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Determining angular frequency and amplitude for SHM

  1. Oct 6, 2016 #1
    1. The problem statement, all variables and given/known data
    A mass "m" is attached to a spring of constant "k" and is observed to have an amplitude "A" speed of "v0" as it passes through the origin.

    a) What is the angular frequency of the motion in terms of "A" and "v0"?

    b) Suppose the system is adjusted so that the mass has speed "v1" at position "x1" and speed "v2" at position "x2". Find the angular frequency, "ω0", and amplitude "A" in terms of these given quantities. Simplify your results.

    2. Relevant equations
    x = A*sin(ω0*t -φ)

    v = A*ω0*cos(ω0*t -φ)

    3. The attempt at a solution
    so for the first part

    x(t0) = A*sin(ω0*t0 -φ) = 0

    therefore φ = ω0*t0

    v(t0) = A*ω0*cos(ω0*t0 -φ) = v0

    therefore ω0 = v0/A

    for part b.)

    x1 = A*sin(ω0*Δt1)
    v1 = A*ω0*cos(ω0*Δt1)

    x2 = A*sin(ω0*Δt2)
    v2 = A*ω0*cos(ω0*Δt2)

    where Δt1 = t1 - t0
    and Δt2 = t2 - t0


    theres four unknowns and four equations so I know I should be able to solve for ω0 and A, but I can't seem to untangle all these trig functions. any hints?
     
    Last edited: Oct 6, 2016
  2. jcsd
  3. Oct 6, 2016 #2

    gneill

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    Staff: Mentor

    I suggest taking a look at an energy conservation approach. Both the KE and PE expressions square their variable so position and velocity directions don't matter there.
     
  4. Oct 6, 2016 #3
    Ok so using conservation of energy

    KE1 + PE1 = KE2 + PE2

    1/2*m*v12 + 1/2*k*x12 = 1/2*m*v22 + 1/2*k*x22

    canceling the halves and dividing by "m" on both sides

    v12 + ω02*x12 = v22 + ω02*x22

    then rearranging for ω0
    ω0 = sqrt [ (v12 - v22)/(x22 - x12)

    for the amplitude I use the fact that the total energy at any time is equal to 1/2*k*A2

    so KE1 + PE1 = 1/2*k*A2

    plugging in and rearranging I get

    A = sqrt [ x12 + v12 * (x22 - x12)/(v12 - v22) ]

    does this look okay?
     
    Last edited: Oct 6, 2016
  5. Oct 6, 2016 #4

    gneill

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    Staff: Mentor

    That looks good.

    For the amplitude you could also have chosen just one of the given (x,v) pairs to find the total energy. The resulting expression might be a bit cleaner...
     
  6. Oct 6, 2016 #5
    I don't quite understand what you mean. I chose just x1 and v1 but x2 and v2 factor in because I had to divide by the mass again to get the angular frequency since they didn't want the answer in terms of "k"
     
  7. Oct 6, 2016 #6

    gneill

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    Staff: Mentor

    You're right, I missed that stipulation. Mind you, if you find ##\omega_o## first, you should be able to use that in the amplitude expression. You should be able to reach:

    ##A = \sqrt{\left(\frac{v}{\omega_o}\right)^2 + x^2}##
     
  8. Oct 6, 2016 #7
    Yes, that's the expression that I arrived at before I plugged in the corresponding values. Thanks for the help :)
     
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