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Giving values to angular velocity

  1. May 31, 2015 #1
    1. The problem statement, all variables and given/known data
    Solve: A*sin(ωt + Θ) = L*i''(t) + R*i'(t) + (1/C)*i(t). Where: A=2, L = 1, R=4, 1/C = 3 and Θ=45°.

    2. Relevant equations
    The system has to be solved by i(t) = ih + ip. I gave the values to A, L, R, 1/C and Θ. I can also give values to ω, but I've come to a doubt when solving the particular part. The teacher said that the initial conditions should not be 0. So I set Θ to 45°, that way when t=0, A*sin(ωt + 45°) = 1.4.

    3. The attempt at a solution

    My homogeneous equation is:
    ih = C1eλ1t + C2eλ2t

    Where λ1 = -1 and λ2 = -3

    My particular solution is ip = a1*sin(ωt) + a2*cos(ωt). So i'p = ωa1*cos(ωt) - ωa2*sin(ωt) and i''p = -ω2a1*sin(ωt) - ω2a2*cos(ωt).

    2*sin(ωt) = i''(t) + 4i'(t) + 3i(t) = -ω2*(a1sin(ωt) + a2cos(ωt)) + 4ω(a1cos(ωt) - a2sin(ωt)) + 3(a1sin(ωt) + a2cos(ωt))

    2 = a1(3-ω2) + 4ωa2
    0 = a2(3-ω2) + 4ωa1

    a1 = (2 - 4ωa2)/(3-ω2)
    0 = a2((3-ω2)2 - 16ω2) + 8ω

    I could solve for a2 and then for a1 but I don't know that's the correct solution. I don't know what to do next, please help. Also, I would appreciate if anyone could recommend a book to read about differential equations.

    Please let me know if you need more information or if I'm not making myself clear enough.

    Thank you in advance.
     
  2. jcsd
  3. Jun 1, 2015 #2

    Simon Bridge

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    Wait - you chose the value of theta? What lead you to pick that particular value?
    Which of the other values did you pick and which are provided in the original problem statement?
    (Aside: angles should be in radians).

    Please provide the problem statement as it was given to you - then talk about your reasoning process.

    You have the equation ##A\sin(\omega t+\theta) = L\ddot\imath(t) + R\dot\imath(t) +\frac{1}{C}\imath(t)## ... which looks like the usual RCL circuit DE: have you tried looking it up online?
    http://en.wikipedia.org/wiki/RLC_circuit#Series_RLC_circuit

    You won't know until after you've found them.
     
  4. Jun 1, 2015 #3

    vela

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    Your teacher probably meant not to assume i(0) = 0 and i'(0) = 0.

    Why'd you stop? It's easy enough to see if you got the right solution. Plug it back into the original differential equation and see if it works.

    You can save yourself from doing some algebra by using Cramer's rule to solve for ##a_1## and ##a_2##.
     
  5. Jun 1, 2015 #4

    Ray Vickson

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    You are heading for disaster if you mix up angular measures like you have done. Calculus on trigonometric functions is always done with angles measured in radians, not degrees. So, wherever you see degrees in your input you should convert first to radians before doing anything else. After all the calculations are done, you can convert the angular velocities from rad/sec to degrees/sec, or rpm (revolutions per minute), or Hertz, or any other system of measurement you want----whatever is convenient and acceptable in the field of study from which the problem arises.
     
    Last edited: Jun 1, 2015
  6. Jun 1, 2015 #5
    Thank you all for answering. It was my mistake to mix radians and degrees. Please assume that Θ = π/4. The teacher gave us this equation:

    [tex]A*cos(\omega t)=\frac{md^2x(t)}{dt^2}+\frac{bdx(t)}{dt}+kx(t)[/tex]

    With
    [itex]x(0) = X_0\neq 0[/itex]
    [itex]\frac{dx(0)}{dt} = V_0\neq 0[/itex]

    We need to set an input (which I set to ##2*sin(\omega t + \theta)## and assign real and positive values to m, b and k. As I'm more familiar to the electrical equivalent, I changed it to L, R and ##\frac{1}{c}##.

    I set Θ = π/4 because if it was 0, the condition
    [itex]x(0) = X_0\neq 0[/itex]
    would be violated. It is a random value that I chose.

    He told us to use

    [tex]X(t)=X_h(t)+X_p(t)[/tex]

    I found the homogeneous equation:
    [itex]X_h=C_1e^{-t}+C_2e^{-3t}[/itex]
    and I've chosen my particular solution:
    [itex]X_p=2sen(\omega t + \frac{\pi}{4}) + 0*cos(\omega t + \frac{\pi}{4})[/itex]

    But I just don't know how to continue. Am I even going through the correct procedure? I'm looking for examples online at the moment with very little success.

