Taking equally spaced *lengths* will *not* reproduce the pendulum wave effect, simply because it's the *periods* that you want to fulfil commesurability conditions, and lengths and periods are not linearly related.

To get a pendulum wave, say you have $N$ pendula with lengths $l_n$. Then, as you know, the $n$th pendulum will have a period $$T_n=2\pi\sqrt{\frac{l_n}{g}}.$$ For the pendulum wave to work you need to prearrange a time $T$ at which all the pendulums come back into step. (Most demonstrations choose $T=60\text{ s}$, but this is not necessary.) For this to happen, $T$ needs to be an integer number of periods for each pendulum, so therefore
$$\boxed{nT_n=T.}$$
(This presupposes that we're using the number of periods in time $T$ to index the pendula. This is not strictly necessary but simplifies things, and people usually choose some relatively narrow range in $n$.)

Given the above relationship between $T_n$ and $l_n$, the commesurability condition reads
$$
l_n=g\left(\frac{T}{2\pi n}\right)^2
$$
in terms of the lengths $l_n$. Here $n$ should be an integer with $n\geq1$, but most demonstrations take $10\lesssim n\lesssim 30$ or so. Using very slow pendula, with low $n$, would mean the ratio of lengths between the first and the last pendulum would come out rather large.

For *relatively small* (see Note) angular twists of the wire, the torque $\tau$ which restores the wire to its untwisted position is proportional to the angle $\theta$ through which the end of the wire is rotated : $$\tau=-\kappa \theta$$

$\kappa$ is called the **torsion constant** of the wire. The minus sign indicates that the direction of the torque is opposite to the direction in which angle $\theta$ is increasing.

The torque causes rotational acceleration $\ddot\theta$ of mass at the end of the wire. Newton's 2nd Law for this acceleration is $$\tau=I\ddot\theta$$ where $I$ is the moment of inertia of the mass. This is the rotational equivalent of $F=ma$.

Combining the above two equations, the equation of motion for the torsional pendulum is $$\ddot\theta+\frac{\kappa}{I}\theta=0$$
This has the form $\ddot x+\omega^2 x=0$ which describes Simple Harmonic Motion. Here $\omega=2\pi f$ is the angular frequency of the periodic motion (radians per second) and $f$ is frequency (cycles per second; one cycle is $2\pi$ radians). Period and frequency are related by $f=\frac{1}{T}$ and $\omega=\sqrt{\frac{\kappa}{I}}$ therefore $$T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{I}{\kappa}}$$

This can be compared with the period of a mass on a spring : $$T=2\pi\sqrt{\frac{m}{k}}$$ The moment of inertia $I$ is equivalent to mass $m$ and the torsion constant $\kappa$ is equivalent to the force constant $k$ or stiffness of the spring.

The final difficult step is to relate torsion constant $\kappa$ to the dimensions of the wire and its shear modulus $\eta$. This calculation is done in Deriving the Shear Modulus S From the Torsion Constant κ. The result is $$\kappa=\frac{\eta \pi r^4}{2L}$$ in which $r$ is the radius of the wire and $L$ its length. Substituting into the equation for the period we get $$T=2\pi\sqrt{\frac{2LI}{\eta \pi r^4}}$$ which is the formula you were given.

**Note**

The angle $\theta$ through which the end of the wire is rotated is not the same as the angle of twist $\psi$ along the length of the wire. Whereas $\theta$ is measured by the rotation of a radius in the base of the wire, $\psi$ is measured by the twist of a line parallel to the axis. The two are related by $$L\psi=r\theta$$

To ensure that the **elastic limit** of the wire is not exceeded $\psi$ should be small, typically no more than $10^{\circ}$. However since $L\gg r$ the angle $\theta$ can be quite large. The base can be rotated a whole circle without exceeding the elastic limit.

## Best Answer

The initial condition affects the amplitude. From a free body diagram and using that the power is null (energy concervation $\frac{\mathrm{d} E_{mec}}{\mathrm{d} t} = 0$ ) you find as solution (forgetting the trivial solution $\dot\theta = 0$, where $\theta$ is the angle of the pendulum)

$$ \ddot\theta = -\omega_0 \sin\theta ,$$

with $\omega_0^2= \frac{g}{l}$.

Then expanding the $\sin$ to first order (linear approximation) you'd find

$$\ddot\theta = \theta$$ which is an harmonic oscillator and has solution

$$\theta(t) = \theta_0\sin{\left(\omega_0 t\right)}$$.

With $\theta_0$ being the drop angle. Now you can already see that the period depends on the gravity and length (so you already found an other one, i.e. gravity).

But one can do more. The $\sin$ approximation is getting valid only when the angle gets small (how small?). One could get a better approximation by expanding to higher orders. $\sin x = x - \frac{x^3}{6} + o(3)$. Then one would look for solutions of the differential equation

$$\ddot \theta = \omega_0^2\left(\theta - \frac{1}{6}\theta^3\right)$$

sometimes called the Duffing equation. It requires advanced tools to be solved (such as Fourier series expansion). I can tell you that the solution would have a period

$$T = \frac{2\pi}{\omega_0}\left(1+\frac{\theta_0^2}{16}\right)$$.

So you see that with this better approximation the period of the pendulum depends on the drop off angle. Which intuitively makes sense.

If you don't want to solve the equations, you could also keep the $\sin$ formulation and solve the system numerically (implicit Euler or Runge-Kutta for instance) and you would see that the oscillations depend on the drop off angle. Or check by yourself with a pendulum (lab style).

If you have friction the energy will not be conserved and the period will not even be a constant of motion. The period would actually decrease in time.

I hope this helps :), Best