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[Calc] First max and min values of an underdamped oscillation

  1. Mar 21, 2017 #1
    1. The problem statement, all variables and given/known data
    Determine the FIRST maximum and minimum values of the underdamped oscillation:

    y=e^(-x/2)(4sin(3x)+3cos(3x)) cm

    2. Relevant equations
    3. The attempt at a solution

    I firstly differentiated the above equation and got:

    (-e^(-x/2)(22sin(3x)-21cos(3x)))/2

    I checked this and it is correct.

    Next I set this equation to = 0

    I crossed off the exponential function (it can't equal 0?) and the /2 on the bottom.

    22sin(3x)-21cos(3x)=0

    21=22tan(3x)

    tan(3x)=21/22

    tan^(-1)(21/22)=0.7621

    I plotted the graph online and checked this point, but it's definitely not a maximum or minimum, so I don't really understand; I thought that I could put dy/dx = 0 and that would tell me the points.

    Also, this has only given me one point. How will I find the next point? Tan^(-1)(a number) only gives one definitive value.

    I know that I can take the second derivative and that will tell me if the point is a maximum or minimum (+for minimum,-for maximum).

    Can anyone help me understand how max and min works for this kind of equation? The oscillation gradually gets smaller, which I haven't ever done before.

    Thank you!
     
  2. jcsd
  3. Mar 21, 2017 #2

    DrClaude

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    Staff: Mentor

    Be careful here. Write an equation for x.


    What is the periodicity of tan?
     
  4. Mar 21, 2017 #3
    Sorry that should be the value of 3x, so x=0.7621/3 = 0.254

    And that does appear to be the first maximum!

    Then I could prove that it is a maximum by taking the second derivative.

    Tan has a period of Pi radians... Maybe I'm being stupid but I don't know how I find what the period is in terms of x.

    When I do work it out though, I add half of one period to 0.254?

    Edit:

    tanx= tan(x+n*π)

    So I put my value of x in here, but which number do I use for n?

    I tried tan(0.254+(0.5*π)) but that wasn't right. n is an integer, but which integer?
     
  5. Mar 21, 2017 #4

    Dick

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    Think about it. You've got ##3x+n\pi=\arctan(\frac{21}{22})##. You've got the ##n=0## solution. What value of ##n## is going to give you the next largest value of ##x##?
     
  6. Mar 22, 2017 #5
    Well l would think it would be n=0.5, seeing as 1π is one period of tan, and I am adding half of a wave length?
     
  7. Mar 22, 2017 #6

    Dick

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    You already said that doesn't work and it gives you a smaller value for ##x##. ##n## needs to be an integer. Hint: ##n## can be negative.
     
  8. Mar 22, 2017 #7
    I found the right answer. I did (3x+1π)/3

    I'm still confused as to how this works though, because looking at the graph that I plotted of y=e^(-x/2)(4sin(3x)+3cos(3x)) it is only half of a wavelength from Maximum to Minimum, so why am I adding 1π, which is one whole wavelength of tan?
     
  9. Mar 22, 2017 #8

    Dick

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    You are confused because you appear to be simply guessing. Your starting point should be ##3x+n\pi=\arctan(\frac{21}{22})##.
    That's the equation that describes your extrema. You found the first one by setting ##n=0##. If you set ##n=1## and solve for ##x## you will find that ##x## is negative, right? That's not it. How about trying ##n=-1##??
     
  10. Mar 22, 2017 #9
    I do understand the equation, so:

    x = (arctan(21/22) + 1π)/3

    I'm just stuck with how to know what n equals.

    I know this is dumb because it's probably the simplest part of the whole task, thanks for being patient.
     
  11. Mar 22, 2017 #10

    Dick

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    That's ok. You may be looking for some deep answer. If you look at ##3x+n\pi=\arctan(\frac{21}{22})## you can see as ##n## increases, ##x## decreases. If ##n## decreases then ##x## increases. The smallest positive solution was at ##n=0##. So the next positive solution will be at ##n=-1##. The next one would be ##n=-2## etc etc. That's all there is to it.
     
  12. Mar 23, 2017 #11
    Ok thank you! I guess I was looking into it too much.
     
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