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Expressions of travelling harmonic wave equation

  1. Jun 23, 2012 #1
    Hi all, apologies if this has been answered elsewhere - I was unable to find an answer using the search function.

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

    "Expressed in terms of wavenumber and angular frequency, the equation for a travelling harmonic wave is: y = Asin(kx-ωt). Express this function in terms of (a) wavelength and wave speed; (b) frequency and wave speed; (c) wave number and wave speed; (d) wavelength and frequency."


    2. Relevant equations

    y = Asin(2∏/λx - 2∏ft)

    v=fλ

    3. The attempt at a solution

    I know that the expression for wavelength is 2∏/λ , and suspect the expression for wave speed is fλ, or (2∏ x 1/τ), although I am not sure on this point. I am not entirely sure what the question is asking; I know how to calculate each of the values given above from the harmonic wave equation, but do not know how to 'express' the equation in these terms.

    Many thanks for your help.

    Edit: Formatting.
     
  2. jcsd
  3. Jun 24, 2012 #2

    Redbelly98

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    Not quite. That is actually the expression for wavenumber, k. Wavelength is λ.
    At any rate, you can substitute your expression for k into the travelling wave equation for y. That will eliminate k, and get the expression in terms of wavelength λ instead.
    You're correct, but you need to get an expression for wavespeed that involves ω instead of f or τ. If you use the relation between f and ω, you can get that expression.

    Instead of y=[expression involving k and ω], they want
    y=[expression involving λ and v] (for part a), etc.
    The idea is to replace k and ω with expressions that use other parameters (λ, v, and/or f)

     
  4. Jun 24, 2012 #3
    Thanks for your reply Redbelly98.

    When you say "you need to get an expression for wavespeed that involves ω instead of f or τ. If you use the relation between f and ω, you can get that expression." (not sure how to quote using the quote reply function yet, apologies) , does 'ω' in that instance equate to wavespeed? I was under the impression that 'ω' meant angular frequency - are they the same thing? As you can see, I am far from grasping the complexities of this equation...

    Also, is the relationship between f and ω simply ω=2pi x f ? Or is there something else I should know?
     
  5. Jun 24, 2012 #4

    Redbelly98

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    Yes, ω is angular frequency. Wavespeed is v.

    You are starting with the equation,
    y = Asin(kx-ωt)​
    It has ω in it, but you want to get rid of ω. So you need an expression for ω to substitute into the equation.

    As you said, you know that
    v=fλ
    But there is no ω there, so that equation, as written here, is useless for finding an expression for ω that can be substituted. What to do? Try using ω=2πf (yes, it is correct).

    To quote what others have written, there are a couple of approaches.

    You can use the https://www.physicsforums.com/Prime/buttons/quote.gif [Broken] button to quote an entire post, then delete the portions you don't want included.

    Or:

    Go to Advanced Edit mode by clicking the "Go Advanced" button.
    Then click the https://www.physicsforums.com/Nexus/editor/quote.png [Broken] button to insert quote tags.
     
    Last edited by a moderator: May 6, 2017
  6. Jun 24, 2012 #5
    I think I'm beginning to see the light.

    ω=2πf

    f = v/λ

    So ω=2πv/λ ?


    And so the expression for part (a) would be y=Asin(2π/λx - 2πv/λt) .


    Thanks so much for your help here. I feel much more confident about approaching these types of problems now.
     
  7. Jun 25, 2012 #6

    Redbelly98

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    Yes, you got it.

    A couple of comments:

    1. An expression like "2π/λx" is interpreted by some as 2π/(λx), when you really mean (2π/λ)·x. So you may want to clarify what you mean, if you submit your answer in this form.

    2. Also, note all the common terms in the expression
    (2π/λx - 2πv/λt).​
    In general it is desirable to factor such an expression. It's not wrong the way you wrote it, but factoring can sometimes provide some insight that you would not otherwise see.
     
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