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Linear Wave Equation

  1. Feb 22, 2008 #1

    ~christina~

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

    As a wave passes through any element of a stretched string under tension T, the element moves perpendicularly to the wave's direction of travel. By applying the laws of physics to the motion of the element, a genral differential equation, called the linear wave equation, can be derived.

    a) Use a small string segment of length)x and mass m= [tex]\mu [/tex])x ([tex]\mu [/tex] is the linear mass density of the string)to derive the linear wave equation. List all physics principles and assumptions used in your derivation.

    b) Waves propagate to the right along a string that has tension T= 2.1N and mass per unit length [tex] \mu= 150g/m [/tex] The maximum displacement onthe string is 12mm. At t= 0 wave peaks occur at x= 5.0cm and every 17cm thereafter. IN SI units, write the wave function for this system in the form [tex] y(x,t)= Asin(kx \pm \ometa t + \phi) [/tex]. This wave function must include the correct sign in front of [tex]\omega[/tex] and the numerical values for A, k, [tex]\omega[/tex], and [tex] \psi [/tex]

    c) Standing waves are produced when ideal harmonic waves also propagate to the left along this string. Find the wave function for these standing waves.

    d) Show that the wave function for the standing waves satisfies the linear wave equation.


    2. Relevant equations
    not so sure about equations..

    v= [tex]\sqrt {T/m} [/tex]

    Other than that an I'm lost.

    3. The attempt at a solution

    First of all how do you do partial derivatives...

    I really need someone to get me started and help me through this problem....
    I'm not too good at derivatives so I'll have to look up that but partial derivatives are new to me totally. I've never seen them even though I took Calculus 2 already. I'll look up partial derivative and such but I still need help :sad:

    Thank you very very much.
     
    Last edited: Feb 22, 2008
  2. jcsd
  3. Feb 22, 2008 #2
    search for d'Alambert derivation of waves equations.. then check some trygonometry to plug you're data in..

    regards
    marco
     
  4. Feb 26, 2008 #3

    ~christina~

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    I STILL don't know how to derive the equation. Note: looked up the d'Alambert derivation of wave equation and got nowhere. :frown: (even more confused)

    my lab professor said that I could start from here but I'm not sure where to go from there....

    a) F= ma

    [tex]F= \mu dx dx/dt[/tex]

    where if I understand correctly

    [tex]\mu dx[/tex] and a is = dx/dt^2

    but other than that i'm not sure where to go.

    Can ANYONE help out with derivation of wave equation??


    b)
     
    Last edited: Feb 26, 2008
  5. Feb 27, 2008 #4
    the way is kind of good... try to find a correct "tangential" frame of reference....

    regards
    marco
     
  6. Mar 1, 2008 #5

    ~christina~

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    Ah...I gave up on that derivation...I'll just show you what I did for the last part of showing that a standing wave equation satisfies the linear wave equation.

    A= 0.012m
    lambda= 0.17m
    k= omega/v = 138rad/s/ 3.74m/s = 37rad
    omega= 2pi*f= 138rad/s
    f= v/lambda= 3.74m/s / 0.17m= 22Hz
    v= sqrt(T/mu) = 3.74m/s

    and for the equation for the standing wave with numbers

    [tex]y(x/t)= (2*0.012m sin37x)cos(138rad/s)t [/tex]

    [tex]\delta y/ \delta x |_t = (0.024m*37rad/s) cos (37x)x cos(138rad/s)t [/tex]

    [tex]\delta^2y/\delta x^2|_t= (-(37^2)(2*0.012m)sin37x)xcos(138rad/s)t [/tex]

    [tex]\delta y/ \delta t|_x = (- 0.024m*138rad/s)sin(138rad/s)t x (sin37 rad/s)t [/tex]

    [tex]\delta^2y/\delta t^2|_x= -(138^2)(0.024)cos(138rad/s)t x(sin 37 rad/s[/tex]

    since

    [tex]\delta^2y/ \delta x^2= 1/v^2 (\delta^2y/ \delta t^2 )[/tex]

    [tex](-32.856 sin37x)cos(138rad/s)t = 1/(3.74)^2(-457.056^2cos(138rad/s)t x(sin 37 rad/s[/tex]

    do I divide the cos and sin over from the right to the left to cancel or something then equate them to the right side (1/v^2)?

