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Maddening problem on dampened harmonic motion

  1. Dec 24, 2009 #1
    Reviewing for the MCAT this Christmas, most of physics is going okay except I run into a few areas I was never familiar with to be begin with.

    1. The problem statement, all variables and given/known data

    By what percentage does the frequency diminish in a horizontal spring mass system where k=1x10^3 N/m and m= 4 kg if the motion is dampened by a frictional force that has a damping coefficient of 2 kg/s?

    k=1000 N/m
    m = 4 kg
    W(d)=2 kg/s

    2. Relevant equations
    w=sqrt(K/m)
    A=A(0)e-(b/2m)t
    w(d)=sqrt(w^2-(b/2m)^2)


    3. The attempt at a solution

    Found omega by taking the square root of 1000/4, which is 15.81. From there I plugged that into my third equation, to give me b=126.12.

    Here is where I'm stuck and have been staring at this for over an hour. I know the solution is 0.0125% so I used that to work backward:

    0.0125A(0)=A(0)e-(126.12/2*4)*t

    and found t to be 0.57 s.

    Thus, apparently if you solve for t you can solve for the percentage. But how the heck can you solve for t? No other harmonics equation gives me that number.


    Uhhhhgggg I never studied damped motion in HS OR college....google isn't helping, can someone else?

    EDIT:

    Okay. Just realized that if it's damped by 0.0125%, then I formatted my equation wrong to find time backwards. It should be

    .999875(0)=A(0)e-(126.12/2*4)*t
    t=7.93e-6

    So this is the proper format I believe, however I'm no closer to solving for time without working backwards!! Help!
     
    Last edited: Dec 24, 2009
  2. jcsd
  3. Dec 24, 2009 #2
    Man the MCAT must have gotten a lot tougher since I took it 25 years ago. What source are you using for the review?

    Like many of the MCAT problems, it is likely trying to fake you out. As far as I know the frequency doesn't change, only amplitude which is an exponential decay.
     
  4. Dec 24, 2009 #3
    But it does. The answer is a 0.0125% change. I just don't understand ~how~ they arrived at that answer. Kaplan physics review.
     
  5. Dec 24, 2009 #4

    ehild

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    ProvidenceNT: you need to know the meaning of the symbols you use. What is the meaning of "b"? What is w(d)? Is not it the angular frequency of the damped oscillator? What is the unit of frequency? Is it kg/s? What is the 2 kg/s?


    denverdoc:

    The frequency of a damped oscillator is different from the one which is free of damping. The damping force is proportional to the velocity and opposite in direction. So the equation of motion is:

    ma = -kx-bv, or

    [tex]x''+(b/m)x'+(K/m)x=0[/tex]

    find the solution in form:

    [tex]x=exp(\lambda t) \rightarrow \lambda^2+(b/m) \lambda +(K/m)=0\rightarrow \lambda= -b/(2m) \pm j\sqrt{K/m -(b/(2m))^2}[/tex]

    The damping factor is b/2m, and the angular frequency is

    [tex]\omega =\sqrt{K/m -(b/(2m))^2}[/tex]

    ehild
     
  6. Dec 24, 2009 #5
    Thanks for the reply. Yeah, I should have probably specified the meanings of the units. I have a vague understanding...I understand that I can find the angular frequency (omega) then use that to find b, which is a constant that describes the damping coefficient used in the equation.

    Still a bit fuzzy about this problem though....How can you find a "percentage" of the frequency diminishing without specifying time? Doesn't it decay with time?

    I understand that I believe you just tried to answer my question, however that explanation ran a bit over my head. The only equation I've even seen is your last one, which describes the damped angular frequency.
     
  7. Dec 25, 2009 #6

    ehild

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    Read the text of the problem:

    "the motion is dampened by a frictional force that has a damping coefficient of 2 kg/s"

    Why do you think that you need to

    "find b, which is a constant that describes the damping coefficient"?

    is not it there?

    The frequency does not change with time, but it changes with respect to the undamped frequency.

    ehild
     
    Last edited: Dec 25, 2009
  8. Dec 25, 2009 #7
    I hate to sound totally ignorant, but I thought the purpose of finding b was just to use it in the other equation, the one that describes the dampened amplitude that does include time as a factor.

    Actually, I just thought of something that has left me even more confused. I understand that only the amplitude is diminishing with time. It makes perfect sense but I didn't think the frequency changes (again, this material is new to me in this review book and was never in my coursework).

    But if the frequency doesn't change...why is the question asking me "By what percentage does the frequency diminish"? Since the frequency does not change with respect to time?? Yet the answer implies it ~does~ change. And so does the equation.

    For amplitude, A=A(0)e-(b/2m)t

    So a larger damping constant will decrease the amplitude faster. And this is with respect to time since time is also a factor in that equation.

