Derivation of wave equation I learned that the wave equation derivation is below.
Suppose $q$ is the displacement on $y$ component, $T$ is string tension, $d$ is the interval of two particles in $x$ direction.
Equation of motion
$$
m \frac{d^2q_j}{dt^2}=\frac {T}{d}(q_{j+1}-q_j)+\frac {T}{d}(q_{j-1}-q_j)  \\q_{j+1}-q_{j}=>q(x+dx)-q(x)
$$
From Taylor expansion
$$
q(x+dx)-q(x)=\frac{\partial q}{\partial x}d+\frac{1}{2}\frac{\partial^2 q}{\partial x^2}d^2+...\\q_{j-1}-q_{j}=q(x-dx)-q(x)=\frac{\partial q}{\partial x}(-d)+\frac{1}{2}\frac{\partial^2 q}{\partial x^2}(-d)^2+...
$$
In continuous situation, I'm looking at a specific position, that is
$$
m \frac{d^2q_j}{dt^2}=>m \frac{\partial^2q_j}{\partial t^2}
$$
Then combine the Taylor expansion to second order
$$
m \frac{\partial^2q_j}{\partial t^2}=\frac {T}{d}\frac{\partial^2 q}{\partial x^2}(d)^2
$$
set u is linear mass density, and m=ud
$$
\frac{\partial^2q_j}{\partial t^2}=\frac {T}{u}\frac{\partial^2 q}{\partial x^2}
$$
My question:
When I calculate the tension force, why don't I need to consider the time term?
since q is the function of x and t, so my taylor expansion might like this?
$$
q(x+dx,t+dt)-q(x,t)
$$
because the left hand side of equation of motion only consider fixing at x=x position. I don't fix the time term.
Although I could think of that the tension force is like I take a photo of this string, and observe the dynamic motion. Won't I consider the tension is changing with time too?
 A: When applying Newton's IInd law, you're relating the acceleration of that particular particle of the string with the forces applied on it at a particular instant in time. The acceleration, at that time, is proportional to the sum of the forces on the particle, at that time.
In this particular case, the tension force is proportional to the separation between the particle and its neighbours.
Essentially,
$$
q_{j+1}-q_{j} \Rightarrow q(x+dx)-q(x)
$$
secretly means
$$
q_{j+1,k}-q_{j, k} \Rightarrow q(x+dx, t)-q(x, t),
$$
where $ k $ is an index labelling time. Note there's no need to Taylor expand in time because there's no $ dt $ to expand in.
If instead you'd had
$$
q_{j+1, k+1}-q_{j, k} \Rightarrow q(x+dx, t+dt)-q(x, t),
$$
that would mean the force between the particles wasn't proportional to their separation. In fact it's not clear what it would mean at all. The force is proportional to the separation between... where the particle is right now, and where its neighbour will be in some small time later?
A: You're interested in the net force on an element of string at a particular time t. That is determined by the curvature of the string at time t. How that curvature will change with time is irrelevant.
