Waves on a string

Figure 11: A stretched string of length $l$ with the ends fixed.

The difference between waves and oscillatory motion is the scale. Waves are the ``continuum'' limit of the problem, or in the jargon ``the long wavelength'' limit. This is because in this limit, all the microscopic details are ``washed out'' and only the long distance behavior survives. In order to understand the transition between these two limits we have to perform a change of scale. The discrete equations of motion, as shown previously, can be written as:

$$\frac{d^{2}u}{dt^{2}}=-\frac{k}{m}(2u_{i}-u_{i+1}-u_{i-1}).$$

We consider the limits

$$N\rightarrow \infty ,a\rightarrow 0.$$

with the length of the chain kept constant. The main result is that in this limit, the discrete equations of motion can me replaced by the continuous wave equation:

$$\frac{\partial ^{2}u(x,t)}{\partial t^{2}}=\frac{1}{v^{2}}\frac{\partial ^{2}u(x,t)}{\partial x^{2}},$$ (54)

where $v$ has the dimension of velocity and it is given by

$$v=\sqrt{k/\rho },$$

where $k$ is the string tension and $\rho$ is the linear density. Observe that the displacement $u$ of the string is the dependent variable, and that the position along the string $x$ and the time $t$ are the independent variables. The existence of two independent variables makes this a Partial Differential Equation (PDE).

There are many solutions to this equation. Examples are:

$$u(x,t) = A\cos \frac{2\pi }{\lambda }(x\pm vt),$$ (55)

$$u(x,t) = A\sin \frac{2\pi }{\lambda }(x\pm vt).$$ (56)

In fact, it is easy to show that any function of the form $f(x\pm vt)$ is a solution. Since the differential equation is a linear equation and hence satisfies the superposition principle, we can understand the behavior of a wave of arbitrary shape using the Fourier's theorem to represent its shape as a sum of sinusoidal waves.

Because both ends of the string are tied down, the boundary conditions are:

$$u(0,t)=u(l,t)=0. \mathrm{(boundary condition)}$$

Since this is a second order PDE, we still need to determine the initial distortion $u(x,t=0)$ and velocity $\partial u/\partial t(x,t=0)$. If the string is released from rest, this reduces to

$$u(x,t) = f(x), \mathrm{(initial condition 1)}$$ (57)

$$\frac{\partial u}{\partial t}(x,t = 0)=0. \mathrm{(initial % condition 2)}$$ (58)