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Solving for Amplitude and delta

  1. Apr 9, 2009 #1
    1. The problem statement, all variables and given/known data
    A simple harmonic oscillator has a frequency of 3.4 Hz. It is oscillating along x, where x(t) = A cos(ωt + δ). You are given the velocity at two moments: v(t=0) = 1.8 cm/s and v(t=.1) = -19.3 cm/s.

    1)Calculate A.
    2)Calculate δ.

    2. Relevant equations
    w= 2pi*f = 21.36 rad/s
    v(t) = -wAsin(wt +δ)


    3. The attempt at a solution

    1) v(0) = -21.36*A*sin(w0 +δ) =1.8 cm/s
    1.8/[-21.36*sin(δ)]=A
    Solved for A and substituted that into 2nd equation A
    v(.1) = -21.36*A*sin(21.36*.1 +δ) =-19.3 cm/s
    -21.36*(1.8/[-21.36*sin(21.36*0 +δ)])*sin(21.36*.1 +δ) =-19.3 cm/s
    1.8*sin(δ)*sin(.1 +δ)=-19.3 cm/s
    sin(δ)*sin(21.36*.1 +δ)=-10.72
    from here is where im not really sure what to do that 21.36*.1+δ throws me off
     
    Last edited: Apr 9, 2009
  2. jcsd
  3. Apr 9, 2009 #2

    rl.bhat

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    Homework Helper

    -21.36*A*sin( +δ) =1.8 cm/s .......(1)
    -21.36*A*sin(w*0.1 +δ) = - 19.3 cm/s.......(2)
    Divide (2) by (1). Expand sin(w*0.1 +δ) and simplify. You will get the value of delta. From that you can get amplitude.
     
  4. Apr 9, 2009 #3
    what exactly do you mean by expand sin(w*0.1 +δ) thats where i got lost last time
     
  5. Apr 10, 2009 #4

    rl.bhat

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    sin(A+B) = sinAcosB + cosAsinB
     
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