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Solving for non moving points of a 1-D wave

  1. Jun 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi!

    I'm suppose to find the points x on the "string" 1-D wave which are not moving during the vibrations, i.e., 0<x<1 such that u(x,t) = 0 for all times t >0

    2. Relevant equations

    [tex]\left.u(x,t) = sin( \pi x)cos(\pi t) + \frac{1}{2}sin(3\pi x)cos(3\pi t) + 3sin(7\pi x) cos(7\pi t)[/tex]

    [tex]\left.u(x,0) = sin( \pi x) + \frac{1}{2}sin(3\pi x) + 3sin(7\pi x) [/tex]

    [tex]\left.\frac{du}{dx}(x,0) = \pi cos( \pi x) + \frac{3}{2}\pi sin(3\pi x) + 21\pisin(7\pi x) [/tex]

    3. The attempt at a solution
    Here's what I have so far,
    I reasoned that if there are stationary points, it doesn't matter at what t I'm at, I'll still achieve the same x values and therefore, I simplified the problem by using t = 0.

    Now the next thing I said was for a x value to be non moving throughout the vibrations, this means that the velocity would be zero therefore, I took the derivative of the function.

    so, to find the x value I simply set the derivative to zero so:

    [tex]\left.\pi cos( \pi x) + \frac{3}{2}\pi sin(3\pi x) + 21\pi sin(7\pi x) = 0[/tex]

    I divided out the [tex]\pi[/tex] to simplify things:

    [tex]\left. cos( \pi x) + \frac{3}{2} sin(3\pi x) + 21sin(7\pi x) = 0[/tex]

    At this point, I'm lost on how to approach this problem in order to find all the non moving points.

    I asked my professor whether I can approximate the points by looking at the graph but he wants to points to be solved in a systematic fashion so...any type of suggestion would be greatly appreciated! Thank you!!!!!
     
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  3. Jun 12, 2009 #2

    gabbagabbahey

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    You seem to be very confused about what [itex]u(x,t)[/itex] actually represents. It represents the amplitude/height of the string at each point [itex]x[/itex] at a time [itex]t[/itex]. For example, at time [itex]t=0[/itex], your string would look like this:

    http://img526.imageshack.us/img526/6533/tzero.th.jpg [Broken]

    While at [itex]t=1[/itex], your string would look like this:

    http://img197.imageshack.us/img197/3382/tone.th.jpg [Broken]

    The points that aren't moving with time are the points that are the same at all values of [itex]t[/itex]. Just by comparing the two graphs above you should see that there are at most 8 such points (including the endpoints of the string).

    To represent this mathematically, you expect both the x values and the amplitudes of any non-moving points to be constant in time. The string is not moving horizontally, so the x-values are all independent of time and so you want to find which points the amplitude is constant in time.

    In other words, you want to find the values of [itex]x[/itex] for which [itex]u(x,t)[/itex] does not depend on [itex]t[/itex].....what would you expect [itex]\frac{\partial u}{\partial t}[/itex] to be for those points?
     
    Last edited by a moderator: May 4, 2017
  4. Jun 12, 2009 #3
    zero right because u(x,t) at those times would just be some kind of constant?
     
  5. Jun 12, 2009 #4

    gabbagabbahey

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    Yes, so find the points where it is zero...
     
  6. Jun 12, 2009 #5
    So if I take the derivative of the function in respect to time, then

    [tex]\left.\frac{du}{du}(u,t) = -\pi sin(\pi x)sin(\pi t) - \frac{3}{2}\pi sin(3\pi x)sin(3\pi t) - 21\pi sin(7\pi x)sin(7\pi t)[/tex]

    so now I set it equal to zero:
    [tex]\left. -\pi sin(\pi x)sin(\pi t) - \frac{3}{2}\pi sin(3\pi x)sin(3\pi t) - 21\pi sin(7\pi x)sin(7\pi t) = 0[/tex]

    am I suppose to solve this assuming t is just some constant integer? Also, how would I approach this in order to solve the problem? Do I need to use trig identity to combine the terms and then solve for x that way? I'm not very good at solving for x with so many trig terms....so sorry for all the questions. Thanks
     
  7. Jun 12, 2009 #6

    gabbagabbahey

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    I would use the multiple angle trig identities to write it as a polynomial in [itex]\sin(\pi x)[/itex] and [itex]\sin(\pi t)[/itex]...then try to factor the polynomial to get it into the form [itex]f(x)g(t)[/itex]....then, you simply find where [itex]f(x)=0[/itex]
     
  8. Jun 12, 2009 #7
    so I'm kind of stuck on converting my function using multiple angle formula. This is the formula I found:
    [tex]\left.sin(nx) = \sum^{\frac{n-1}{2}}_{k = 0}(-1)^{k}\left(\stackrel{n}{2k+1}\right)sin^{2k+1}xcos^{n-2k-1}x[/tex]

    so if my n is [tex]\left.7\pi[/tex], how does the summation work since [tex]\left.\frac{n-1}{2} = \frac{7\pi - 1}{2}[/tex] which is not an integer value?
     