    I'm not sure what to do since I did divide them over but the equaled 0.7 and 0.2 but they are not perfectly = 0= 0 ....(not sure if the numbers are same this time since I handed in the paper but I did it over and that's above)

    Thanks Marco 84
     
  7. Mar 1, 2008 #6
    Sorry but isnt it easier to work out your formula first with letters (constant in this case A,f,omega blah blah) and then plug numbers at the end??

    Then
    your question switched to something else, you were asking about the derivation of wave equation (1D) and now you are trying to verify I DONT KNOW WICH BC that you didint tell...

    My suggestion is:
    try first to understand what you are deriving, then do the calc.... for me is the best rule....
    good night
    here is 4.30 am
    marco
     
  8. Mar 1, 2008 #7

    ~christina~

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    Ah...oops..I was doing the last part (part d)

    Well I'm trying to prove that the standing wave equation satisfies the linear wave equation. I hope that helps you see what I was doing.

    good night to you too :smile:

    you didn't have to answer it at 4:30am in the morning though...The due date for this question was a week ago but I'm trying to see how it's done so I know how to do this for an exam that's all.
     
  9. Mar 1, 2008 #8

    cepheid

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    Hi Christina,

    First, a quick primer on partial derivatives. I, too, was introduced to them in this context in my first year, before being formally taught about them in second year. That having been said, I think that a wave on a string is a great context to learn about partial derivatives. Here's why:

    Let's say you have a string that's vibrating in such a way that there's a sine wave "passing across" it (a travelling wave). Let's call vertical displacement of the string y. What does y depend upon? First, let's freeze time. In other words, take a snapshot of the wave at a single instant. Let t = t0 = a constant. If you look at the picture, you'll see the string in the shape of a sine wave that varies as a function of your horizontal position along the string, x. We could call this function f(x) if we wanted. Evidently, the vertical displacement is a function of x.

    Now, what happens if we turn time back on again? The sine wave moves. Another way to think of that, is that now, instead of freezing time, letting t be constant, we're going to let x be constant, and confine ourselves to a single point along the string. Let's call this point x0. If we sit on this point on the string, our vertical position will oscillate. At one moment, we might be as high as the peak of the wave, but then we'll gradually move down past the equilibrium position (y=0) and way down to the lowest point (the trough). Evidently, at this point on the string, our vertical displacement is a function of time. We could call it g(t) if we wanted. g(t) looks like a sine wave, but unlike f(x), g(t) is NOT a picture of the wave at a single moment in time. Rather, g(t) is a PLOT of the vertical displacement of the point on the string at x0 as a function of time.

    I think you can see that the most general case is that neither variable is frozen. The vertical displacement of the string depends not only on WHERE along the string you look (x), but also on WHEN you look (t). We say that y is a function of two independent variables and we write, y = y(x,t). In second year, you learn about functions of more than one independent variable. This is called multivariable calculus. By definition, a partial derivative of y(x,t) with respect to time is just the derivative of y(x,t), keeping x constant. In other words, you are finding the rate of change with time of the vertical displacement of the string g(t) at a single point in space x0. Similarly, the partial derivative of y(x,t) with respect to x is the derivative of y(x,t) keeping t constant. In other words, you're freezing time and taking the derivative of the that snapshot of the wave f(x) with respect to x. Mathematically...

    [tex] \frac{\partial}{\partial t}y(x,t) = \frac{d}{dt} y(x_0, t) = \frac{d}{dt}g(t) [/tex]

    [tex] \frac{\partial}{\partial x}y(x,t) = \frac{d}{dx} y(x, t_0) = \frac{d}{dx}f(x) [/tex]
     
  10. Mar 1, 2008 #9

    cepheid

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    Now, as far as the derivation goes, it's not the sort of thing you'd easily think of on your own. As you already stated, you can get one side of the wave equation by noting that the net transverse (vertical) force on the string element is given by:

    [tex] F = ma = \mu dx \frac{\partial^2}{\partial t^2}y(x,t) [/tex]

    That's just because the vertical acceleration is the second derivative of the vertical position with respect to time.

