# D'Alembert Equation and standing wave

As we know the d'Alembert Equation is

$$\frac{\partial^2\psi}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2\psi}{\partial t^2}$$

for an undimentional string.

Now if we seek standing wave solution, we put $\psi(x,t)=f(x)g(t)$ can you tell me a physical argument which shows if $\psi$ is a standing wave then $\psi(x,t)=f(x)g(t)$.

I was speaking of this with my maths teacher when he said me he never understood why we use that. My physics teacher has no idea and I don't find anything about it in Feynman lessons.

And I have no idea for tags so I take "waves"

Thank you

Suppose $x_0$ is one of the nodes at a particular time instant $t_0$, then we have $\psi(x_0,t_0)=0$. If position and time can be separated, then we have $f(x_0)g(t_0)=0$, which gives us two cases:
1. $f(x_0)=0$, then we have $\psi(x_0,t)=f(x_0)g(t)=0$ for all $t$, i.e. $x_0$ is always a node of the wave. It doesn't move with time.
2. $f(x_0)\neq 0$, then $g(t_0)=0$, then we have $\psi(x,t_0)=f(x)g(t_0)=0$ for all $x$, including $x_0$. This only tells us that at this particular time $t_0$ all the $x$ are nodes. Since we need to look at the time evolution characterized by a function $g(t)$ that is not constantly zero, this case is not so interesting.
Using the same argument, you can easily check that if we have functions where $x$ and $t$ are combined in such way that they are not able to be separated, a typical example is the general travelling wave solution $f(x\pm vt)$, then the nodes of the wavefunction will move with time.
• you only show that $f(x)g(t)$ describes standing wave, but not that standing wave is described by $f(x)g(t)$ – aaaaa says reinstate Monica May 14 '15 at 4:13