Let's apply Newton's second law in the vertical y direction:

Fy  =  may

The sum of forces in the y direction is

Fy  =  T sin θ2 − T sin θ1

Using the small angle approximation, sin θ   tan θ = ∂y/∂x≅ . So we may write:

The mass per unit length is μ, so its mass dm = μdx. The acceleration in the y direction is the rate of change in the y velocity, so ay  =  ∂vy/∂t  =  ∂y2/∂t2. So we can write Newton’s second law in the y direction as

Rearranging this gives

Now we have been using the subscript 1 to identify the position x, and 2 to identify the position (x+dx).  So the numerator in the last term on the right is difference between the (first) derivatives at these two points. When we divide it by dx, we get the rate of change of the first derivative with respect to x, which is, by definition, the second derivative, so we have derived the wave equation:

So the acceleration (on the left) is proportional to the tension T and inversely proportional to the mass per unit length μ. It is also proportional to ∂y2/∂x2. So the a greater curvature in the string produces a greater acceleration and, as we have seen, a straight portion is not accelerated.

Now there is nothing special about this function, or the point we chose, so any function y = f(x−vt) is a wave travelling to the right with speed v and with unchanging shape f(x').Why is there a minus sign before vt? Consider for a moment the point on the wave where x'  =  0, which is the peak of the function used here. As the peak travels to the right, x is increasing and time is increasing, so that x'  =  x − vt  =  0.Conversely, consider what happens if we set x"  =  x + vt. In this case, to keep x + vt   =  0, we need x to decrease as time decreases, which is a wave travelling to the left. So remembery  =  f(x − vt ) is a wave travelling in the positive x direction;y  =  f(x + vt ) is a wave travelling in the negative x direction.

let's write the stationary sine wave like this:

when x' increases by λ, the argument of the sine function increases by 2π, so the sine function goes through one complete cycle. λ is called the wavelength

let's now write the equation in terms of the stationary coordinate x, where x'  =  x − vt :

angular frequency ω  =  2πv/λ

T  =  2π/ω =  λ/v

v  =  λ/T

y  =  A sin (kx − ωt) 

The wave equation is a partial differential equation. We know that sine waves can propagate in a one dimensional medium like a string. And we know that any function f(x − vt) is a wave travelling at speed v. General expression for a sine wave travelling in the positive x direction is

y  =  A sin (kx − ωt + φ). A suitable choice of x or t axis allows us to set φ to zero, so let's look at the equation

y  =  A sin (kx − ωt)to see whether and when this is a solution to the wave equation

So we have seen that the second partial derivatives have the correct shape, which means we are on the right track. However, to be a solution to

the partial derivatives

must not only have the correct shape, but also the correct ratio. In other words, y  =  A sin (kx − ωt) is a solution, provided that

ω/k was the wave speed v, so we now have an expression for the speed  of  a wave in a stretched string:

http://www.animations.physics.unsw.edu.au/jw/wave_equation_speed.htm