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DiffyEQs - Undamped motion of spring

  1. Jan 26, 2012 #1
    1. The problem statement, all variables and given/known data
    A 1 kg mass is attached to a spring with constant k = 16 N/m. Find the motion x(t) in amplitude-phase form (2.37) if x(0) = 1 and x′(0) = −1.
    Ignore damping forces.


    2. Relevant equations
    combined with 3


    3. The attempt at a solution
    So I know m=1, k=16, c=0, x0=1, and v0=-1
    w=sqrt(k/m)=sqrt(16)=4
    md^2x/dt^2+kx=0
    characteristic equation
    mr^2+k=0
    r=+/-iw=+/-4i
    x(t)=c1*cos(4t)+c2*sin(4t) (e^0t = 1 and is excluded)
    initially x0=1 so
    1=c1
    v=x'=-4c1sin(4t)+4c2cos(4t)
    use intial value of -1 for v
    -1=4c2cos(0)=4c2
    c2=-1/4
    A=sqrt(c1^2+c2^2)=sqrt(17)/4
    tan(phi)=c2/c1=-.245, add pi =2.90
    so now we have
    x(t)=Acos(wt-phi)=sqrt(17)/2*cos(4t-2.90)

    the book says the answer should be sqrt(17)/16*cos(4t-6.038)
    so I all have right is the angular frequency
    where did I mess up?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 26, 2012 #2

    ehild

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    Hi Pi Face

    Tangent pi= -1/4 which is not -0.245.

    Find the inverse tangent before adding pi or 2pi.

    Look in the unit circle in which quadrant the angle is when the sine is negative, the cosine is positive.

    ehild
     
  4. Jan 26, 2012 #3
    Woops, sorry I missed a step, but the answer is still the same.
    tan phi=c2/c1
    phi=tan-1(c2/c1)=-.245
     
  5. Jan 26, 2012 #4

    ehild

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    OK, phi = -0.245±k(2pi). If the book says 6.038 as the phase constant it should be positive (-0.245+2pi.). The standard amplitude-phase form is Acos(wt+phi).

    ehild
     
    Last edited: Jan 26, 2012
  6. Jan 26, 2012 #5
    ah i see.

    but what about the amplitude?

    A=sqrt(c1^2+c2^2)=sqrt(17)/4, not sqrt(17)/16 right? could it be a typo? maybe its sqrt(17/16)?
     
  7. Jan 26, 2012 #6

    ehild

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    You are right, it is sqrt(17/16)=sqrt(17)/4. I did not notice.

    Never trust in the solutions of books....

    ehild
     
  8. Jan 27, 2012 #7
    Just one more question, and thank you for your help so far:

    I still don't understand exactly when to add pi/2pi to the phase angle. I know we want the phase angle to be between 0 and 2pi. Why did I have to add 2pi in this example and not just pi? Wouldn't have adding pi still led to a value within the 0-2pi range?

    On the quiz I had today, I ended up with something like tan^-1 (-1)=-pi/4. by plugging in -pi/4 for phi in sinphi and cosphi, I know that it is in the 4th quadrant. From here, I wasn't sure whether to add pi or 2pi, both would've resulted in values within the desired range. So between 3pi/4 and 7pi/4, I guessed 7pi/4. I know I got a 95 on the quiz so maybe it was 3pi/4 and I added one pi too many...but how do you know how much to add?
    Thanks.
     
  9. Jan 27, 2012 #8

    ehild

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    If you know both the sine and a cosine of an angle you exactly know in which quadrant it is. You give the angle+2pik as the whole solution.

    When the x,y components (ax=Acos(θ), ay =Asin(θ)) of a vector are given you can decide in which quadrant the angle is from the sign of the components. Draw the unit circle. (I always draw it in case of similar problems.) First quadrant (0,pi/2): both components positive. Second quadrant (pi/2, pi): ax<0, ay>0. Third quadrant(pi,3pi/2): both components negative. Fourth quadrant (3pi/2, 2pi): ax>0, ay<0. You need to add either 0, pi or 2pi to make the angle positive and place it into the appropriate quadrant.

    Example: A=5, ax=5cosθ, ay=5sinθ.

    ax=4, ay=3: tan(θ)=0.75, arctan(0.75)=0.6435, first quadrant, θ=0.6435

    ax=-4, ay=3. tan(θ)=-0.75, the angle is in the second quadrant. arctan(-0.75)=-0.6435, we need to add pi: θ=-0.6435+pi=2.498.

    ax=-4, ay=-3. The angle is in the third quadrant: tan(θ)=0.75,arctan(0.75)=0.6435, we add pi to place the angle into the third quadrant: θ=0.6435+pi=3.785.

    ax=4, ay=-3: fourth quadrant. tan(θ)=-0.75, arctan(-0.75)=-0.6435, we need to add 2pi to get a positive angle in the fourth quadrant. θ=-0.6435+2pi=5.6397.

    ehild
     
  10. Jan 27, 2012 #9
    Thanks for your help.

    Also I did get the phase angle right on the quiz, but on one of the problems on the last few steps solving for c2, I accidentally wrote down c2 as c1. Oh well, at least I know how to do them now.
     
  11. Jan 27, 2012 #10

    ehild

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    Silly mistakes are the worst... but they occur at everybody. Always check your test.

    ehild
     
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