# Wave equation for a string

1. Feb 6, 2005

### makris

Assume the well known PDE of an infinite length string

D^2(y)/Dt^2 = c^2* ( D^2(y)/Dx^2)

where y=y(x,t) is the transverse displacement of the string.
D/Dx= partial derivative with respect to x
D/Dt= partial derivative with respect to t
c= velocity of the wave

According to Morse and Ingard's Theoretical Acoustics (page 97), if the shape of the string is described by the function y(x,t)=F(x-ct), then Dy/Dt= -c*F'(x-ct)= -c*Dy/Dx (Where F'(z)= the derivative of F with respect to z).

I found the last statement a little bit confusing. Could anyone explain why F'(x-ct)=Dy/Dx ? Obviously if F'(x-ct)=Dy/Dt then a constrain is put on c which is not correct...

2. Feb 6, 2005

### Galileo

What is $\frac{\partial}{\partial x}F(x-ct)$?

Does it equal $F'(x-ct)$?

3. Feb 6, 2005

### makris

I have concluded that Dy/Dx=F'(x-ct) after working with some examples.
So I wrote down several functions with (x-ct) as an argument and I took the partial derivative with respect to x. Then I took the derivative with respect to the argument as a whole (z=(x-ct) ) to form F'(x-ct). They are equal. Done.

4. Feb 6, 2005

### dextercioby

Did u write everything in one line like i did?? :tongue2:

$$\frac{\partial F}{\partial x}=\frac{dF(x-vt)}{d(x-vt)}\frac{\partial (x-vt)}{\partial x}=\frac{dF(x-vt)}{d(x-vt)}$$

Daniel.

5. Feb 6, 2005

### makris

Exactly!

I have done the same both for Dy/Dx and Dy/Dt to verify that Dy/Dt=-c*Dy/Dx

Thanks.