Let’s derive the wave equation for a stretched string. This is the same as saying, let’s show that waves will propagate on a string.

The string lies along the x-axis, with a uniform tension and a uniform mass density (for convenience). We restrict our derivation to nonviolent shapes, that is, the angle q any part of the string makes with the horizontal is small. Now, if we call the displacement from the x-axis of any part of the string f(x), we can see that the slope of the string is df/dx, and that slope is of course tan(q).

If we’re considering a straight portion of the string, then the net force on a tiny piece of it, of length dx, is zero – the tension forces pulling on each side of this piece simply cancel each other. However, if we look at a portion of the string that is curved, then the tension forces on either end of our tiny piece do not point in exactly opposite directions. Therefore there is a tiny net force, which can be significant because this is a tiny piece of stuff.

Let’s calculate the components of this tiny force. If the piece of rope lies between x and x+dx, then the directions of the two tension forces are: to the left and downward at q(x) below the horizontal, and to the right and upwards at q(x+dx)= q(x)+dq.

The horizontal component of the left-side tension force is –Tcos(q), and that of the right-side tension force is Tcos(q(x)+dq). The net vertical force on that little chunk of string is therefore T[cos(q(x)+dq)-cos(q(x)]=-Tsin(q)dq.

The vertical component of the left-side tension force is –Tsin(q), and that of the right-side tension force is Tsin(q(x)+dq). The net vertical force on that little chunk of string is therefore T[sin(q(x)+dq)-sin(q(x)]=Tcos(q)dq.

Notice that, because q is very small, the horizontal force is really small compared to the vertical force; therefore, we shall neglect it in what follows.

The vertical force is approximately equal to Tdq. Now, by Newton’s second law, we know that this force is equal to the mass of our tiny chunk of string times its vertical acceleration. The mass is µdx, where µ is the mass per unit length of the string. So we have µa dx=Tdq, which means the vertical acceleration is a=(T/µ)dq/dx. But let’s recall that tan(q)=df/dx, and that q is very small, so that q=df/dx. Let’s put this into our equation for a, and we find a=(T/µ)d²f/dx². One last observation is that a, being the vertical acceleration, must be d²f/dt², and so we have

d²f/dt²=(T/µ) d²f/dx²

which is the (one-dimensional) wave equation, as long as we acknowledge that those derivatives are partial derivatives.

This equation is satisfied by any function f(x-vt), as long as v=±squareroot(T/µ).