N-oscillator system with two sides attached

[Miles123K]

Homework Statement

The system is shown in the image. In the beaded string shown in Figure 1, the interval between neighboring beads is a, and the distance from the end beads to the wall is a/2. All the beads have mass m and are constrained to move only vertically in the plane of the paper. The strings are massless with constant string tension T.

Homework Equations

I am sure that I'm supposed to construct an analogous infinite system first. That system would have standard translation symmetry matrix and should be solved in that way. Boundary conditions need to be applied later to make the system limited.

The Attempt at a Solution

I made a wild attempt to try to add imaginary oscillators to the system so that there will be one oscillator every a/2 length. I am pretty sure I am very wrong since an N-oscillator system should not have 2N normal modes.

 

[Miles123K]

I gave some extra thoughts and managed to get sub-question A and B, and have a potential solution (don't know right or wrong) for the normal mode frequency, but I couldn't solve the the normal modes amplitude.

 

[TSny]

It would help if you wrote out some of your expressions explicitly so that we know what notation you are using, etc. When you have the usual case where the "zeroth" particle is held fixed, the normal mode displacement, $u_n$, of the beads can be written as $u_n = \cos (\omega t) \sin (np)$ where n is an integer labeling the beads and $p$ is to be determined from the boundary conditions. The reason for using the sine function here is to satisfy the boundary condition for the case where the $n = 0$ bead is fixed.

In your case, you want the $n = 0$ bead to "mirror" the $n = 1$ bead: $u_0 = -u_1$ at all times. Try writing $u_n = \cos (\omega t) \sin (np+\phi)$ where $\phi$ is a phase consant. Can you choose $\phi$ so that the boundary condition $u_0 = -u_1$ is satisfied at all times?

 

[Miles123K]

It would help if you wrote out some of your expressions explicitly so that we know what notation you are using, etc. When you have the usual case where the "zeroth" particle is held fixed, the normal mode displacement, $u_n$, of the beads can be written as $u_n = \cos (\omega t) \sin (np)$ where n is an integer labeling the beads and $p$ is to be determined from the boundary conditions. The reason for using the sine function here is to satisfy the boundary condition for the case where the $n = 0$ bead is fixed.

In your case, you want the $n = 0$ bead to "mirror" the $n = 1$ bead: $u_0 = -u_1$ at all times. Try writing $u_n = \cos (\omega t) \sin (np+\phi)$ where $\phi$ is a phase consant. Can you choose $\phi$ so that the boundary condition $u_0 = -u_1$ is satisfied at all times?

The problem is for some reason I couldn't choose any $\phi$ that also satisfies the conditions of the other wall ($u_N = -u_(N+1)$). I did get $\omega$ though, If I am correct that should be $\omega_n = 2 \omega_0 \sin((n\pi)/(N+1))$ where n varies from 1 to N+1?

 

[TSny]

The problem is for some reason I couldn't choose any $\phi$ that also satisfies the conditions of the other wall ($u_N = -u_(N+1)$).

Use the boundary condition at the left end to find $\phi$. Then use the boundary condition at the right end to determine the allowed values of $p$ in $u_n = \cos (\omega t) \sin (np+\phi)$

.I did get $\omega$ though, If I am correct that should be $\omega_n = 2 \omega_0 \sin((n\pi)/(N+1))$ where n varies from 1 to N+1?

Did you mean to write that n varies from 1 to N, instead of 1 to N +1? There should only be N independent modes.

I don't quite get the same expression for $\omega_n$. I believe the fact that the phase constant $\phi$ is nonzero in your problem will modify the expression for $\omega_n$ compared to the case where the n = 0 and n = N +1 beads are fixed in place.

 

[Miles123K]

Use the boundary condition at the left end to find $\phi$. Then use the boundary condition at the right end to determine the allowed values of $p$ in $u_n = \cos (\omega t) \sin (np+\phi)$

Did you mean to write that n varies from 1 to N, instead of 1 to N +1? There should only be N independent modes.

I don't quite get the same expression for $\omega_n$. I believe the fact that the phase constant $\phi$ is nonzero in your problem will modify the expression for $\omega_n$ compared to the case where the n = 0 and n = N +1 beads are fixed in place.

Yes I mean from 1 to N, my bad. I feel the same way actually. That's why I remain in doubt of my answers...

 

[TSny]

So far, you haven't shown how you are getting your answers. If you show your work, it is more likely that we can help.

