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More Differential Equations

  • #1
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From d'Alembert's solution for the one dimensional wave equation

A string of length l=1 is initially fixed i the position [itex] u = \sin{\pi x} [/itex] and is released at time t = 0 . Find its subsequent motion

What do they mean by the subsequent motion?? DO they mean the velocity? Do i differnetiate u by time? the answer i know is [itex] u = \sin{\pi x} \cos{ \pi ct} [/itex]

ANd where di they give the c from anyway? The intial u represents the soltuion with the integral part of d'Alembert's soltuion equal to zero, doesn't it?

Please help! Thank you!
 

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  • #2
Physics Monkey
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Find the subsequent motion just means write down the displacement [tex] u(x,t) [/tex] as a function of time given the initital condition [tex] u(x,t=0) = \sin{\pi x} [/tex].
 
  • #3
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i am not quite sure what that means...

does this mean i jhave to use the original wave equation and plug u in there?
 
  • #4
Physics Monkey
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Something like that. The general solution to the wave equation can be written in terms of normal modes. The initial condition you have there is a very special one becuase it corresponds to a particular normal mode of the string. Does this help or have you not heard these concepts before?
 
  • #5
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Physics Monkey said:
Something like that. The general solution to the wave equation can be written in terms of normal modes. The initial condition you have there is a very special one becuase it corresponds to a particular normal mode of the string. Does this help or have you not heard these concepts before?
i have not heard these concepts before
at least not relating to this class - i have done this is physics a while ago
 
  • #6
saltydog
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stunner5000pt said:
From d'Alembert's solution for the one dimensional wave equation

A string of length l=1 is initially fixed i the position [itex] u = \sin{\pi x} [/itex] and is released at time t = 0 . Find its subsequent motion

What do they mean by the subsequent motion?? DO they mean the velocity? Do i differnetiate u by time? the answer i know is [itex] u = \sin{\pi x} \cos{ \pi ct} [/itex]

ANd where di they give the c from anyway? The intial u represents the soltuion with the integral part of d'Alembert's soltuion equal to zero, doesn't it?

Please help! Thank you!
Subsequent motion? How about if it were a rubber band and pulled up into the form of a sine wave and then let go?

The c comes from various constants in the derivation of the wave equation. How about formulating it this way:

[tex]\text{DE:}\quad u_{tt}=c^2u_{xx},\quad 0\leq x\leq 1,\quad -\infty < t <\infty[/tex]

[tex]\text{BC:}\quad u(0,t)=0,\quad u(L,t)=0[/tex]

[tex]\text{IC:}\quad u(x,0)=f(x),\quad u_t(x,o)=g(x)[/tex]

Now, using D'Alembert's formula by periodically extending the odd-extensions of f(x) and g(x) to the new functions:

[tex]\Tilde{f_0}(x)[/tex]

[tex]\Tilde{g_0}(x)[/tex]

respectively, we obtain for the solution:

[tex]u(x,t)=\frac{1}{2}\left[\Tilde{f_0}(x+ct)+\Tilde{f_0}(x-ct)\right]+\frac{1}{2a}\int_{x-ct}^{x+ct} \Tilde{g_0}(\tau)d\tau[/tex]

So if [itex]f(x)=Sin[\pi x][/itex]

it is already the periodic odd-extension of itself in (-1,1) so we obtain the solution you indicated and is shown in the attached plot. Of course the plot is 2-D and the string is well 1-d. The plot just shows a continuous "movie" of what the string does as it vibrates off towards the right.
 

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  • #7
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err...

im still unsure as to what to do
so F(x) = sin pi x?

