]\sqrt{3}[/tex]

    [tex]\sqrt{3}[/tex]*4*cos(2t)+4*B*sin(2t)=-4[tex]\sqrt{3}[/tex]

    dividing 4 out on both sides of the equation I am left with:

    [tex]\sqrt{3}[/tex]*cos(2t)+B*sin(2t)=-[tex]\sqrt{3}[/tex]

    Not sure how to obtain B . Wait a minute, doesn't t=0 at |x'| so B= -1 since particle is projected towards origin. In addition , finding the period seems so trivial to me: tau=2*pi/omega and omega = 2 , therefore the period is tau= pi right?

    Now isn't this the equation for amplitude: C=sqrt(B^2+A^2)

    My book says the period is pi/6 , rather than pi. I know from the principal root and Euler Formula's I know that 2(cos(pi/6)-i*sin(pi/6))= sqrt(3)-i. not sure what that has to do with the time since the equation for period is : tau= 2*pi/omega
     
  2. jcsd
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