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Frequency Problem

  1. May 10, 2006 #1
    Walking beside you, your friend takes 54 strides per minute while you take 26 strides per minute. If you start in step, when will you be in step again?

    I am assuming that the 2 friends stay next to each other, but one of the friends is walking faster.

    I am assuming that "in step" means that the friends are on the same foot.

    I created a spreadsheet and found at 30 seconds that both have taken an odd number of steps (whole numbers) and at 60 seconds that both have taken an even number of steps (whole numbers).

    But both of these answers are incorrect.

    Have I made some errors in my assumptions? Please guide me in the right direction.

    Thanks
     
  2. jcsd
  3. May 10, 2006 #2
    You could do a graphic representation of the walking paces of the two friends. Let's think of a function that oscilates 54 times in a minute and a function that oscilates 26 times in a minute. You could take, for example sin(54(2 pi)x) and sin(26(2 pi)x) , x being the time unit (in this case minute). Now all you have to do is see when these two functions take the same value, meaning you have to solve the equation
    sin(54(2 pi)x) = sin(26(2 pi)x).
    Knowing that sinx is periodical with a period of 2 pi I think you can solve it from here
     
    Last edited: May 10, 2006
  4. May 10, 2006 #3
    Thanks for your quick response. Is there another way to solve the problem that a Physics I student could understand?
     
  5. May 10, 2006 #4
    I'm not familiar with the learning system in other countries than my own, so I don't know what Physics I is about, but in what grade are you? Maybe I can explain better so that you can understand.
     
  6. May 10, 2006 #5
    11th grade
     
  7. May 10, 2006 #6

    Curious3141

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    The question is really pretty simple. You can think of one step as a complete cycle. You start in phase (at the same point in the cycle) then go out of phase as you walk. Obviously after any amount of time, your friend and you have taken a different number of steps. The question is, when are you in phase again ?

    Note that at this time (let's call it [tex]t_0[/tex], neither of you has to be at the start or end of a step. Your respective feet could be in mid-air. The important criterion is that your feet must be matched in the position of the step cycle, so that you can truly be said to be in step again.

    So how to find that time? Well, when two cyclical processes are in phase, it stands to reason that the difference in the number of cycles elapsed is a whole number. Do you see why this should be so?

    In this case, you can find out the point by finding an expression for the number of steps taken by each person at time [tex]t_0[/tex]. Then take the difference between those expressions and relate it to the smallest positive whole number to find [tex]t_0[/tex].
     
  8. May 10, 2006 #7
    Ok. So you know how the sinx graph looks like. The sinx function does a complete oscilation during an interval with a length of 2pi. This means that the function sin(2pi x) will do a complete oscilation during an interval with a length of 1. Now this is a little harder to illustrate in your mind but try to imagine how the graph of sin(k 2pi x) would look like. It will oscilate k times during an interval with a length of 1.

    If you think of the walking process as of an oscilating process, than a walking pace wich has 54 strides per minute can be represented by a function that oscilates 54 times in a minute: sin(54 2pi x). Using the same logic, a walking pace with 26 strides per minute can be represented by a function that oscilates 26 times in a minute: sin(26 2pi x). All you have to do is see when the two oscilating processes (the walkings of the 2 friends) are in the same stage of the oscilation. This means that you equal the 2 functions: sin(54 2pi x) = sin(26 2pi x). Knowing that sinx is periodical with a period of 2pi, solving the equation resumes to solving:
    54 2pi x = 26 2pi x + k 2pi (where k is any natural number) which gives the result x = k/28 minutes.
     
  9. May 10, 2006 #8

    Curious3141

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    I have to raise a caveat here. A cyclical function is not necessarily a sinusoidal one. This function representing walking could well assume some other shape, for example, a triangular waveform.

    The point is, the shape of the function is not specified, so we shouldn't assume it's sinusoidal. If there is some reason to pin this down as simple harmonic motion then it's OK, but we can't justify it here.

    So just approaching the problem as a whole number difference in elapsed cycles is the best approach, since it doesn't make assumptions about the shape of the function (the only assumption is periodicity, which is of course, justified). It also simplifies the approach a great deal without having to bring the sines, 2 pi, etc. into it. :smile:
     
  10. May 10, 2006 #9
    Curious3141,
    Thanks....I did what you suggested using a spreadsheet and the time would be 30 seconds, but this is the wrong answer. One walker will have taken 13 steps and the other 27 steps. We were asked to give the answer in seconds.
     
  11. May 10, 2006 #10

    Curious3141

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    No, don't use a spreadsheet, use a paper, pencil and the good ol' noggin. Oh, and a calculator for getting a numerical answer. To help you along, I've given some steps, fill in the blanks and work out the answer.

    First : Guy walking 54 steps/min. After [tex]t_0[/tex] minutes, how many steps has he covered? Answer : ____ (call this expression 1)

    Second: Guy walking 26 steps/min. After the same amount of time, how many steps has he covered? Answer : ____ (expression 2)

    Third : For them to be in step, the difference between the no. of steps has to be a whole number (0, 1, 2, ...etc.)

    "0" of course corresponds to the situation at the outset, when they both start walking. So we're looking for "1" when they're next in step.

    So (expression 1 - expression 2) = 1

    Hence [tex]t_0[/tex] = ___ minutes = ____ seconds.

    Can you proceed with that?
     
  12. May 10, 2006 #11
    We could use any periodical function, that's why I said "You could take, for example sin(54(2 pi)x) and sin(26(2 pi)x)". The shape of the function doesn't interest us because we only care about the frequency of the oscilation and not the absolute intensity at any specific moment, so I don't think it is wrong to use any periodical function.
    It is true that your approach is easier for a 11-th grade student, but if he is familiar with a little trigonometry mine isn't so hard either :smile:
     
  13. May 10, 2006 #12
    Curious3141,
    Thanks...I submitted the answer using your method and it was correct.
     
  14. May 10, 2006 #13

    Curious3141

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    You were comparing two equally normed sine functions to identify the points where they were in phase. Without any further clarification, the tacit assumption is that the *actual* "walking function" is considered to have that form, which is incorrect.

    I'm sure what you meant was using the sine functions as a proxy for the *phase* of the two ill-defined walking functions, but that wasn't clearly stated. In any case, doing so adds no benefit because essentially you are comparing (walking function 1 to sine function 1) AND (walking function 2 to sine function 2) THEN (sine function 1 to sine function 2) to draw your conclusions, when you might just as well directly compare walking function 1 to 2.

    I'm saying your approach is unnecessarily complicated, redundant, and if not carefully qualified, conceptually flawed.
     
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