What are the approximations used in these equations of motion for two coupled oscillators?

[zenterix]

Homework Statement

The questions in this post are about the lecture "Beat Phenomena" from MIT OCW's 8.03 "Vibrations and Waves".

Relevant Equations

Consider the coupled oscillators below.

Here is the actual lecture.

The equations of motion for $m_1$ are

$$m_1\ddot{x}_1=-T_1\sin{\theta_1}+k(x_2-x_1)\tag{1}$$

$$m_1\ddot{y}_1=T_1\cos{\theta_1}-m_1g\tag{2}$$

The lecture says that we are using small angle approximations

$$\sin{\theta_1}\approx\theta_1\tag{3}$$

$$\cos{\theta_1}\approx 1\tag{4}$$

and proceeds to write the equations of motion as

$$m_1\ddot{x}_1=-m_1g\frac{x_1}{l}+k(x_2-x_1)\tag{5}$$

$$T_1=m_1g\tag{6}$$

I have two questions.

1) If we use the small angle approximation for $\cos$ in (2) then we get

$$m_1\ddot{y}_1=T_1-m_1g\tag{2a}$$

It seems they are making an additional assumption that $\ddot{y}=0$. Is this so, and if so why?

2) If we use the small angle approximation for $\sin$ in (1) then we get

$$m_1\ddot{x}_1=-T_1\theta_1+k(x_2-x_1)\tag{1a}$$

However, it seems that this is not what was done. No approximation was used because $\sin{\theta}=\frac{x_1}{l}$.

Is this so?

 

[anuttarasammyak]

Though I have not read the text, I observe an approximation in (1) which says the spring keeps horizontal.

 

[kuruman]

It seems they are making an additional assumption that y¨=0. Is this so, and if so why?

To add to the post above by @anuttarasammyak, the vertical displacement is $y=l(1-\cos\theta_1)$. This is zero to first order. I think we've been there in another post.

Is this so?

This is so. The approximation is in considering the motion strictly horizontal to first order.

 

[zenterix]

To add to the post above by @anuttarasammyak, the vertical displacement is $y=l(1-\cos\theta_1)$. This is zero to first order. I think we've been there in another post.

This is so. The approximation is in considering the motion strictly horizontal to first order.

We have been there in another post it is true. Ok, so that is the reason then. I will remember now.

 

[zenterix]

I understand at this point most of what I have seen about coupled oscillators. I still have a few doubts about some of the fundamental/basics of the math involved.

Suppose that in the system in the original post we have $m_1=m_2$ for simplicity.

The system of equations is

$$\begin{bmatrix}\ddot{x}_1 \\ \ddot{x}_2 \end{bmatrix}=\begin{bmatrix}-\left (\frac{k}{m}+\frac{g}{l} \right )&\frac{k}{m} \\ \frac{k}{m} & -\left (\frac{k}{m}+\frac{g}{l}\right )\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix}$$

It seems there are various ways to go about solving this system.

One way is to use the trial solution $\vec{z}=e^{i(\omega t+\phi)}\vec{A}$.

Is it possible to omit the phase and use $\vec{z}=e^{i\omega t}\vec{A}$?

In both ways, the eigenvalue equation is the same so we reach the same eigenvalues and the same eigenvectors (ie, normal modes).

In the first way we reach

$$\vec{z}_1=e^{i(\omega_1 t+\phi_1)}\vec{A}_1\tag{1}$$

$$\vec{z}_2=e^{i(\omega_2 t+\phi_2)}\vec{A}_2\tag{2}$$

We take the real part

$$\vec{x}_1=\text{Re}(\vec{z}_1)=\cos{(\omega_1 t+\phi_1)}\begin{bmatrix}1\\1 \end{bmatrix}\tag{3}$$

$$\vec{x}_2=\text{Re}(\vec{z}_2)=\cos{(\omega_2 t+\phi_2)}\begin{bmatrix}1\\-1 \end{bmatrix}\tag{4}$$

We construct a general solution by taking a linear combination

$$\vec{x}=\alpha\vec{x}_1+\beta\vec{x}_2\tag{5}$$

There are four unknowns: $\alpha, \beta, \phi_1, \phi_2$.

My question is about the second way.

$$\vec{z}_1=e^{i(\omega_1 t)}\vec{A}_1\tag{6}$$

$$\vec{z}_s=e^{i(\omega_2 t)}\vec{A}_2\tag{7}$$

$$\vec{x}_1=\text{Re}(\vec{z}_1)=\cos{(\omega_1 t)}\begin{bmatrix}1\\1 \end{bmatrix}\tag{8}$$

$$\vec{x}_2=\text{Re}(\vec{z}_2)=\cos{(\omega_2 t)}\begin{bmatrix}1\\-1 \end{bmatrix}\tag{9}$$

If we now take a linear combination, we have only two unknowns, $\alpha$ and $\beta$.

I think we can fix this problem by using the imaginary part of (6) and (7).

$$\vec{x}_1=c_1\text{Re}(\vec{z}_1)+c_2\text{Im}(\vec{z}_1)=(c_1\cos{(\omega_1 t)}+c_2\sin{(\omega_1 t)})\begin{bmatrix}1\\1 \end{bmatrix}\tag{8a}$$

$$\vec{x}_2=c_3\text{Re}(\vec{z}_2)+c_4\text{Im}(\vec{z}_2)=(c_3\cos{(\omega_2t)}+c_4\sin{(\omega_2 t)})\begin{bmatrix}1\\-1 \end{bmatrix}\tag{9a}$$

And then we sum these to get a general solution with four unknowns.

Equivalently, we have four different independent solutions

$$\cos{\omega_1 t}\begin{bmatrix}1\\1 \end{bmatrix}\tag{10}$$

$$\sin{\omega_1 t}\begin{bmatrix}1\\1 \end{bmatrix}\tag{11}$$

$$\cos{\omega_2 t}\begin{bmatrix}1\\-1 \end{bmatrix}\tag{12}$$

$$\sin{\omega_2 t}\begin{bmatrix}1\\-1 \end{bmatrix}\tag{13}$$

and we take a linear combination.

So my question is, why is it (perhaps in linear algebra terms) that using phase $\phi$ eliminates the need to find four independent solutions?

It seems that when we use a phase in the trial solution we only need two independent solutions.

 

[anuttarasammyak]

The system of equations is

The simplest observation of the system would be
\ddot{(x_1+x_2)}=-\frac{g}{l}(x_1+x_2)
\ddot{(x_1-x_2)}=-(\frac{2k}{m}+\frac{g}{l})(x_1-x_2)
, two independent oscillations. Two amplitudes and two phases are expressed by two initial positions and two initial velocities.