Why is vertical displacement second order in horizontal displacement for coupled oscillator?

[zenterix]

Homework Statement

Consider a system of coupled pendulums.

Relevant Equations

Below we see a picture of this

The masses are $m_1$ and $m_2$.

We measure the displacements of each mass as follows

When the blocks move horizontally, they will move vertically as well, because the length of the pendulums remains fixed. Because vertical displacement is second order in the $x_j$'s,

$$y_j\approx \frac{x_j^2}{2}\tag{1}$$

we can ignore it in thinking about the spring.

Where does (1) come from?

 

[kuruman]

If $L$ is the pendulum length and angle $\theta$ is relative to the vertical, write an expression for $y$ as a function of $\theta$. Apply the small angle approximation. Note that $x\approx L\theta.$ What do you get?

 

[TSny]

$$y_j\approx \frac{x_j^2}{2}\tag{1}$$Where does (1) come from?

There's a typographical error in equation (1). Note that the dimensions of the two sides are not the same. The pendulum's length $l$ was left out of the expression on the right side.

You can derive the correct expression by following @kuruman's suggestion. However, I think he meant $x \approx L \theta$ instead of $y \approx L \theta$.

 

[zenterix]

I did the following

$$T\cos{\theta_i}-m_ig=m_i\ddot{y}$$

$$\cos{\theta_i}=\frac{l-y_i}{l}=1-\frac{y_i}{l}\approx 1$$

$$\sin{\theta_i}=\frac{x_i}{l}\implies x_i\approx l\theta_i$$

We can also consider the radial and tangential components

$$m_ig\cos{\theta_i}-T+F_s\sin{\theta_i}=-m_il\dot{\theta_i}^2$$

$$m_ig\sin{\theta_i}-F_s\cos{\theta_i}=m_il\ddot{\theta_i}$$

 

[kuruman]

You can derive the correct expression by following @kuruman's suggestion. However, I think he meant $x \approx L \theta$ instead of $y \approx L \theta$.

That's what I meant. Good catch - typo corrected.

 

[kuruman]

I did the following

View attachment 348821

$$T\cos{\theta_i}-m_ig=m_i\ddot{y}$$

$$\cos{\theta_i}=\frac{l-y_i}{l}=1-\frac{y_i}{l}$$

$$\sin{\theta_i}=\frac{x_i}{l}\implies x_i\approx l\theta_i$$

We can also consider the radial and tangential components

$$m_ig\cos{\theta_i}-T+F_s\sin{\theta_i}=-m_il\dot{\theta_i}^2$$

$$m_ig\sin{\theta_i}-F_s\cos{\theta_i}=m_il\ddot{\theta_i}$$

Start from $y=L(1-\cos\theta)$. This expression is exact. What is $\cos\theta$ for small values of $\theta$?

 

[zenterix]

Start from $y=L(1-\cos\theta)$. This expression is exact. What is $\cos\theta$ for small values of $\theta$?

$$y_i=l(1-\cos{\theta_i})\approx 0$$

since $\cos{\theta}\approx 1$.

 

[zenterix]

If we do a second order approximation of cosine we get

$$\cos{\theta_i}=1-\frac{y_i}{l}\approx1-\frac{\theta_i^2}{2}$$

$$\implies y_i\approx \frac{l\theta_i^2}{2}\approx\frac{lx_i^2}{2l^2}=\frac{x_i^2}{2l}$$

 

[kuruman]

Now that you have answered your title question "Why is vertical displacement second order in horizontal displacement for coupled oscillator?" by doing the actual math, you see that could have answered this question immediately without doing the algebra. The potential energy is a symmetric function in $x$, $U(-x)=U(x)$, and has a minimum at $x=0$. This means that the first derivative of the potential vanishes at $x=0$. The zeroth, constant term, is the zero of potential energy which can be chosen arbitrarily. The first term with a dependence on $x$ is to second order.

By the way, this idea is used for modeling solids as very long chains of spring-mass systems. We know that the atoms in solids are in stable equilibrium which means that they are in a potential with a local minimum. We don't know the exact form of the the potential function but, for small displacements from equilibrium, its expansion will be of the form $U(x)=U(x_0)+\frac{1}{2}kx^2.$ So we can imagine that each atom is connected to its neighbors with tiny springs and proceed to model wave propagation in solids.