    Now, draw this string element. It's a little curved piece of string. One end is at x, and the other is at x+dx a short distance away. The tension is the same in magnitude at both ends, but it could be in totally different directions, because it points in the same direction as the end of the string at each end. Draw the string in such a way that the tension vector points up and to the right at the rightmost end (at x + dx), and down and to the left at the leftmost end (at x). Then, let the angle between the tension vector on the right end and the horizontal be given by [itex] \theta' [/itex], and the angle between the tension vector at the left end and the horizontal be given by [itex] \theta [/itex]. Now, I think you can see that the net vertical force on the string is given by the upward d component of the tension force at the right end, minus the downward component of the tension force at the left end. If the tension is T, then what I just said is that:

    [tex] F = T\sin(\theta') - T\sin(\theta) [/tex]

    Now, a major hint for you is that the wave equation is only valid for small vertical displacements, because we're going to make the assumption that those angles are small. If that's true, then:

    [tex] \sin \theta \approx \tan \theta [/tex]

    and

    [tex] F = T[\tan(\theta') - \tan(\theta)] [/tex]

    We can rewrite tangent of theta as the slope of the string at each end:

    [tex] F = T\left(\left \frac{\partial y}{\partial x}\right |_{x + dx} - \left \frac{\partial y}{\partial x}\right |_{x}\right) [/tex]

    Now, can you think of a way to simplify

    [tex] \left(\left \frac{\partial y}{\partial x}\right |_{x + dx} - \left \frac{\partial y}{\partial x}\right |_{x}\right) [/tex]

    using a SECOND derivative? Hint: use a first-order approximation.
     
    Last edited: Mar 1, 2008
  11. Mar 8, 2008 #10

    ~christina~

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    Hi cepheid :smile:

    Hm..okay
    so Is the x_0 representing initial x? (I know it's constant x but if it's x initial...)

    same as above question but this time it pertains to the t_0 and whether it's the initial time or the constant t or both which I suspect it is.

    Thanks cepheid
     
  12. Mar 8, 2008 #11

    ~christina~

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    okay.

    drew that.

    I've heard of this before. alright..

    Sorry..I'm not sure about after this part.

    I didn't mention it before but my book has a derivation but I don't understand it.
    It's similar to your method but for the simplification they just replace the equation with this:

    [tex]F \Sum = ma_y= \mu \Delta x( \partial^2 y/ \partial t^2 ) [/tex]

    I'm not sure if this is the simplification you were speaking of but I don't know how they simplified it to this (the previous part they showed was like the equation you found)

    The first order aproximation I'll have to look up since either I don't remember what that is or I've never learned it since nothing comes to mind at the moment.

    Thanks alot cepheid
     
  13. Mar 8, 2008 #12

    cepheid

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    Hi Christina,

    No actually, x0 is not the initial position. I hope I didn't confuse things by using that notation. The point is that x0 is some number. It's a constant. It's an arbitrary constant though, because we could sit at ANY point along the string and measure how the vertical displacement changes with time. You could have plugged in a or c or a smiley face in place of x0, as long as those things were meant to represent a constant. In reality, you don't explicitly plug in any constant at all. All you do is this (bottom line):

    When taking the partial derivative of a function y(x,t) with respect to t, treat x as IF it were a constant. Here's a random example:

    [tex] y(x,t) = x^2 \sin t [/tex]

    [tex] \frac{\partial y}{\partial t} = -x^2 \cos t [/tex]

    You can see that the partial derivative with respect to time is STILL a function of BOTH x and t. This is true for our wave on a string as well. The partial derivative with respect to time is still a function of both x and t, because the rate of change with time of the vertical displacement still depends on WHERE you sit on the string and measure it. I hope that clears up any misconceptions. Everything I've said is likewise true for the partial derivative with respect to x. Just treat t as if it were a constant, and differentiate normally.
     
  14. Mar 8, 2008 #13

    cepheid

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    If you need convincing, draw a really "skinny" right triangle (one in which theta is quite small). Call the side opposite to theta 'o', the side adjacent to theta 'a', and the hypotenuse 'h'. By definition, the sine of theta is o/h. But since theta is quite small, a is approximately equal to h (as you can see in the picture). Therefore o/h is approximately equal to o/a. In other words, the sine of theta is about the same as the tangent of theta. Another way to look at it is that if a is nearly the same as h, then the cosine of theta, which is equal to a/h, is approximately 1. Since tangent = sine/cosine, it follows that tangent is approximately sine divided by 1.

