A simple question about the stationary wave and fundamental frequency

Question

Suppose we have 2 fixed end connected with a wire and now we insert a vibrator in the middle of the wire, and resonance occur. How would the fundamental frequency looks like?

I know the case when the vibrator is at one ends and another ends are fixed while in this case, there are 2 fixed point and the vibrator is at the middle.

Is the fundamental frequency like this? I imagine half of the original wire acts like a wire with a vibrator at one end and get the result.

Would the fundamental frequency be different if the vibrator which is on the string is vibrating with a very large amplitude?

Answer 1

Presumably the vibrator works by grabbing on to the wire and shaking it up and down. In that case, each half of the original wire acts like a wire with a vibrator at one end. The fundamental frequency will be that of half the wire.

You're basically solving the wave equation

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

with three boundary conditions:

$$\begin{align}y(0,t) &= 0 & y(L,t) &= 0 & y\biggl(\frac{L}{2},t\biggr) &= y_D(t)\end{align}$$

where $y_D(t)$ is the transverse position of the vibrator support as a function of time. Contrast this with the usual two boundary conditions if you put a vibrator at the end:

$$\begin{align}y(0,t) &= 0 & y(L,t) &= y_D(t)\end{align}$$

Answer 2

If both ends of the wire are fixed then the fundamental will have a wavelength twice the distance between the fixed ends, that is:

displacement of wire = $A sin(\frac{\pi x}{d})$

where $d$ is the distance between the ends, $x$ is the distance along the wire and $A$ is the maximum displacement.

As usual, Wikipedia has an excellent article on this, so if you want more details see http://en.wikipedia.org/wiki/Vibrating_string