Reflection in stationary waves

Question

A stationary wave is formed by the superposition of two waves with the same frequency and amplitude travelling at the same speed in opposite direction.

Consider a wave generated by a vibrator is travelling to the right and approaching a fixed end.

The wave is then reflected at the fixed end.

The reflected wave is travelling to the left while the incident wave is travelling to the right.

My question:

Why the reflected wave from the right is not reflected again at the left when it arrives to the position of the vibrator? If it does, shouldn't there be two waves travelling to the right after the second reflection and hence amplitude of the stationary wave should increase along with time? If it doesn't, where will be the reflection wave?

Answer 1

For sure it does ! In textbooks,one usually assumes that the incident wave comes from infinity ($x\to -\infty$) and then there is not reflection of the reflected wave.

If oppositely, you have two ends at finite distance, you have a more subtle system described as a "cavity". In such case, one could consider that there are an infinity of waves propagating to the left and to the right. To make finite that infinity you'd assume that there is some damping, due to dissipation along the travel or to energy losses at the reflection.

You will also observe some resonances, in the sense that the stationary wave has quantified possible wavelength.

In fact the main point is that stationary waves an not only stationary in the sense that the phase does not propagate (it is constant on each half wavelength between two nodes), but also that it describes a steady state where the origin in time or the "origin" of the waves does not play any role.

By the way, I'm afraid to tell that, IMO, your drawing do not really describe a standing wave,especially because the reflection coefficient on a fixed end is $-1$.

Answer 2

Let us assume a driving force $F=F_{0}e^{j\omega t}$ at $x=0$ excites a transverse wave between $x=0$ and $x=L$ with a fixed end condition at $x=L$, the general solution will write as a sum of propagating waves in two opposite directions (namely, $+x$ and $-x$) : $$ u(x,t)= Ae^{j(kx+\omega t)} + Be^{-j(kx+\omega t)} \tag{1}$$ In the remaining, the $e^{j\omega t}$ term will be dropped. The boundary conditions yield: $$ F_{0}= \frac{\partial{u}}{\partial x}(0,t)= jkA-jkB$$ $$ Ae^{jkL}+Be^{-jkL}=0$$ After substitution we find $B=-A e^{j2kL}$ and $A=\frac{F_{0}}{jk(1+e^{j2kL})}=\frac{F_{0}}{2jk\cos(kL)}e^{-jkL}$, $B=-\frac{F_{0}}{2jk\cos(kL)}e^{jkL}$

So for particular $k$ such that $kL=(2n+1)\frac{\pi}{2}$, $A,B \to \infty$, a resonance phenomenon occur.

The general solution eventually writes (substituting $A$, $B$ in (1), and taking the real part): $$\frac{F_{0}}{k\cos(kL)}\sin(k(x-L))\cos(\omega t) $$

which is a standing wave pattern wich has (in general) a constant and finite energy. As to what happens, my understanding is that in steady state the energy produced in the forward direction is dissipated at the source when it's coming back in the reverse direction, except at resonance, when energy builds up (theoretically) at infinity.