    Another thing we can conclude if theta is small, is that the sine of theta is approximately equal to theta itself (in radians). This is because if a is approx. equal to h, you can consider them to be both radii of a circle and call them both 'r'. Do that, and draw a small curved arc in between them, representing the portion of the full circle that lies in between those two radii. If that arc has length 's', then the definition of theta (in radians) is theta = s/r. But s/r, from the diagram, is about the same as o/h, meaning sine of theta is approximately theta if theta is very small. (The short arc 's' connecting the radii is not much longer than the *straight line* 'o' connecting the radii).

    I went over this "small angle approximation" in quite a bit of detail because it's quite commonplace and handy to know. This was geometric argument. There is also a calculus-based argument that makes use of the first-order approximation I was talking about before.
     
  15. Mar 8, 2008 #14

    cepheid

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    It's pretty simple. Say you're driving along and your horizontal position is given by a function x(t). If you're driving along at a constant speed, then v(t) = dx/dt = constant. The graph of x(t) is a straight line with a slope given by the constant v. If you want to figure out how FAR you've gone in a certain time interval, then you just use the fact that the TOTAL change in position is given by the rate of change of position with time multiplied by the time interval:

    [tex] x(t_f) - x(t_0) = \Delta x = \frac{dx}{dt} \Delta t = v\Delta t [/tex]

    In calculus you learn how to deal with the more complicated example of driving at a speed that is NOT constant, so you can't just multiply the time interval by the rate of change, because the rate of change also changes. Your graph of position vs. time x(t) is now curved, not linear. BUT, a good approximation is that, if the time interval is very small, then the rate of change (ie the speed) is APPROXIMATELY constant OVER that time interval. And so we can say that

    [tex] x(t_f) - x(t_0) = \Delta x \approx \left \frac{dx}{dt}\right |_{t=t_0} \Delta t = v(t_0)\Delta t [/tex]

    In other words, we're saying that if the time interval is small, we can ignore the curve (the changing slope) and assume we were going MORE OR LESS at a constant speed. If you ZOOM IN far enough into the graph of a curve, it begins to start looking like a straight line. Hence, under these conditions, this LINEAR approximation (also called a first-order approximation) to x(t) is appropriate. We're saying that over this very small time interval, x(t) is approximately linear.

    Getting back to your wave equation derivation, we had that the net force on the string element was given by:

    [tex]
    F = T\left(\left \frac{\partial y}{\partial x}\right |_{x + \Delta x} - \left \frac{\partial y}{\partial x}\right |_{x}\right) [/tex]

    Let define the partial derivative of y wrt x as a new function:

    [tex] \frac{\partial y}{\partial x} = u(x,t) [/tex]

    The equation reduces to

    [tex] F = T(u(x+\Delta x, t) - u(x,t)) [/tex]

    Now since delta x is small, we can use the first-order approximation:

    [tex] F \approx T \left( \left\frac{\partial u}{\partial x}\right|_x \Delta x\right) [/tex]

    [tex] = T \left( \left\frac{\partial }{\partial x}\left(\frac{\partial y}{\partial x}\right)\right|_x \Delta x\right) [/tex]

    [tex] = T\left(\frac{\partial^2 y}{\partial x^2}\right)\Delta x [/tex]

    To summarize, using Newton's second law on the string element, we found that the net vertical force F was:

    [tex] F = ma_y= \mu \Delta x\left( \frac{\partial^2 y}{\partial t^2} \right) [/tex]

    and using the sum of the vertical components of the forces due to tension at either end of the string, we found that the net vertical force was ALSO equal to

    [tex] F = T\left(\frac{\partial^2 y}{\partial x^2}\right)\Delta x [/tex]

    Equating the two, we have the final wave equation:

    [tex] \frac{\partial^2 y}{\partial x^2} = \frac{\mu}{T} \frac{\partial^2 y}{\partial t^2} [/tex]
     
    Last edited: Mar 8, 2008
  16. Mar 8, 2008 #15

    ~christina~

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    Okay
     
  17. Mar 10, 2008 #16

    ~christina~

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    getting back to this...

    I did draw this and found that I got the same result that I was supposed to for all of them.
    I'm assuming that wasn't the calculus based argument and also that the geometric argument is sufficient for problems unless asked right?
     
  18. Mar 10, 2008 #17

    ~christina~

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    ah..yep

    not sure why you have the [tex]t_o[/tex] on the last part of the equation.

    ok
     
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