    Thanks again.
     
    Last edited: Jun 1, 2015
  7. Jun 1, 2015 #6

    SteamKing

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    Everything looks OK so far.

    You use the initial conditions to find the values of C1 and C2. As a check, once this task is done, you can substitute the general solution back into the original DE and see if it is satisfied. That's how you check your work. :wink:
     
  8. Jun 1, 2015 #7
    Thank you, it is good to know that I'm on the track.

    I made a mistake on the particular solution part, I chose it to be:
    [itex]x_p(t)=a_1sin(\omega t + \frac{\pi}{4})+a_2cos(\omega t + \frac{\pi}{4})[/itex] so:
    [itex]\frac{dx_p(t)}{dt}=\omega a_1cos(\omega t + \frac{\pi}{4})-\omega a_2sin(\omega t + \frac{\pi}{4})[/itex]

    [itex]\frac{d^2x_p(t)}{dt^2}=-\omega^2 a_1sin(\omega t + \frac{\pi}{4})-\omega^2 a_2cos(\omega t + \frac{\pi}{4})[/itex]

    Then:

    [itex]2sin(\omega t + \frac{\pi}{4})=\frac{d^2x(t)}{dt^2}+4\frac{dx(t)}{dt}+3x(t)=-\omega^2[a_1sin(\omega t + \frac{\pi}{4})+a_2cos(\omega t + \frac{\pi}{4})]+4\omega[a_1cos(\omega t + \frac{\pi}{4})-a_2sin(\omega t + \frac{\pi}{4})]+3[a_1sin(\omega t + \frac{\pi}{4})+a_2cos(\omega t + \frac{\pi}{4})][/itex]

    [itex]2=-\omega^2a_1-4\omega a_2+3a_1[/itex]
    [itex]0=-\omega^2a_2+4\omega a_1+3a_1[/itex]
    [itex]a_1=\frac{2-4\omega a_2}{3-\omega^2}[/itex]
    [itex]a_2=\frac{8\omega}{-\omega^4+22\omega^2-9}[/itex]

    And this is where I'm stuck, should I just give a value to omega and solve it? If I do I would get rad/seg at different powers and I won't be able to add them and get to a single value. I don't know what to do next. Please help.

    Thank you very much in advance.
     
  9. Jun 2, 2015 #8

    vela

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    I disagree that everything looks fine. First, you are given an input to the system. That's what the ##A \cos \omega t## is. Rewriting it as ##A \sin(\omega t+\theta)## only makes you write more. Second, your reasoning is flawed. Just because the input is 0 at t=0 doesn't mean that ##X_0## and ##V_0## are automatically 0. In fact, you could have no forcing function, i.e., input=0, and still satisfy non-zero initial conditions. Arbitrarily setting ##\theta=\pi/4## doesn't guarantee anything about the initial conditions.

    The constants depend on the parameter ##\omega##. You don't get to arbitrarily choose a value for ##\omega##. Just leave everything in terms of ##\omega##.
     
  10. Jun 2, 2015 #9
    Thanks for answering.

    Yes, but we were told to change the input to one of our choosing. And I chose sin() to have a simple input.

    So, if I leave Θ=0 and without a forcing function, I could still do something like:

    [itex]x(0)=C_1e^{-0}+C_2e^{-3(0)}=C_1+C_2[/itex] and with an input=0, C1=-C2, right?

    How? I don't understand how ##2sin(\omega t)## at t=0 can give ##X_0\neq 0## or were you referencing the cosine input? I'm almost sure that I'm not understanding properly what the term "Initial conditions" is and which part of the system it acts on. I assume that any book of control of dynamic systems can help me understand that, right?
     
  11. Jun 2, 2015 #10

    vela

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    You have ##x(t) = x_h(t) + x_p(t)##, right? The initial conditions are
    \begin{align*}
    x(0)=X_0 = x_h(0) + x_p(0) \\
    \dot{x}(0)=V_0 = \dot{x}_h(0) + \dot{x}_p(0).
    \end{align*} They simply specify the state of the system at t=0 — where the mass is and its velocity. If there's no forcing function, you have ##x_p(t) = 0##, so the initial conditions become ##x_h(0) = X_0## and ##\dot{x}_h(0)=V_0##. For this particular problem if there were no input, you'd end up with
    \begin{align*}
    C_1 + C_2 &= X_0 \\
    -C_1-3C_2 &= V_0.
    \end{align*} Clearly, you can find values of ##C_1## and ##C_2## to satisfy non-zero initial conditions. Non-zero initial conditions don't require a forcing function at all.
     
  12. Jun 2, 2015 #11
    It is all clear now. Thank you very much.
     
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