# Coupled pendulums at half height Suppose we have the system described below (poor quality but it'll do the trick). We have two pendulums of mass $$m$$ coupled by a string of constant $$k$$ placed at a height $$a$$ from the top (as shown). My job is to find the normal modes of vibration.

My attempt: For a small angle the movement of the masses is, to a very close approximation ,linear, the force exerted by the string is proportional to the height wich could be modeled by $$F=k\frac{a}{L}$$ and the restoring force by the weight of the pendulum is $$F_1=\frac{mg}{L}$$. With this we can write the equations of motion as follows:

$$m\ddot{x}_1=-\frac{mg}{L}x_1-\frac{ka}{L}(x_1-x_2)\\ m\ddot{x}_2=-\frac{mg}{L}x_2-\frac{ka}{L}(x_2-x_1)$$

With this we just plugin in the matrix equation $$M\ddot{X}=KX$$ and find the eigen values (where $$M$$ is the mass matrix, $$X$$ the position matrix and $$K$$ the constants matrix) giving the following modes of vibration: $$\omega^2_+=\frac{g}{L}+\frac{2ka}{mL}\\\omega^2_-=\frac{g}{L}$$

And this made snese and seems correct to me, but then I saw a solution that considered the motion in terms of angle displacement where the equations of motion would be written as: $$mL^2\ddot{\theta}_1=-mgL\theta_1-ka^2(\theta_1-\theta_2)\\ mL^2\ddot{\theta}_2=-mgL\theta_2-ka^2(\theta_2-\theta_1)$$

with the normal frequencies: $$\omega^2_+=\frac{g}{L}+\frac{2ka^2}{mL^2}\\\omega^2_-=\frac{g}{L}$$

There is a clear ressemblance between both answers, though are both correct? The first one seems right for small angles, and the second one seems flawed to me because I don't think that the force exerted by the spring (though it is modeled by Hookes Law) is directly proportional to the angle displacement. Could someone shed a light into this?

The problem with deriving the equation of motion from a balance of forces is that you need to include the reaction forces acting at the pivot. However, these reaction forces are unknowns that are to be eliminated, and so the balance of forces is “wasted” in solving for these reaction forces. It is for this reason that a balance of moments is to be considered, where moments are measured about the pivot: this way, the unknown reaction forces are not involved.

The 1st set of equations contain an error in one of the terms, where $$\frac{ka}{L}(x_1 - x_2)$$ should be replaced with $$\frac{ka^2}{L^2}(x_1-x_2)$$. The 2nd set is correct. Note that the terms in the 2nd set correspond to the moments exerted by the forces.

For given rotations $$\theta_1$$ and $$\theta_2$$, the deflection of the spring is $$a(\theta_2-\theta_1)$$, and so the spring force acting on pendulum 1 is $$k a(\theta_2-\theta_1)$$. With the lever arm of the spring force about the pivot being $$a\cos{\theta_1}$$, the moment that the spring force exerts on pendulum 1 is equal to $$ka^2(\theta_2-\theta_1)\cos{\theta_1}$$. Linearised, this becomes $$ka^2(\theta_2-\theta_1)$$. This is the term that appears in the 2nd set of equations.

Both sets of equations will be equivalent to one another. The corrected first set can be obtained by the substitutions $$x_1 = L\theta_1$$ and $$x_2 = L\theta_2$$.

Regarding the spring force, in response to comments

Let's look at the spring force more carefully.

To do so, an extra dimension must be included (which will disappear once the system is linearised). This is the distance between the pivots of the pendulums, which I will call $$b$$. We can see that the positions of the connection points $$A_1$$ and $$A_2$$ of the springs are at $$(a\sin{\theta_1},-a\cos{\theta_1})$$ and $$(b+a\sin{\theta_2},-a\cos{\theta_2})$$ respectively, with the first pivot being treated as the origin. The position of the first point relative to the second can be written as

$$\vec{A_1 A_2} = \begin{bmatrix}a\sin{\theta_1} - a\sin{\theta_2} - b\\-a\cos{\theta_1}+a\cos{\theta_2}\end{bmatrix}$$

The length of this vector gives the length of the stretched string (natural length $$b$$ + extension $$\Delta x$$). By noting the following Taylor expansions,

$$\sin{x} = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \dots = x + O(x^3)$$

$$\cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \dots = 1 + O(x^2),$$

the relative position vector can be represented as

$$\vec{A_1 A_2} = \begin{bmatrix}a(\theta_1 - \theta_2) - b + O(\theta_1^3)+ O(\theta_2^3)\\O(\theta_1^2)+ O(\theta_2^2)\end{bmatrix}$$

When the system is linearised by small angle approximations, we find that $$O(\theta_1^2)$$ and $$O(\theta_2^2)$$ can be neglected in the relative position vector, and so it approximates to

$$\vec{A_1 A_2} = \begin{bmatrix}a(\theta_1 - \theta_2) - b \\0\end{bmatrix}$$

We can see that the above relative position vector is approximately horizontal, i.e. the spring is approximately horizontal, and so the force it exerts is approximately horizontal - the vertical component of the force can be neglected.

The length of the relative position vector is $$a(\theta_1 - \theta_2) + b$$, and so therefore, the extension of the spring is $$\Delta x = a(\theta_1 - \theta_2)$$. We know that the spring force is $$F = k\Delta x$$, and so the spring force is $$F = ka(\theta_1 - \theta_2)$$

• I hadn't noticed the substitution, but with it, we can see again that $\theta_1=\frac{x_1}{L}$ and $\theta_2=\frac{x_2}{L}$, which is equivalent to a small angle approximation ($\sin \theta_1 \approx \theta_1$, which makes sense, now the second set, the problem I see ( I may be wrong) is that the force is proportional to the displacement "transversal to the twisting of its wire", that is, if $L\theta_1$ corresponds to the distance traveled by the pendulum from $(0,0)$ to $(x,y)$ the the final force on the spring should be be $F=k\sqrt{x^2+y^2}$ and not $F=kL\theta_1$. – Bidon Dec 26 '18 at 9:26
• It's this "implied" linearity that's confusing me – Bidon Dec 26 '18 at 9:27
• Then again, just thought a bit more, and that substitution won't do the trick... – Bidon Dec 26 '18 at 11:12
• If the spring force on pendulum 1 was $F = k\sqrt{x_1^2 + y_1^2 }$, that would correspond to a spring being attached between the pendulum bob and a fixed point at the lowest point of travel of the bob. I will edit my answer to include a note regarding the spring force in more detail. – Involutius Dec 26 '18 at 13:15