Normal mode of an infinite spring pendulum system

[Miles123K]

Homework Statement

The problem description is as follows. I am expected to find the frequency of the standing wave apparently.

Relevant Equations

F = ma

First I worked out the dispersion relations, which is pretty easy:
$M \ddot x_j = K x_{j-1} + K x_{j+1} - 2K x_j -mg \frac {x_j} {l}$ (All t-derivatives)
We know $x_j$ will be in the form $Ae^{ijka}e^{-i\omega t}$
so the above becomes:
$-\omega^2M = K (e^{-ika}+e^{ika}-2)-\frac {g} {l}$
Use trig identities to simplify and we get:
$\omega^2 = \frac {4K} {M} sin^2(\frac {ka} {2}) + \frac {g} {l}$
Now I think I am supposed to consider the forces on Block 0, so:
$F_{left} + F_{right} = F_0 = M \ddot x_0$
$K x_{j-1} + K x_{j+1} - 2K x_j = M \ddot x_0 = -M \omega^2 x_0$ I think this is the subsystem mentioned in the hint?
In the question, it mentioned considering a normal mode of the form:
$A(x) = A_0 e^{-k \left| x \right |}$
So, I just plugged that into the above equation and got something like:
$2Ke^{-ka} = 2K - m \omega^2$
$e^{-ka} = 1 - \frac {m \omega^2} {2K}$
Log the above:
$ka = - ln(1-\frac {m \omega^2} {2K})$
However, I am unsure whether this is the correct solution. Do I just plug this $ka$ into the dispersion relations we got earlier?
Could someone check my answer?

 

[TSny]

First I worked out the dispersion relations, which is pretty easy:
$M \ddot x_j = K x_{j-1} + K x_{j+1} - 2K x_j -mg \frac {x_j} {l}$ (All t-derivatives)
We know $x_j$ will be in the form $Ae^{ijka}e^{-i\omega t}$

The factor $e^{ijka}$ implies that the amplitudes of the blocks vary sinusoidally in space. But in this problem, you are looking for a mode for which the amplitudes exponentially decay as you move away from the origin. Replace $e^{ijka}$ by an appropriate expression and rederive the dispersion relation. Use the dispersion relation in conjunction with the equation of motion of block 0.

 

[Miles123K]

The factor $e^{ijka}$ implies that the amplitudes of the blocks vary sinusoidally in space. But in this problem, you are looking for a mode for which the amplitudes exponentially decay as you move away from the origin. Replace $e^{ijka}$ by an appropriate expression and rederive the dispersion relation. Use the dispersion relation in conjunction with the equation of motion of block 0.

Oh okay, the new dispersion relations I obtained are as follows:
$\omega^2=\frac {K} {m} (2 - e^{ka} - e^{-ka}) + \frac {g} {l}$
which simplifies to:
$\omega^2=\frac {2K} {m} (1 - cosh(ka)) + \frac {g} {l}$
If I plug in the $ka$ I got above I get the answer?

 

[TSny]

Oh okay, the new dispersion relations I obtained are as follows:
$\omega^2=\frac {K} {m} (2 - e^{ka} - e^{-ka}) + \frac {g} {l}$
which simplifies to:
$\omega^2=\frac {2K} {m} (1 - cosh(ka)) + \frac {g} {l}$

That looks good.

If I plug in the $ka$ I got above I get the answer?

That should work. Off hand, I don't see a way to avoid some messy algebra. You could simplify the writing a bit by defining $\omega_s = \sqrt{\frac{2K}{M}}$ and $\omega_p = \sqrt{\frac{g}{l}}$.