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Homework Help: Qustion on sinewave

  1. Dec 18, 2011 #1
    If the expression forsinewave is y=Asin(ωt±α)
    then α is the phase angle.
    But what is time displacement looking at that expression?
    Thank you
  2. jcsd
  3. Dec 18, 2011 #2


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    Time displacement? I assume you mean how long it takes for the sine wave to repeat a cycle, or its [STRIKE]frequency[/STRIKE].

    All sine waves repeat every [itex]2\pi[/itex] units. So what we're looking for is


    to be equal to


    and we want to find t2 in terms of t1, t2>t1

    So equating each expression and simplifying:




    So clearly from this, we can see that the time it takes from the first point in a cycle (t1) to the next (t2) takes [itex]2\pi/w[/itex] time.

    Also to get a more intuitive understanding of this, just think about the length of a complete cycle for sin(x), then sin(2x), sin(x/3) - which is the same as sin(1/3*x) etc.

    edit: meant to say period, not frequency.
    Last edited: Dec 18, 2011
  4. Dec 18, 2011 #3
    Thank you for your really good explanation.
    But I am not sure this is the answer to the question I was given. I am not sure.
    Can frequency and time period T be called time displacement? Is it not about that phase angle?
    Thanks any way
  5. Dec 18, 2011 #4
    Is it ω that is worrying you?
    ω is an 'angular velocity' it is the number of cycles (given in radians) completed per second
    1 cycle is 2∏ radians and therefore ω = 2∏/T where T is the time for 1 cycle.... the time period

    So T = 2∏/ω

    and frequency f = 1/T = ω/2∏

    Hope this helps
  6. Dec 18, 2011 #5


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    You haven't given the definition of "time displacement". If you write[tex]
    sin(\omega t \pm \alpha) = \sin(\omega(t \pm \frac \alpha \omega))[/tex] is it the [itex]\pm\frac \alpha \omega[/itex] you want? (Notice that is a question; I'm just guessing here).
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