Equation describing a string on an elastic foundation

In summary, the conversation discusses a problem involving a string on an elastic foundation. The main focus is on understanding the eigenfrequencies, which are the frequencies of the normal modes of the system. The conversation also addresses a sign mistake and the role of eigenfrequencies in stability and bifurcations.
  • #1
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0

Homework Statement


I'm given an equation describing a string on an elastic foundation:
utt+w20u = c2uxx
with the boundary conditions u(0,t)=u(l,t)=0.


The Attempt at a Solution


I think I understand how to do this problem up to a certain point, but that point occurs when I find the actual eigenfrequencies. I'm not exactly sure what they are, in a general meaning, however I have attached my solution nonetheless.

http://img695.imageshack.us/img695/628/scan2j.jpg" [Broken]

Thank you for your help.
 
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  • #2


I think you have a sign mistake, so you should have [itex]\omega_n^2 = \omega_0^2-c^2k_n^2[/itex]. Anyway, what don't you understand about the eigenfrequencies? They're the frequencies of the normal modes of the system.
 
  • #3


Ah, I do see where I made the sign mistake. I forgot to carry the negative sign through when I transferred T(w[tex]^{2}_{0}[/tex]-k[tex]^{2}_{n}[/tex]c[tex]^{2}[/tex]) to the other side.

As to eigenfrequencies, I'm afraid I don't quite understand their role or how they're derived. We didn't discuss normal modes in class at all, and it seems as if the eigenfrequencies are just equal to the terms associated with my time portion of the equation with respect to the solution of kn granted by my X(x,t) boundary conditions. Is this correct?

For instance, if my boundary conditions were ux(0,t)=ux(l,t)=0 instead, I come up with X(x,t) = Acos(knx) with kn=n*pi/l. Do I again just allow w2n=w20-c2k2n?
 
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  • #4


EsponV said:
As to eigenfrequencies, I'm afraid I don't quite understand their role or how they're derived. We didn't discuss normal modes in class at all, and it seems as if the eigenfrequencies are just equal to the terms associated with my time portion of the equation with respect to the solution of kn granted by my X(x,t) boundary conditions. Is this correct?

For instance, if my boundary conditions were ux(0,t)=ux(l,t)=0 instead, I come up with X(x,t) = Acos(knx) with kn=n*pi/l. Do I again just allow w2n=w20-c2k2n?
Your kn is wrong, but other than that, yes, what you said is correct. The allowed values are dictated by the differential equation and the boundary conditions.
 
  • #5


Hum, do you mind telling me where I went wrong on my kn value? This is how I solved for it:

if kn2 > 0
Then X(x,t) = Acosh(kx)+Bsinh(kx)
X'(x,t) = 0 = -Aksinh(kx)+Bkcosh(kx)
X'(0,t)= Bkcosh(0) and since kn2 > 0, B = 0
X'(l,t) = -Aksinh(kl) = 0 so kn2 = 0, but kn2 > 0

So kn2 is not > 0

So kn2 can be <0

Therefore, I have X(x,t) = Acos(kx)+Bsin(kx) with X'(x,t) = -Aksin(kx)+Bkcos(kx)
and with the same boundary conditions,
X'(0,t) = 0 = Bkcos(kx) so B=0
X'(l,t) = 0 = -Aksin(kx) so k = (n*pi)/l
so X(x,t) = Acos(kx) with k=(n*pi)/l

and therefore our kn2 = ((n*pi)/l)^2

Granted, if kn2=0, then we find another arbitrary constant not being zero, but that is placed out in front of the summation when creating a Fourier series, or so I thought, leading me to believe it didn't have relevance. Is this correct?
 
  • #6


I'm sorry. You're right. I was trying to do it in my head and mixed up a few terms.
 
  • #7


Ah, thank you very much for you help. I just have one more question involving the stability of the eigenfrequencies. Does the stability depend on the sign of the eigenfrequency? Such as if our eigenfrequency is w2n=w20-c2k2n, then if c2k2n > w20 we have stability or does it work differently than that?
 
  • #8


If ω2n<0, you have an imaginary frequency ωn. What happens to your solution then?
 
  • #9


So, if wn2<0, does my T(t) then become T(t)=aewn2t+be-wn2t meaning that my total equation will become:

u(x,t)=B0+[tex]\sum(ae^{w^{2}_{n}t}+be^{-w^{2}_{n}t})Bsin(npix/l)[/tex]

I'm not sure if that is entirely correct, and if it is, this is the point where I get stuck at. I'm just not quite sure what I need to do to find the points of stability/bifurcations in the system.
 
  • #10


The solutions should be of the form [itex]\exp(\pm |w_n| t)[/itex]. The frequency shouldn't be squared.
 

What is an equation describing a string on an elastic foundation?

An equation describing a string on an elastic foundation is a mathematical model that predicts the behavior of a string or cable when it is placed on top of an elastic surface, such as a rubber mat or a trampoline. It takes into account factors such as the tension in the string, the elasticity of the foundation, and the weight of the string itself.

What is the significance of studying the behavior of a string on an elastic foundation?

Studying the behavior of a string on an elastic foundation is important in various fields such as engineering, physics, and mathematics. It provides insight into the mechanics of structures and materials, and can help with the design and optimization of structures that involve strings or cables.

What are some real-life applications of the equation describing a string on an elastic foundation?

The equation describing a string on an elastic foundation has many practical applications. It is commonly used in the design of suspension bridges and cables, as well as in the study of seismic activity and the behavior of strings or wires in musical instruments.

How is the equation describing a string on an elastic foundation derived?

The equation describing a string on an elastic foundation is derived using principles of mechanics, such as Hooke's law and the Euler-Bernoulli beam theory. It also takes into account the boundary conditions of the string, such as fixed or free ends, and the properties of the elastic foundation.

Are there any limitations to the equation describing a string on an elastic foundation?

Like any mathematical model, the equation describing a string on an elastic foundation has limitations. It assumes certain simplifications, such as a perfectly elastic foundation and a perfectly straight string, which may not always hold true in real-life situations. It is important to understand these limitations when applying the equation to practical problems.

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