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Rearranging equation

  1. Jan 12, 2015 #1
    1. The problem statement, all variables and given/known data
    The rate at which the mass–spring system loses energy to its surroundings is referred
    to as the Q-value for the oscillator. The Q-value is defined as:

    Q=2π(E/ΔE)

    I need to find ΔE/E, which is the fractional energy loss per cycle of the oscillation.

    How would I rearrange this so that ΔE/E = 2π/Q

    2. Relevant equations
    Q=2π(E/ΔE)

    3. The attempt at a solution
    I've no idea how to rearrange it so that E/ΔE is swapped round
     
  2. jcsd
  3. Jan 12, 2015 #2
    $$Q=2\pi(\frac{E}{\triangle E})$$
    then
    $$\frac{Q}{2\pi} =(\frac{E}{\triangle E})$$
    $$\frac{2\pi}{Q}=(\frac{ \triangle E}{ E})$$
     
    Last edited: Jan 12, 2015
  4. Jan 12, 2015 #3
    Hi Maged, thanks for your reply. Apologies, this is all new to me. What do you mean when you say /frac?
     
  5. Jan 12, 2015 #4
    I have tried to write my answer in Latex preview
    to make those staff clear for you ,,
    But it seems there is something went wrong

    any way ,,

    I have written it again

    $$Q=2\pi(\frac{E}{\triangle E})$$
    $$\frac{Q}{2\pi}=(\frac{E}{\triangle E})$$
    $$\frac{2\pi}{Q}=(\frac{\triangle E}{E})$$
     
  6. Jan 12, 2015 #5
    What it looks like now
    is it understandable ?
    (:
     
  7. Jan 12, 2015 #6
    I have fixed my problem though,
    Reload the page ,
     
  8. Jan 12, 2015 #7
    Indeed it is, thanks very much. Appreciate your help.
     
  9. Jan 12, 2015 #8

    NascentOxygen

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    Now, take the reciprocal of each side ...
     
  10. Jan 12, 2015 #9
    I have missed this point ,, it should be there ,,,

    :)
     
  11. Jan 12, 2015 #10

    BvU

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    Did you read the later postings in your thread 79065 ?

    Point is that the 'definition' doesn't hold unless Q is very large.
    Energy in an oscillator has to do with amplitude squared (e.g. with a spring: E = 1/2 k x2).

    A damped oscillation is represented (see here ) by $$ z(t) = A e^{-\zeta\omega_0t}\; sin \left ( \sqrt{1-\zeta^2}\; \omega_0t + \phi \right )$$ and with

    ##\zeta \ll 1## so ##T \approx {2\pi\over \omega_0}## for ##\ \left( z(t+T)\right )^2 / \left( z(t)\right )^2\ ##

    you get ##\ e^{-2 \zeta\; 2\pi} = e^{-2\pi / Q}\ ## if ##\ Q={1\over 2\zeta}##.

    Again with Q large enough, this is ##1-{2\pi \over Q}\ .\ \ ## So ##{\Delta E\over E} = {2\pi \over Q}##


    Now to your question: how to rearrange Q = 2π(E/ΔE) to get ΔE/E = 2π/Q :

    multiply left and right by ΔE
    divide left and right by E
    divide left and right by Q

    but is this really what you were asking :nb) ?
     
  12. Jan 12, 2015 #11
    Thanks
    Indeed, it was what I was asking. With reference to the earlier post, it turns out the graph was difficult to read and I was advised to read it so that it fitted the equation.

    Thanks for your reply though, appreciated,
     
  13. Jan 12, 2015 #12

    NascentOxygen

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    Good catch! Ryan McDonald has yet to answer my question in my last post there. :(

    Hopefully he will find time to do so shortly.

    That restriction wasn't clear to me in that thread. Discussion wasn't exhaustive, but I concluded that in a system showing stable oscillation, ΔE would represent the energy deficit the system must add during each cycle to maintain constant amplitude of oscillation, regardless of Q. Are you saying this would not be correct unless Q is "large"?
     
  14. Jan 12, 2015 #13

    BvU

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    @NascentOxygen: I've been looking at this thing as an undriven oscillator all the time. Mainly because of the way the OP in thread Calculating energy left in oscillator after 3 cycles was formulated.

    However, in post #5 there it becomes clear that Ryan found his Q from the resonance curves, so what rude man states in his post #15 applies.

    I must admit I have a hard time (=lazy me) coming up with a simple and elegant derivation for ΔE/E in the case of a driven damped oscillator.

    @Ryan McDonald: 790685 post #13 should still be OK


    @moderator: this whole thread should be appended to (or referred to in) thread 790685
     
  15. Jan 12, 2015 #14
    Thanks BvU. I found the question to be irritating and extremely ambiguous. Anyways, hopefully I've arrived at the correct solution now.
     
  16. Jan 12, 2015 #15

    BvU

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    If you share it we can look at it -- and perhaps comment :)

    There's no reason to find the question irritating or ambiguous if there is a context.

    And: any investment in this subject pays off for a lifetime in a vast number of disciplines, so cheer up!
     
  17. Jan 12, 2015 #16

    NascentOxygen

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    I'm unsure whether fo/b.w. can be equated to Q for low Q's. I'd like to see that demonstrated as being true; meanwhile I'm regarding it as an approximation which holds for high Q's. After all, I expect it's only going to be exact for 2nd order systems, in any case.
     
  18. Jan 12, 2015 #17
    Ok, here is my attempt.

    I've taken Q to be 7 for plot a and 10 for plot b.

    Rearranging Q=2π (E/ΔE)

    ΔE/E = 2π/Q

    For plot a:

    After one cycle 2π/Q = 0.1

    After two cycles (0.9*0.9) = 0.01

    After three cycles = (0.81*0.9) = 0.001


    For plot b:

    After one cycle 2π/Q = 0.37

    After two cycles = (0.63*0.63) = 0.23

    After three cycles = (0.4*0.63) = 0.15

    I'm sure there must be a more elegant way of calculating the energy left after 3 cycles but unfortunately, it evades me.

    Thoughts welcome

    Would it be 1-(2π/7)n for plot a? Which would show the amount of energy lost after n cycles
     
    Last edited: Jan 12, 2015
  19. Jan 12, 2015 #18

    BvU

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    Are these still from the picture 790685 post #5 ?
    In that case I expect a to have a higher Q than b !!!

    And what is ##\Delta\omega/\omega## ?

    This one says "the bandwidth over which the power of vibration is greater than half the power at the resonant frequency, ωr = 2πfr is the angular resonant frequency, and Δω is the angular half-power bandwidth" but you are looking at an amplitude plot !

    Then: your calculation: if q = 7, 2π/Q = 0.9, so after 1 cycle there is only 0.1 left. After 2 cycles 0.01 and after 3 cycles 0.001 !
     
  20. Jan 12, 2015 #19
     
  21. Jan 12, 2015 #20
    Ah, silly me! I see my error now.

    Yes they are from picture 790685 post #5

    Q can be calculated from ω/Δω

    where ω is the value at the peak of the curve

    and Δω is the width of its peak at the halfway point

    The graph is difficult to read but for plot b i originally took that to be 50/(55-45) which didn't work as q was less than 2π. I was advised to use a value of 7 to make the calculation work.
     
    Last edited: Jan 12, 2015
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