Equation describing a string on an elastic foundation

EsponV

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]

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Staff Emeritus
Homework Helper

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

EsponV

Ah, I do see where I made the sign mistake. I forgot to carry the negative sign through when I transferred T(w$$^{2}_{0}$$-k$$^{2}_{n}$$c$$^{2}$$) 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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Staff Emeritus
Homework Helper

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.

EsponV

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?

Staff Emeritus
Homework Helper

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

EsponV

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?

Staff Emeritus
Homework Helper

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

EsponV

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+$$\sum(ae^{w^{2}_{n}t}+be^{-w^{2}_{n}t})Bsin(npix/l)$$

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.

Staff Emeritus
The solutions should be of the form $\exp(\pm |w_n| t)$. The frequency shouldn't be squared.