[tex]u(x,t)=\frac{1}{2}\left[\sin[\pi(x+ct)]+\sin[\pi(x-ct)]\right] [/tex]
both the sine arguments are equal since ct is the speed of the wave (correct?)
[tex] u(x,t)=sin[\pi(x+ct)] [/tex]
if i expanded the sine using the sum formula for the sine i could obtain part of the solution in that
[tex] u(x,t) = \sin{\pi x} \cos{\pi ct} + \cos{\pi x} \sin{\pi ct} [/tex]
but why would the second term disappear?

im sorry i ts just not sinkin in
thank you for the help so far
 
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  • #8
saltydog
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stunner5000pt said:
im still unsure as to what to do
so F(x) = sin pi x?
Yes, f(x) up there is [itex]Sin[\pi x][/itex] and g(x)=0

Read carefully my post up there.

[tex]u(x,t)=\frac{1}{2}\left[\Tilde{f_0}(x+ct)+\Tilde{f_0}(x-ct)\right]+\frac{1}{2a}\int_{x-ct}^{x+ct} \Tilde{g_0}(\tau)d\tau[/tex]

is the solution to the problem Ok? Just so happens that g(x) is zero so that integral drops out. Those "extensions" I know are a bit confussing but it just so happens that f(x) meets them so you don't have to do anything about them in this particular case so:

[tex]u(x,t)=\frac{1}{2}\left[f(x+ct)+f(x-ct)\right][/tex]

is the solution to the problem. Use trig identities and you're left with:

[tex]u(x,t)=Sin[\pi x]Cos[\pi ct][/tex]

Can you plot that solution (I used c=1)? Do you understand what's going on with the plot I posted? Remember the plot is just a "movie" of the actual string as it vibrates up and down. Note how it starts as a sine wave at the beginning (the plot at t=0 on the far left side).

Edit: Oh yea, take the partial derivatives twice with respect to x and t of the solution and back-substitute it into the PDE to verify that the left side=right side. Can you do that?

Also, you know why g(x) is zero right? That represents the initial velocity of the string which in this case is zero: just pick it up and let it go. However, if you picked it up then smacked it with a hammer, then the initial velocity would be greater than zero and so g(x) would be non-zero. Ok?
 
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  • #9
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why does the integral of the g function drop out? G represents teh velocity, does it not, so the subsequent motion doesnt care about the velocity?

also f(x) satisfies the conditions such that [itex] \sin[\pi(x+ct)] = \sin[\pi(x-ct)] [/itex] because of the periodicity of the function?

also why does the second term taht is - [itex] \cos{\pi x} \sin{\pi ct} [/itex]
drop out from the solution??

Thank you for the help!
 
  • #10
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does the sin[ pi (x+ct) = sin [pi (x-ct)] because ct is the speed of the wave. As well why does the second part of the trigonometric expansion cancel out? Or why is it irrelevant? Does it represent the velocity or what?
 
  • #11
saltydog
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stunner5000pt said:
why does the integral of the g function drop out? G represents teh velocity, does it not, so the subsequent motion doesnt care about the velocity?
Stunner . . . Your problem stated that that string, say a violin string is initially fixed in the [itex]Sin[\pi x][/itex] position. That means the initial velocity of the string is zero right? It's fixed in that position initially right so it's not moving so the velocity is zero. If you look at the IBVP problem I stated (initial-boundary value problem) above, the initial conditions required us to know the initial velocity which we defined as g(x). Well that's zero so then the integral part of the solution drops out.

also f(x) satisfies the conditions such that [itex] \sin[\pi(x+ct)] = \sin[\pi(x-ct)] [/itex] because of the periodicity of the function?
Don't know what you mean by that. Again, f(x) is just the initial position of the string. It could be anything. I could have drawn it out to look like a triangle or whatever. Whatever it's initial shape, we define it as f(x) since again the IBVP requires an initial condition which in the case of the wave equation, that initial condition is the initial shape of the string.

also why does the second term taht is - [itex] \cos{\pi x} \sin{\pi ct} [/itex]
drop out from the solution??
[tex]Sin[a+b]+Sin[a-b]=Sin(a)Cos(b)+Sin(b)Cos(a)+Sin(a)Cos(b)-Sin(b)Cos(a)=2Sin(a)Cos(b)[/tex]
 

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