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Numberator or denometer?

  1. Jul 5, 2006 #1
    numberator or denometer??

    good lord. im working on a lesson about instruments. and the the formula has a bunch of different fractions (eg/ tension fraction, length fraction, density fraction, etc.)

    so they have one list of the "initial conditions" and then the other list of "final conditions".

    and there seems to be no logic for the putting the initial or final conditions into the fractions.

    like on question it'll suggest "since an increase in density results in a decrease in frequency, the density fraction will have the smaller density in the numerator"

    and then on another question "since a decrease in density results in an increase in frequency, the density fraction will have a larger density in the numerator"

    :confused:

    ~Amy
     
  2. jcsd
  3. Jul 5, 2006 #2

    Andrew Mason

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    It would be helpful if you could tell us what it is you are measuring and explain more clearly what your question is.

    AM
     
  4. Jul 6, 2006 #3
    so if the initial conditions are:
    fi = 147 Hz
    Fi = 289 N
    di = 1.40 mm
    li = 60 cm

    and then one of the questions says "find the final frequency (ff) if the tension (Fi) is reduced to 196N".
    whats the numerator, 289N or 196N?

    so i'd go 147Hz * (289/196)? or
    147 * (196/289)?

    ~Amy
     
  5. Jul 6, 2006 #4

    nrqed

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    I am *guessing* you are using the formula
    [tex] f_n = { n \over 2L} {\sqrt {T \over \mu}} [/tex]
    right??

    Let's say "A" refers to the initial values and "B" refers to the final values . Then write the equation for the initial values and the equation for the final and take the ratio of the two equations:
    [tex] {f_A \over f_B} = { n_A \over n_B} {L_B \over L_A} {\sqrt { T_A \over T_B}} {\sqrt {\mu_B \over \mu_A}} [/tex]
    If everything is kept the same except for the tension which is changed, then all the ratios give one except for the ratio with tensions so that gives
    [tex] {f_A \over f_B} = {\sqrt { T_A \over T_B}} [/tex]

    Hope this makes sense

    Patrick
     
  6. Jul 6, 2006 #5
    hi Patrick

    thanks, but im not sure what you mean :uhh:

    i just need to know whats the numerator, 196 or 289.

    ~Amy
     
  7. Jul 6, 2006 #6

    nrqed

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    Your initial frequency was what? It was 147 Hz, right? That's [itex] f_A[/itex]. What was your initial tension? It was 289 N, right? That's [itex] T_A[/itex]. What's your final tension? It's 196 N, right? That's [itex] T_B [/itex]. You want the final frequency, right? So you are solving for [itex] f_B[/itex]. Just plug everything in the equation and solve for the final frequency.

    (I am assuming that all the other variables are kept the same (same harmonic, same length, same linea mass density).

    (I showed you all the steps so that you could do any problem of that type. For example if it woul dbe the length that would be changed while keeping the same tension... or changing both the tension an dthe length, etc)

    Patrick
     
  8. Jul 6, 2006 #7
    update: i think i figured it out. according to my book "a decreased force = a decreased frequency"

    so they only way to come up with a decreased frequency answer is to the the 196N as the numerator.

    so my answer ended up being 99.695 Hz :biggrin:

    ~Amy
     
  9. Jul 6, 2006 #8
    Patrick, i think i see what you're saying now. but in my workbook it's different. for example, the Ta and Tb switch numerators/denometors depending on the question. each question is different and they dont follow a pattern.

    ~Amy
     
  10. Jul 6, 2006 #9

    nrqed

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    You should not have to rely on the book to see this if you know the equation. Look at my very first equation for the frequency. You should see that if the tension is decreased the frequency is decreased, indeed. It is clear from the *equation*
    You never told me if the equation I gave is what you are using (frequency of a vibrating string attached at both ends). That must be what you are using.

    If it is, then you answer is wrong. I gave you all the steps to get t he correct answer but I don't know if you paid any attention to what I wrote. In any case, everything is spelled out clearly in my previous posts so I won't add anything else.

    Best luck.

