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