Should the Wave Equation for a Longitudinal Wave Include Time?

[dyn]

Hi.
I am working through " A Student's guide to waves " by Fleisch. In deriving the wave equation for a longitudinal wave it uses
dψ = (∂ψ/∂x) dx

where ψ is the displacement but ψ is a function of x and t ; so shouldn't this equation be
dψ = (∂ψ/∂x) dx + (∂ψ/∂t) dt

Thanks

 

[kuruman]

Strictly speaking you are correct. However, in these derivations one takes a snapshot of the wave at some particular instant and analyzes perhaps a free body diagram of a piece of the medium. (I assume this is the kind of derivation you are looking at.) So "time" is frozen, and one considers the spatial variation only.

 

[dyn]

thanks. It does look like a snapshot taken at a particular time but it also involves acceleration taken as ∂²ψ/∂t² so it looks like when its convenient t is taken as a constant and when its convenient t is not a constant !

 

[kuruman]

Acceleration is always involved in the use of a FBD the purpose of which is the apply Newton's 2nd Law. Snapshots of the wave at different times will yield the same equation m(d²ψ/dt² )= F_net. Think of d²ψ/dt² as "acceleration" not as an instruction to you to take the second time derivative and apply the chain rule as you do so.

 

[dyn]

in this case I'm following a derivation from a book but in general if I'm faced with a function of 2 or more variables I would always apply the chain rule. How would I know when this is not to be applied ?

 

[kuruman]

How would I know when this is not to be applied ?

It depends on the context, what you are doing and where you are going with it. For example, if you write $r^2=x^2+y^2$, then
$$d(r^2)=\frac{\partial (x^2)}{\partial x}dx+\frac{\partial (y^2)}{\partial y}dy=2xdx+2ydy$$
You can see what this is saying: when you move in a general direction both $x$ and $y$ change. However, if you move along only one of the independent variables, say $x$ only, then $dy=0$. Similarly here, when you "freeze" time to study the FBD of the string segment, it is implied that $dt=0$ because you "move" along independent variable $x$ only when you consider $d\psi$.

 

[dyn]

thank you. That helps