    Patrick
     
  11. Jul 6, 2006 #10

    nrqed

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    This is why I gave you the *general* equation! With the equation I gave you (and I showed where it comes from), just call A= initial situation and B= final situation and you can do *any* problem without having to memorize any rule about things being in the numerator or denominator! You just identify your variables and plug in the equation.

    From the very first equation it should be clear that an increase of tension gives an increase of frequency, an increase of length gives a decrease of frequency and so on. But that's not enough to do calculation, you need to know how things depend on each other. For example the frequency is proportional to the *square root* of the tension, not to th etension itself (which is what you assumed when you got an answer of 99 Hz).
    The formula with the ratios I gave will solve *any* problem, even those where *two* variables or more are varied.

    Patrick
     
  12. Jul 6, 2006 #11
    nevermind my prior post about 99Hz. i forgot to square root it. :blushing:

    so the correct answer should be 12,75 Hz?

    ~Amy
     
  13. Jul 6, 2006 #12

    nrqed

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    Did you mean 127.5??
    In any case, that's not right
     
  14. Jul 6, 2006 #13

    nrqed

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    Did you mean 127.5??
    In any case, that's not right
     
  15. Jul 6, 2006 #14
    Patrick

    im looking over all the sample questions in my workbook, and they all have the final tension (Tb) (in newtons) as the numerator..

    ~Amy
     
  16. Jul 6, 2006 #15
    i meant 12.75 Hz. but according to your equation, the answer should be 0.0867 Hz, but that doesnt sound right either. :eek:

    ~Amy
     
  17. Jul 6, 2006 #16
    according to what i've pieced together, the equation for everything is:
    ff = fi * (a square rooted tension fraction) * (a length fraction) * (a dimeter fraction) * (a density fraction)

    so

    ff = fi * (Ff/Fi square rooted) * (li/lf square rooted) * (di/df) * (pi/pf square rooted).

    sorry for aggravating you, but my workbook is contradicting your formula.

    ~Amy
     
  18. Jul 6, 2006 #17

    nrqed

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    How did you get that with my equation?:surprised
    Show me how you got this from my equation?:uhh:

    If you read my post #4 and my post #6, I gave all the numbers (I even identified all the variables) and the formula. You just have to plug in.

    Patrick
     
  19. Jul 6, 2006 #18

    nrqed

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    No, it's not contradicting my formula. Do you know what [itex] \mu [/itex] means in my formula? The "p" in your formula is, I assume, the volume density [itex] \rho[/itex], right?

    Before saying that two formula are different, you should express them in terms of the same variables! The linear mass density [itex] \mu [/itex] is given by the cross sectional area of the wire times the volume mass density so [itex] \mu = \pi {d^2 \over 4} \rho [/itex]. Therefore,
    [tex] {\sqrt {\mu_i \over \mu_f}} = {\sqrt {\rho_i \over \rho_f}} {d_i \over d_f} [/tex]
    and you can check that my eqaution will agree with yours *except* that your equation does not contain the harmonic number "n" so it's your equation wich is incorrect, unless they always look at the *special* case of keeping the same harmonic. So my equation is more general.

    It's important to understand equations and the meaning of the terms instead of just trying to get a quick answer by plugging in things.

    Regards

    Patrick
     
  20. Jul 6, 2006 #19
    but why do i need the density fraction (p) and what not? based on the question, it sounds like the only variable is the tension, so wouldnt everything else just = 1? so why even plug in the other numbers?

    im just trying to find the tension in newtons.

    im going to ask the teachers who run this course and see what they say.. i'll let you know
    :smile:
    thanks

    ~Amy
     
  21. Jul 6, 2006 #20

    nrqed

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    I did not introduce the volume density, YOU did!!:cry:
    The only reason why we talked about [itex] \rho [/itex] is that you said that the equation of your book does not agree with mine. I then showed why my equation is the same as in the book.
    That was a side issue. Going back to the specific question you askes, of course you don't need the volume density because it cancels out in the ratio.
    You are welcome. But I already provided everything you need to get the answer in my posts 4 and 6. Everything is spelled out very clearly there. I don't know why you don't use those two posts to get the answer you are looking for.

    Best luck.

    Patrick
     
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