  9. Jun 12, 2009 #8

    gabbagabbahey

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    It is easier if you use a different variable in your trig identity:

    [tex]\left.sin(n\theta) = \sum^{\frac{n-1}{2}}_{k = 0}(-1)^{k}\left(\stackrel{n}{2k+1}\right)\sin^{2k+1}\theta \cos^{n-2k-1}\theta[/tex]

    Then, use this with [itex]\theta=\pi x[/itex] :wink:
     
  10. Jun 12, 2009 #9
    Thanks, so after expanding a bit I get this:

    [tex]\left. -\pi sin(\pi x)sin(\pi t) + \frac{3}{2}\pi (3cos^{2}(\pi x)sin(\pi x) - sin^{3}(\pi x))(3cos^{2}(\pi t)sin(\pi t) - sin^{3}(\pi t)) - 21\pi (3sin(\pi x)cos^{6}(\pi x)[/tex]
    [tex]\left. - 21sin^{3}(\pi x)cos^{4}(\pi x) + 35sin^{5}(\pi x)cos(\pi x))(3sin(\pi t)cos^{6}(\pi t) - 21sin^{3}(\pi t)cos^{4}(\pi t) + 35sin^{5}(\pi t)cos(\pi t))[/tex]

    So I mean this term is very long already so should I be trying to use more trig identities on the polynomials or should I multiply the terms and hope there are like-terms that I can factor out?(I've never worked with so many trig terms before heh) Thanks!
     
  11. Jun 12, 2009 #10

    gabbagabbahey

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    That doesn't look right, you shouldn't have any odd powers of cosine in there....this gets a little ugly, so you'll want to use a program like Mathematica or Maple (something with a built in function for factorizing).
    After you fix up your expression use the fact that [itex]\cos^2\theta=1-\sin^2\theta[/itex] to write everything in terms of powers of [itex]\sin(\pi x)[/itex] and [itex]\sin(\pi t)[/itex]....then define two new variables [itex]\alpha\equiv\sin(\pi x)[/itex] and [itex]\beta\equiv\sin(\pi t)[/itex] and write your expression as a polynomial in them, then factor it and find any values of [itex]\alpha[/itex] for which the expression is zero
     
  12. Jun 13, 2009 #11
    I've only started using maple until recently but I entered my original derivative, and I right clicked on it and told maple to expand it?

    I got the following:
    [tex]\left.-\pi sin(\pi x)sin(\pi t) - 6\pi sin(3\pi x)sin(\pi t)cos^{2}(\pi t) + \frac{3}{2}\pi sin(3\pi x)sin(\pi t)[/tex]
    [tex]\left.- 1344\pi sin(7\pi x)sin(\pi t)cos^{6}(\pi t) + 1680\pis in(7\pi x)sin(\pi t)cos^{4}(\pi t)[/tex]
    [tex]\left. - 504\pi sin(7\pi x)sin(\pi t) cos^{2}(\pi t) + 21\pi sin(7\pi x)sin(\pi t)[/tex]

    So now you said to use the trig identity of [itex]
    \cos^2\theta=1-\sin^2\theta
    [/itex]

    but so what do I do about the other powers of cosine? Do I need to apply trig identities to bring them down to powers of 2 as well? Thanks!
     
  13. Jun 13, 2009 #12

    gabbagabbahey

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    [tex]\cos^6 (\pi t)=(\cos^2 (\pi t))^3=(1-\sin^2 (\pi t))^3[/tex]

    You will also need to use the same multiple angle trig identity to write the [itex]\sin(7\pi x)[/itex] and [itex]\sin(3\pi x)[/itex] terms as powers of [itex]\sin(\pi x)[/itex]
     
  14. Jun 13, 2009 #13
    Hm alright thanks, do you know why maple only applied multiple angle rule to the t variable and not the x? Is there a way to specify that in maple?
     
  15. Jun 13, 2009 #14

    gabbagabbahey

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    Sorry, I haven't used Maple in many years....thee should be some documentation that will help you though.
     
