Waves on a String

Suppose we have a uniform string (such as a guitar string) that is stretched so that it is under tension $T$. The string is characterized by its mass per unit length $\mu$ - thick guitar strings have more mass per unit length than thin ones. It is fairly harmless at this point to imagine that the string is fixed to pegs at the ends that maintain the tension.

Now imagine that we have plucked the string somewhere between the end points so that it is displaced in the $y$-direction from its equilibrium (straight) stretched position and has some curved shape. If we examine a short segment of the string of length $\Delta x$, we can write Newton's 2nd law for that segment. If $\theta is small$, in the $x$-direction the components of the tension nearly perfectly cancel. Each bit of string therefore moves more or less straight up and down, and a useful solution is described by $y(x,t)$, the $y$ displacement of the string at position $x$ and time $t$. The permitted solutions must be continuous if the string does not break.

In the $y$-direction, we find write the force law by considering the difference between the $y$-components at the ends:

$$F_y = T \sin(\theta_2) - T \sin(\theta_1) = \mu \Delta x a_y$$ (10.10)

We make the small angle approximation: $\sin(\theta) \approx \tan(\theta) \approx \theta$, divide out the $\mu \Delta x$, and note that $\tan(\theta) = \frac{dy}{dx}$ (the slope of the string is the tangent of the angle the string makes with the horizon). Then:

$$\frac{d^2y}{dt^2} = \frac{T}{\mu} \frac{\Delta(\frac{dy}{dx})}{\Delta x}$$ (10.11)

In the limit that $\Delta x \to 0$, this becomes:

$$\frac{d^2y}{dt^2} - \frac{T}{\mu} \frac{d^2y}{dx^2} = 0$$ (10.12)

The quantity $\frac{T}{\mu}$ has to have units of $\frac{L^2}{t^2}$ which is a velocity squared.

We therefore formulate this as the one dimensional wave equation:

$$\frac{d^2y}{dt^2} - v^2 \frac{d^2y}{dx^2} = 0$$ (10.13)

where

$$v = \pm \sqrt{\frac{T}{\mu}}$$ (10.14)

are the velocity(s) of the wave on the string. This is a second order linear homogeneous differential equation and has (as one might imagine) well known and well understood solutions.

Note well: At the tension in the string increases, so does the wave velocity. As the mass density of the string increases, the wave velocity decreases. This makes physical sense. As tension goes up the restoring force is greater. As mass density goes up one accelerates less for a given tension.