    But how is this relating to frequency at all? That's the connection I'm sorry I'm still not quite getting. I don't conceptually understand how the frequency can change, nor do I understand mathematically how the answer is 0.0125%. In the meanwhile I'm moving my review onto optics but I'm keeping a tab on this page in case you still have the patience to explain this to me.
     
  9. Dec 25, 2009 #8
    The question asks you to compare the damped frequency with the undamped frequency.

    [tex]\omega[/tex] is the angular frequency of the oscillation.
    I recommend the following lecture if you want a bit more insight into damped harmonic oscillators, especially to know the names and units of everything:
    http://ocw.mit.edu/OcwWeb/Physics/8-03Fall-2004/VideoLectures/detail/embed22.htm [Broken]
     
    Last edited by a moderator: May 4, 2017
  10. Dec 25, 2009 #9
    I was confused by the wording as well until rereading: you are right that the frequency doesn't change with time in either case, just as Ehild states; it is comparing the natural frequency of the system with and w/o dampening that the question is seeking.

    Maybe one way to think is a pendulum--the period (and hence frequency) will be very different if the bob of the pendulum is in air or water. Or a spring-mass system on ice versus sandpaper. In this case the dampening is small.

    BTW, I'd be surprised if you see anything this detailed on the MCAT--for one you don't have a calculator! Best of luck.
     
  11. Dec 25, 2009 #10

    ehild

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    Imagine you have a spring with spring constant K, and attach a mass m to it, and pull the mass out, and release: the mass will vibrate with the angular frequency
    [tex]\omega =\sqrt{\frac{K}{{m}}[/tex]

    Now you put the whole thing into water or honey or into any medium which damps the motion of the mass: it will still vibrate unless the damping is to high, but with lower frequency . The spring is the same, the spring constant the same as before, the damping is given, can you plug in the value of K=1000 N/m, m=4 kg and b=2 kg/s into the formula that you you have written in your first post:

    [tex]\omega_d =\sqrt{\frac{K}{m}-(\frac{b}{2m})^2}[/tex]?
     
  12. Dec 25, 2009 #11
    By the way, what were the multiple answers for this question?

    If none were on the scale of 0.0125% then you could've gone straight for that answer.
    For systems that still oscillate when damped (If the damping is too great, sometimes they won't oscillate), the ratio of [tex]\frac{\omega _d}{\omega}[/tex] is very very close to one.

    If, however, you had several answers on that scale, then I think the method they wanted you to go for is the following:

    [tex]\Delta \% = 100(1-\frac{\omega_d}{\omega_{natural}})[/tex]

    Some algebra:

    [tex]\Delta \% = 100(1-\sqrt{1-\frac{b^2}{4mk}})[/tex]

    Taking the binomial approximation:

    Where [tex]x<<1[/tex]

    [tex](1+x)^a\approx 1+ax[/tex]

    We find (In this case [tex]a[/tex] is [tex]\tfrac{1}{2}[/tex]):

    [tex]\Delta \% \approx 100(1-(1-\frac{b^2}{8mk}))[/tex]

    [tex]\Delta \% = \frac{100\cdot b^2}{8mk}[/tex]

    [tex]\Delta \% = \frac{100}{8}\cdot 10^{-3}[/tex]

    The numbers are very reasonable for the question to be done by hand without a calculator.

    Best of luck with the exam!

    Heh, denverdoc, I wish I knew. I'm from another country altogether!
     
    Last edited: Dec 25, 2009
  13. Dec 25, 2009 #12
    This may be but I would be very surprised to see such a question on the MCAT assuming this level of mathematical sophistication and a knowledge of physics deeper than a non-calculus physics I/II course (unless the equations were provided).

    Unless the test has really, really changed. I haven't seen any exam questions for several years so a bit of a guess on my part, and 20+ since I took it.
     
  14. Dec 25, 2009 #13

    ehild

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    So MCAT is some kind of intelligence test, to check the ability of the student to read and understand the problem, be familiar with very basic Science and Maths formulas, and estimate numerical values? Then no use of trying to explain how to solve the differential equation of a damped oscillator, this problem was percent calculation...

    ehild
     
  15. Dec 25, 2009 #14
    ....wow.

    There were no multiple choice answers here. It was only a question, with the correct answer in the back of the book. The "high yield" sections of this book deal with more calculation-based concepts that are supposed to reinforce the concepts previously introduced.

    Thank you for that ~highly~ detailed explanation. My girlfriend is a physics major and when she comes back in town I'll ask her to go over some of what you posted to clarify a few things (I don't want to keep bugging you with minor details).

    Thanks again everyone! And I'll be bookmarking the YT link...
     
  16. Dec 25, 2009 #15
    Yea,I think thats a pretty good assessment. On one hand a survey of science knowledge in bio, chem, and physics and then an assessment of aptitude/problem solving ability where data may be provided along with formulae possibly looking at analytic skills.
     
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