  16. Jun 13, 2009 #15
    Ok so for the x terms I ended up double checking my previous expansion(fixed the mistakes) and then substituted for the x terms. I also used [itex]
    \cos^2\theta=1-\sin^2\theta
    [/itex]

    and combined some like terms after multiplying stuff through and got the following:

    [tex]\left.-\pi sin(\pi x)sin(\pi t)[/tex]
    [tex]\left.-\frac{3}{2}\pi (3sin(\pi x)-4sin^{3}(\pi x))(3sin(\pi t)-4sin^{3}(\pi t))[/tex]
    [tex]\left.-21\pi (7sin(\pi x)-21sin^{3}(\pi x)+35sin^{5}(\pi x)-21sin^{7}(\pi x))(7sin(\pi x)-21sin^{3}(\pi t)+35sin^{5}(\pi t)-21sin^{7}(\pi t))[/tex]

    So am I suppose to find the roots of the x terms? Thanks!!!
     
  17. Jun 13, 2009 #16

    gabbagabbahey

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    That's closer, but still a little off:

    [tex]\sin(7\theta)=7 \cos^6(\theta) \sin(\theta)-35 \cos^4(\theta) \sin^3(\theta)+21 \cos^2(\theta)\sin^5(\theta)-\sin^7(\theta)[/tex]
    [tex]=7 [1-\sin^2(\theta)]^3 \sin(\theta)-35 [1-\sin^2(\theta)]^2 \sin^3(\theta)+21 [1-\sin^2(\theta)]\sin^5(\theta)-\sin^7(\theta)[/tex]
    [tex]=7\sin(\theta)-56\sin^3(\theta)+112\sin^5(\theta)-64\sin^7(\theta)[/tex]

    After you correct your above expression, substute [itex]\alpha=\sin(\pi x)[/itex] and [itex]\beta=\sin(\pi t)[/itex]....then use Maple to factorize the resulting polynomial.....are there any factors that are polynomials in [itex]\alpha[/itex] only?
     
  18. Jun 13, 2009 #17
    Is there a thread where I can ask questions about maple because I've been searching the internet trying to find out how to factor the polynomial and was unable to find an answer. Apparently the Factors command is suppose to factor multivariate polynomials but when I type in the command to factor the expression, it just placed a bracket around the expression and then wrote Factors next to it.....

    But I assume there would be factors that are only in alpha which means they are independent of time? Would I take simply those factors, set them to zero and solve for x? By solving for x, I should get the six roots(excluding the ends)? Thanks
     
  19. Jun 13, 2009 #18

    gabbagabbahey

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    You can start a thread in the appropriate place yourself, but for now I'll just quote Mathematica's output for the factored form:

    [itex]-\frac{1}{2} \pi \alpha \beta \left(2087-16500 \alpha ^2+32928 \alpha ^4-18816 \alpha ^6-16500 \beta ^2+131760 \alpha ^2 \beta ^2-263424 \alpha ^4 \beta ^2+150528 \alpha ^6 \beta ^2+32928 \beta ^4-263424 \alpha ^2 \beta ^4[/itex]
    [itex]+526848 \alpha ^4 \beta ^4-301056 \alpha ^6 \beta ^4-18816 \beta ^6+150528 \alpha ^2 \beta ^6-301056 \alpha ^4 \beta ^6+172032 \alpha ^6 \beta ^6\right)[/itex]

    Now you want to find values of [itex]x[/itex] where this expression is zero for all [itex]t[/itex]....[itex]\alpha[/itex] is a function of [itex]x[/itex] only, and [itex]\beta[/itex] is a function of [itex]t[/itex] only; so you are looking for values of [itex]\alpha[/itex] for the above expression is zero, for all values of [itex]\beta[/itex]....what are those value(s)?
     
  20. Jun 13, 2009 #19
    I understand the reasoning but I am having troubles translating the meaning into math. I mean I set the expression to zero but mathematically, how does "for all values of t" fit in.

    This is what I'm thinking currently:

    [tex]\left. 0 = 2087-16500\alpha^{2}+32928\alpha^{4}-18816\alpha^{6}[/tex]

    This is because all the other terms contain beta while these do not. In addition, the non moving x values are independent of t and so these are the only terms that matter?

    On a side note, do you know which thread I should post in to ask the question about maple? Would it be in general discussion?
     
  21. Jun 13, 2009 #20

    gabbagabbahey

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    No, for example, [itex]3\alpha^2+2-6\beta=0[/itex] does not mean that [itex]3\alpha^2+2=0[/itex]...it means that there are no values of [itex]\alpha[/itex] where [itex]3\alpha^2+2-6\beta=0[/itex] is true for all [itex]\beta[/itex]....On the otherhand if you had [itex]f(\alpha)(3\alpha^2+2-6\beta)=0[/itex] then any values of [itex]\alpha[/itex] that made [itex]f(\alpha)=0[/itex] would make the entire expression [itex]f(\alpha)(3\alpha^2+2-6\beta)=0[/itex] true for any value of [itex]\beta[/itex]....
     
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