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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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2 Answers 2

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

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

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is the diagram correct??? –  Mathematics Feb 3 '12 at 13:03
    
The info in wikipedia is a vibrating strong with no vibrator at the middle –  Mathematics Feb 3 '12 at 13:04
    
Are you assuming the vibrator holds the string still, and is that why you've drawn a node in the middle of the string? If the vibrator was, for example, a violin bow then it works by moving the string so there would not be a node in the middle. In that case the fundamental is as I've described above. If the vibrator does hold the string still then the fundamental will be as you've drawn. maybe you need some clrification about the question. –  John Rennie Feb 3 '12 at 14:03
    
What do you mean by " vibrator holds the string still'??Isn't it the vibrater always treated as a node in this case? –  Mathematics Feb 3 '12 at 15:48
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If this is a homework question then you need to ask your professor what they meant by "vibrator". I would interpret it as "something that vibrates the string" and by definition something that vibrates the string can't hold the string still because vibration always implies motion. I would describe a violin bow as a classic example of a "vibrator" and it works by moving the string. –  John Rennie Feb 3 '12 at 17:18

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