Physical manifestation of mechanical waves for a non-defined value of the cotangent function

Question

Consider a horizontal rod that is fixed on one of its ends. The other end is under sinusoidal excitation. The wave propagation in the rod can be described with a wave function which is a solution to the following wave equation with the boundary conditions:

$$c^2{\partial ^2u(t,x) \over \partial x^2}={\partial ^2u(t,x) \over \partial t^2} \tag1$$

$$ u(0,x)=0\tag2$$ $$ {\partial u(t,x) \over \partial t}\Bigg|_{t=0}=0 \tag3$$ $$ u(t, 0) = A \sin (w_ot)\tag4$$ $$ u(t, L)=0 \tag5$$ $$ t \geq 0 \tag6$$ $$ L \geq x \geq 0 \tag7$$

where $u(t, x)$ is the wave function, $c$ is the wave speed, $w_o$ is circular frequency, $t$ is time, $x$ is space and $L$ is the rods length. The solution to this problem is derived here. The wave function is formulated as:

$$ u(t,x) = {A \over 2} \Bigg[ \sin \Biggl(w_0\bigg(t+ {x\over c}\bigg)\Bigg) + \cot \bigg( w_0 {L \over c} \bigg) \cos \Bigg( w_0\bigg(t+ {x\over c}\bigg)\Bigg) \Bigg] + \Bigg[A \sin \Bigg(w_0\bigg(t - {x\over c}\bigg)\Bigg)- {A \over 2} \Bigg( \sin \Bigg(w_0 \bigg(t- {x\over c}\bigg)\Bigg) + \cot \bigg( w_0 {L \over c} \bigg) \cos \Bigg(w_0 \bigg(t- {x\over c}\bigg)\Bigg)\Bigg)\Bigg] \tag{8} $$

In theory, we can choose the material and length of the rod so the argument in the cotangent function is $n\pi$, where $n$ is a whole number. Since the cotangent function is not defined for $n\pi$, what would actually happen if we designed the rod so that $w_0 {L / c} = n\pi$? Would the rod explode?

Answer 1

Short Answer: Real life would kick in, and the rod would vibrate in a slightly different manner than this theory would suggest.

Longer Answer:

The wave equation is a mathematical idealization that neglects a variety of other effects. (As far as I know, the only place in physics that the linear wave equation shows up without any approximation is for electromagnetic waves in a vacuum; everywhere else approximations are being made.) For the bar in question, there are at least two simplifications that need to be made to get the wave equation you cite. First, the bar is perfectly elastic, or that no mechanical energy can be lost to heat or plastic deformation. This assumption is valid as long as you are looking at short time scales; if the system is left on for a long enough period, these effects will become important. The second simplification is that the bar responds linearly to acoustic waves. For example, if you apply one newton of force and get a displacement of 1 mm, then you would expect that applying two newtons of force would get a displacement of 2 mm. For very low-amplitude signals this assumption is valid, but as the amplitude of the pressure increases all materials begin to deviate from this approximation.

If you choose your frequency such that $w_0L/c=n\pi$, you are pushing the system into resonance. The amplitude will steadily grow in this case, and at some point nonlinear effects would kick in. Also, as the time wore on, more energy would be taken from the system. Thus, the amplitude would be large, but it would not be likely to explode.

Answer 2

You would be at resonance. The amplitude of the wave would grow with time to the point at which the linear wave equation is no longer a good approximation.