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Homework Help: Simple Harmonic Motion homework

  1. Apr 18, 2007 #1
    I came across a physics problem for homework that I was able to work out on my own, however when I got to class, my teacher said that my numbers were not what the book had calculated, but couldn't tell if there was a flaw in my logic that would prevent me from getting the correct phi and therefore the correct answers.

    The question:

    A simple harmonic oscillator consists of a block of mass 4.20 kg attached to a spring of spring constant 200 N/m. When t = 4.00 s, the position and velocity of the block are x = 0.129 m and v = 3.415 m/s.

    My approach

    a.) What is the amplitude of the oscillations?

    I used conservation of mechanical energy to solve this part of the problem, and got the correct amplitude this way:

    (1/2)kx^2 + (1/2)mv^2 = (1/2)k(amplitude)^2

    amplitude = .5114 m

    b.) What was the position of the mass at t = 0 s?

    For this part, I plugged the position I was given and the time I was given as well as the amplitude and omega, which I solved for, into this formula:

    x(t) = amplitude * cos(omega * t + phi)

    I solved for omega:

    omega = (k/m)^(1/2)
    omega = 6.9 rad/s

    and used the amplitude from part a) and the given x and t to solve for phi

    .129 = .5114 * cos(6.9 * 4 + phi)
    .252 = cos(27.6 + phi)
    1.316 = 27.6 + phi
    phi = -26.28 rad

    then I use this phi in the x(t) function to solve for x at t = 0:

    x(0) = .5114 * cos(-26.28)
    x(0) = .21 m

    c.) What was the velocity of the mass at t = 0 s?

    I use the same phi, amp, and omega from before, except in the equation:

    v(t) = -omega * amplitude * sin(omega * t + phi)
    v(0) = -(6.9) * (.5114) * sin(-26.28)
    v(0) = 3.217 m/s

    When my teacher does the problem another way, however, he got a phi that was different from mine.

    Teacher's Approach

    v(4)/x(4) = [-omega * amplitude * sin(omega * t + phi)] / [amplitude * cos(omega * t + phi)]
    v(4)/x(4) = -omega * tan(omega * t + phi)
    26.473 = -6.9 * tan(6.9 * 4 + phi)
    -3.837 = tan(27.6 + phi)
    -1.316 = 27.6 + phi
    phi = -28.916 rad

    He then proceeded to solve for the amplitude using x(t) = amplitude * cos(omega * t + phi) and x(0) and v(0) in pretty much the same way that I did.

    My question is, why did using his method get the equation:

    -1.316 = 27.6 + phi
    phi = -28.916 rad

    when my way instead gave me:

    1.316 = 27.6 + phi
    phi = -26.28 rad
    ?

    In order to get the phi that my teacher got, I would have to alter the formula that I used to solve for phi to:

    x(t) = amplitude * cos(-omega * t - phi)

    which I don't think is correct. I would greatly appreciate any insight into this matter.
     
    Last edited: Apr 18, 2007
  2. jcsd
  3. Apr 18, 2007 #2
    very nice post. What I think happened, is in your soln, amplitude could be
    +/-.5114.
    Assume you chose the negative root,
    then -.252=cos(27.6+phi) which is same as since cos(pi-x)=-cos(x)
    hence .252=cos(pi-27.6-phi)
     
  4. Apr 18, 2007 #3
    Thanks for the clarification denverdoc, I guess neither me nor my teacher realized that by taking the sqrt we had a +/- .5114. For my approach, and not knowing the answer that my teacher got, is there any other way to tell that I should have used the negative amplitude instead of the positive one?
     
  5. Apr 18, 2007 #4
    maybe a wave guru can help out, technically speaking the teachers approach was the more rigorous approach since it used d(y(t)/y(t) along with a snap shot of conditions at t=4, but yours would have worked as well just from knowing the argument is usually given as wt+/-phi, and then testing the soln to make sure of the sign. Overall I think you get an A+ for creative approach. Again, very nice job.
     
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