0
$\begingroup$

What is the sound of a solution $y(x,t)$ to the Wave Equation

$\frac{\partial^2y}{\partial t^2}=c^2\,\frac{\partial^2y}{\partial x^2}\enspace$?

To be clear, I don't look for a solution of the Wave Equation, I have a concrete solution $y(x,t)$. I only want to describe the sound of this concrete solution.

In my case I want to describe the sound of a vibrating string, so the ends of the string are fixed, i.e. $y(0,0)=y(L,0)=0$ where $L$ is the length.

With a solution $y(x,t)$, I get perfectly the shape of the string between $x=0$ and $x=L$ for every time $t$.

I can show that $y(x,t)$ must be also $2L$-periodic according to $x$, i.e. $y(\lambda+2L,t)=y(\lambda,t)$, and $(2L/c)$-periodic according to $t$, i.e. $y(x,\lambda+2L/c)=y(x,\lambda)$.

To describe the sound, I would like to

  1. find a frequency spectrum (or magnitude/amplitude spectrum) or
  2. produce a sound file with a series of samples $x(i)$ to hear the vibrating string.

For 1. I don't know how to do that. Should I develop the Fourier-Series according to $x$ or to $t$ (a Fourier-Series exists, because I've showed the periodicity) or do complety other things? For a Fourier-Series according to $x$, the Fourier-Coefficients $a_k,b_k$ depends on $t$ and for a Fourier-Series according to $t$, the Fourier-Coefficients depends on $x$.

Any idea?


Edit:

Here is an example for a string that get plucked at $x_p$ with a displacement of $h$. Its shape is a triangle as follows:

$y_0(x)= \begin{cases} x(h/x_p)&\text{if}\enspace 0\leq x < x_p\\[1ex] h\left(1-\frac{x-x_p}{L-x_p}\right)&\text{if}\enspace x_p \leq x \leq L \end{cases}$

The initial velocity is 0 so finally the string is described as: $y(x,t)=y(x,t)=\frac{1}{2}\left(y_0(x+ct)+y_0(x-ct)\right)$

In the following picture, you can see the process of the string starting at $t=0$ Process of a Plucked String

$\endgroup$
1
  • $\begingroup$ all depends on your initial conditions. what $f(x)$ was at $t=0$. the period (in time) will be $\frac{2L}{c}$ and the pitch will be $ log_2\left( \frac{c}{2L f_0} \right)$ measured in octaves from your reference pitch that has fundamental frequency $f_0$. $\endgroup$ Mar 27, 2015 at 0:20

1 Answer 1

0
$\begingroup$

Since you already have the solution to the wave equation, what you need to do now is define a "pick-up" point somewhere along the length of the string (some $x$) and obtain its trace in time, i.e. look at what that specific $x$ value is "doing" across all the $t$ instances that you have. In other words, draw a straight line down your vibrating string diagram and take the vertical displacement of your string at each time point as the amplitude of your sound signal at that time point.

In reality, when you hear an electric instrument you hear exactly that, you hear what the vibrating string is doing exactly above the pick-up coil. Furthermore, electric guitars (and other instruments) have more than one pick-up points along the length of a string and a switch that is used to change the way these pick-up points are combined to offer different qualities of sound. For example, the coil closest to the neck (closest to the middle of the distance between the neck and bridge) produces sound with more low frequencies, while the coil closest to the bridge (the fixed point at the lower end of the instrument) produces sound with more high frequencies. The placement of the coils is a very important issue. Imagine for example what would happen if you placed the coil at a nodal point on the string (when it is excited to produce a standing wave).

Similarly, in acoustic instruments, the sound that reaches the ear is the projection of the vibrating instrument (notice now we are talking about the whole instrument) towards the direction of the ear or microphone. So, in effect, you are still applying this 'slice of vibrating element in time' concept. By the way, this is why mic placement is important when you record acoustic instruments (see for example: http://www.soundonsound.com/sos/1995_articles/sep95/mikinginstruments.html)

Hope this helps.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.