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Deriving the Frequency of Beats

  1. Apr 9, 2013 #1
    1. The problem statement, all variables and given/known data

    attachment.php?attachmentid=57703&stc=1&d=1365592186.png


    2. Relevant equations



    3. The attempt at a solution

    I've done all the parts easily. I just don't know how to derive the beat frequency fbeat.

    I cannot provide any workings because I don't know where to begin. I am not asking for an answer.

    I am asking for a push in the right direction. Give me something to get my brain working on this.

    I hope the mods will understand my intentions.
     

    Attached Files:

    Last edited by a moderator: Apr 10, 2013
  2. jcsd
  3. Apr 9, 2013 #2
    The photo doesn't seem to be working.

    f (beat) = |f' - f(initial)|

    Since the photo isn't working, can you explain the problem?
     
  4. Apr 9, 2013 #3

    sophiecentaur

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    If you think just in terms of time variations for a fixed value of x (say 2nπ) then the equation becomes easier and it reduces to the trig identity for Cos(A)XCos(B), which gives you an A+B term and an A-B term (giving the beat frequencies when there is a non-linearity involved). The difference doesn't 'have' to be small - except when you're talking about audio tones and the sense that the ear / brain makes when two are heard at once. Sum and difference frequencies can always occur.
     
  5. Apr 10, 2013 #4
    I don't follow your explanation at all. Treat me like an idiot.
     
    Last edited by a moderator: Apr 10, 2013
  6. Apr 10, 2013 #5

    sophiecentaur

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    Uninformed does not mean daft! :wink: I know nothing about premiership football or horse racing.

    Trig is one of those subjects that can just get more and more complicated. Many people just use sin, cos and tan to work out right angled triangles - but that's only the start of things. There are a whole lot of so-called identities which you start off with in A level maths and just go on and on getting more and more complicated. If you did any trig at school, you may have come across the fact that sin2θ+cos2θ=1, which can be shown to follow from Pythagoras' theorem.

    Trig is also involved with a lot of difficult Integrations. I guess you don't want to get into that stuff BUT it's not really avoidable if you want an answer to your question, I think. If you take two cos (or sin) functions and multiply them together (which is what happens when two signals go through a non-linearity) the maths can show that there will be products at beat frequencies (sum and differences).
    If you want to batter your brain, take a look at this link. You may spot the sort of thing I was talking about near the top of the page.
    Treat it like a black box that works!
     
  7. May 14, 2015 #6
    sophiecentaur has the right idea. Look at your function (ii) at t=0 for simplicity. We see that there are two frequencies in play here.
    1. [itex]\frac{\omega_1+\omega_2}{2} [/itex]
    2. [itex]\frac{\omega_1-\omega_2}{2} [/itex]
    We see that the the first frequency is much larger than the second. This corresponds with having a shorter period. The second frequency with its longer period represents the frequency of the envelope of the superposition. Try looking at the time difference between when the envelope function is zero. In other words ...
    [tex] cos(\frac{\omega_1-\omega_2}{2}t) = 0 [/tex]
    This will give you some [itex]t_n[/itex]. The period of your beat will this be the difference between [itex]t_{n+1}[/itex] and [itex]t_n[/itex]. You can then easily see the beat frequency.
     
    Last edited: May 14, 2015
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