Speed $v$ of a wave on a tightrope Starting from D'Alembert's general wave equation we can find the propagation speed of an impulse along a tightrope as a function of linear mass density,
$$\boxed{v= \sqrt{\frac \tau \mu}} \tag 1$$
where $\tau$ it is tension of the rope (in Newton) and $\mu=m/\ell \,$ it is the mass of the rope per unit of length.

Are there simple methods for students of an high school question who don't know calculus (derivatives and partial derivatives, of course) to be able to get the relation $(1)$?

 A: If you know (just using trig) the formula for centripetal acceleration $v^2/r$, and (again just using trig) how to get the centripetal force as $T/r$ (analogous to the excess pressure in a soap bubble being $2T_{\rm surface}/r$) then you can see that if the rope moves  at speed $v= \sqrt{T/\mu}$ then the  curvature $\kappa=1/r$  of the rope can be anything. Now go to the frame in which the rope is stationary. In this frame  the "anything" shape will be moving  at $v= \sqrt{T/\mu}$.
A: You can use unit analysis to get the form of the formula.
The properties that completely define a thin string are the tension ($\tau$) with unit $N = kg\cdot{}m/s^2$ and linear mass density ($\mu$) with unit $kg/m$. From these two quantities, we need to derive a velocity. As a guess, we assume that the form of the formula is of a product of the two quantities raised to some powers:
$$v=\tau^a\mu^b$$
We can assume this because multiplication, division, and powers are the only operations that give results with different units. Addition and subtraction require the inputs to have the same units and the output will also have the same units (meters plus meters gives meters). Other operations ($\sin$, $\cos$, $\exp$, $\log$, etc.) require unitless inputs and outputs, and we don't have other quantities to cancel the units of the values we have.
By considering just the units, we can figure out what the exponents $a$ and $b$ must be.
$$\left[\frac{m}{s}\right] = \left[\frac{kg\cdot{}m}{s^2}\right]^a\left[\frac{kg}{m}\right]^b$$
The exponent $a$ must be $1/2$ to get seconds in the denominator, which means $b$ must be $-1/2$ so that we get meters in the denominator and kilograms to cancel. This results in a formula for the wave speed:
$$v = \tau^{1/2}\mu^{-1/2} = \sqrt{\frac{\tau}{\mu}}$$
The thing to watch out for is that this only gives the form of the equation. There may be unitless constants that won't show up in this kind of derivation. If you do the same derivation for the period of a pendulum ($T$) using the length of the pendulum ($L$) and the gravitational acceleration ($g$), then you will get the expression
$$T=\sqrt{\frac{L}{g}}$$
when the actual formula is
$$T=2\pi\sqrt{\frac{L}{g}}$$
The $2\pi$ is missed because it doesn't have any units. It can be derived through measurements of a real pendulum.
A: Instead of a sine wave I consider a single symmetric pulse like the one in the following image. We choose a reference system in which the pulse remains stationary (i.e. we chase the pulse in so that we always have it in view). In this way the string will appear to flow in front of us from right to left with velocity $v$.

Consider a small segment of length $Δl$ of the string crossed by the impulse that forms an
arc of circle of radius $R$ and which subtends an arc of amplitude $2θ$.

It results that
$$Δl = (2θ)R \iff 2\theta=\frac{\Delta l}{R}.$$
A force of magnitude equal to the tension $\mathbf F$
acts tangentially on this segment from
both sides. The horizontal components of these two forces add up to cancel each other out, but the
vertical components add to form the
radial force.

The intensity of the resultant force $\mathbf F_{\text{radial}}$ will be
$$F_{\text{radial}}=2F\sin(\theta)\sim_0 F(2\theta)=\frac{F\Delta l}{R}$$
The mass of the segment is given by $Δ m = μ Δl \ $
where $μ$ is the linear density of the string.
We know that the string element $Δl$ is moving on an arc of a
circumference. Therefore it has a centripetal acceleration toward the center of the circle given by
$$a_{\text{cp}}=v^2/R$$
From the second law of dynamics we know that
$$\frac{F\Delta l}{R}=Δ m\frac{v^2}R= \frac{μ Δl v^2}R \tag 1$$
which solving the $(1)$ respect to the speed we get:
$$v=\sqrt{\frac{F}{\mu}}\quad .\square$$
