Equation describing a string on an elastic foundation

[EsponV]

Homework Statement

I'm given an equation describing a string on an elastic foundation:
u_tt+w²₀u = c²u_xx
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"

Thank you for your help.

 

[vela]

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 k_n granted by my X(x,t) boundary conditions. Is this correct?

For instance, if my boundary conditions were u_x(0,t)=u_x(l,t)=0 instead, I come up with X(x,t) = Acos(k_nx) with k_n=n*pi/l. Do I again just allow w²_n=w²₀-c²k²_n?

 

[vela]

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 k_n granted by my X(x,t) boundary conditions. Is this correct?

For instance, if my boundary conditions were u_x(0,t)=u_x(l,t)=0 instead, I come up with X(x,t) = Acos(k_nx) with k_n=n*pi/l. Do I again just allow w²_n=w²₀-c²k²_n?

Your k_n 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 k_n value? This is how I solved for it:

if k_n² > 0
Then X(x,t) = Acosh(kx)+Bsinh(kx)
X'(x,t) = 0 = -Aksinh(kx)+Bkcosh(kx)
X'(0,t)= Bkcosh(0) and since k_n² > 0, B = 0
X'(l,t) = -Aksinh(kl) = 0 so k_n² = 0, but k_n² > 0

So k_n² is not > 0

So k_n² 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 k_n² = ((n*pi)/l)^2

Granted, if k_n²=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?

 

[vela]

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 w²_n=w²₀-c²k²_n, then if c²k²_n > w²₀ we have stability or does it work differently than that?

 

[vela]

If ω²_n<0, you have an imaginary frequency ω_n. What happens to your solution then?

 

[EsponV]

So, if w_n²<0, does my T(t) then become T(t)=ae^w_n2t+be-^w_n2t 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.

 

[vela]

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