 

[Miles123K]

I did another attempt using the methods from this lecture:
Provided boundary conditions $y_0 = - y_1$, $y_N = - y_{N+1}$,
Incorporate the system as if it's a part of an infinite system, and the infinite system is subject to translation symmetry matrix $S$, and $S \vec A = \beta \vec A$
I wrote the motion equation $m \ddot y_p = \frac T a y_{p-1} - \frac 2T a y_p + \frac T a y_{p+1}$ Note that $T$ is tension and $a$ is distance between the beads
We know that any system subject to translation symmetry matrix $S$ has normal mode amplitude $\vec A = \begin{bmatrix} \beta \\ \beta^2 \\ \beta^3 \\ \vdots \end{bmatrix}$ whereas $\beta = e^{ika}$
define $\omega_0^2 = \frac T {ma}$ $m$ as mass
$M^{-1}K \vec A = \beta \vec A$ so, for individual terms of $\vec A$
$\omega^2 A_{j} = \omega_0^2 (-A_{j-1}+2A_j-A_{j+1})$
$\omega^2 \beta^j = \omega_0^2 (-\beta^{j-1}+2\beta^j-\beta^{j+1})$
so $\omega^2 = \omega_0^2 (-\beta^{-1}+2-\beta^{1})$
Recall that $\beta = e^{ika}$
$\omega^2 = 2 \omega_0^2 (1-\cos(ka))$
because of the nature of this equation, $k$ and $-k$ have identical solutions of $\omega$
Following the step of the professor, amplitude should be a linear combination of $\beta$ for $k$ and $-k$
$\beta^j = A_j = c_1e^{ijka} + c_2e^{-ijka}$
To apply the boundary conditions here, I would have for $A_0 = A_1$, $A_N = - A_{N+1}$,
So, $c_1+c_2 = -c_1 e^{ika} - c_2e^{-ika}$ (eq.1)
and $c_1e^{iNka}+c_2e^{-iNka} = -c_1 e^{i(N+1)ka} - c_2e^{-i(N+1)ka}$ (eq.2)
My attempt at solving this weird system of equations is to use the fact that since both equations can be rearranged be equal to zero and equation 2 can be reduced, I made the equations like this with all the $c_2$ terms canceled out. Next, I just rearranged to make the equation zero.
$c_1 +c_1 ce^{ika} = c_1 e^{i(2N)ka} + c_1e^{i(2N+1)ka} = c_1e^{i2Nka}(1+e^{ika})$
Keep rearranging and I got:
$(1-e^{i2Nka})c_1(1+e^{ika})=0$
To solve for k that can take different values, $(1-e^{i2Nka})=0$
so $2Nka = n2\pi$ and $k = \frac {n\pi} {Na}$, substituting $k$ back to $\omega^2 = 2 \omega_0^2 (1-\cos(ka))$ would give me the normal modes frequencies but I don't know how to solve for the amplitudes...

I'm actually in High School so I am really doing this all by my own. I sent emails to MIT asking for solutions of their pset and they said this course material is too recent that they are still using them so solutions aren't uploaded. Please don't judge if everything is screwed up.

 

[Miles123K]

So far, you haven't shown how you are getting your answers. If you show your work, it is more likely that we can help.

Ah Yes. I just uploaded my solution.

 

[TSny]

OK, Miles. You are to be commended for tackling this material while still in high school!

So, you are working with complex exponentials. Your equations (1) and (2) below look good:

$c_1+c_2 = -c_1 e^{ika} - c_2e^{-ika}$ (eq.1)
$c_1e^{iNka}+c_2e^{-iNka} = -c_1 e^{i(N+1)ka} - c_2e^{-i(N+1)ka}$ (eq.2)

I also believe you now have the correct result for the allowed values of $k$:

$k = \frac {n\pi} {Na}$

The denominator has a factor of $N$, rather than $N+1$. Good. So, I think this will give you the correct values for $\omega$.

To get the amplitudes of the beads for a particular mode, use your (eq.1) for the boundary condition at the left end. See if you can show that this yields $c_2 = -c_1 e^{ika}$.

Then use this result in $\beta^j = c_1e^{ijka} + c_2e^{-ijka}$. Can you manipulate it into the following?

$\beta^j = c_1 e^{ika/2} \left[e^{i(j-1/2)ka} - e^{-i(j-1/2)ka} \right] = c_1e^{ika/2} 2i \sin\left[(j-1/2)ka \right]$

Since $c_1$ is an arbitrary complex constant, you can just lump together all the stuff in front of the sine function as a new constant, $C$.

 

[Miles123K]

OK, Miles. You are to be commended for tackling this material while still in high school!

So, you are working with complex exponentials. Your equations (1) and (2) below look good:

I also believe you now have the correct result for the allowed values of $k$:
The denominator has a factor of $N$, rather than $N+1$. Good. So, I think this will give you the correct values for $\omega$.

To get the amplitudes of the beads for a particular mode, use your (eq.1) for the boundary condition at the left end. See if you can show that this yields $c_2 = -c_1 e^{ika}$.

Then use this result in $\beta^j = c_1e^{ijka} + c_2e^{-ijka}$. Can you manipulate it into the following?

$\beta^j = c_1 e^{ika/2} \left[e^{i(j-1/2)ka} - e^{-i(j-1/2)ka} \right] = c_1e^{ika/2} 2i \sin\left[(j-1/2)ka \right]$

Since $c_1$ is an arbitrary complex constant, you can just lump together all the stuff in front of the sine function as a new constant, $C$.

Thanks so much! That perfectly resolves my problem.