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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?

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1 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:

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

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Is the diagram correct? –  Mathematics Feb 3 '12 at 13:03
    
If the vibrator works the way I guessed it does, then yeah, it's something like that. –  David Z Feb 3 '12 at 18:27

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