The idealised model of a stretched string assumes that it is perfectly flexible, and stretched between two fixed points. It is allowed to vibrate transversely with a (small) displacement from equilibrium given by $w(x,t)$. Suppose the string has tension $T$, mass per unit length $m$, and length $L$. To obtain the equation of motion, consider a small element of the string between positions $x$ and $x+ \delta x$ as sketched below. (Note that the displacement of the string is hugely exaggerated in the plot, for clarity.)

Newton’s law for this small element requires

$$m~\delta x \frac{\partial^2 w}{\partial t^2} =-T \sin \theta_1 + T \sin\theta_2 \tag{1}$$

But the angles $\theta_1$ and $\theta_2$ are both very small, so

$$\sin \theta_1 \approx \theta_1 \approx \tan \theta_1 = \left[ \frac{\partial w}{\partial x} \right] _x \tag{2}$$

and

$$\sin \theta_2 \approx \theta_2 \approx \tan \theta_2 = \left[ \frac{\partial w}{\partial x} \right] _{x + \delta x}. \tag{3}$$

Thus

$$m \frac{\partial^2 w}{\partial t^2} \approx T \left[ \dfrac{\left[ \frac{\partial w}{\partial x} \right] _{x + \delta x} – \left[ \frac{\partial w}{\partial x} \right] _{x} }{\delta x} \right] \rightarrow T \dfrac{\partial^2 w}{\partial x^2} \tag{4}$$

as $\delta x \rightarrow 0$. Thus the equation of motion is

$$m \frac{\partial^2 w}{\partial t^2} – T \dfrac{\partial^2 w}{\partial x^2} =0 . \tag{5}$$

If the motion were not free because a force $f(x,t)$ per unit length was applied to the string, $f$ would replace the zero on the right-hand side of this equation.

A vibration mode of the string is a free motion in which all points move sinusoidally at some frequency $\omega$. So to find the natural frequencies and vibration modes, we need to look for solutions of the form

$$w(x,t) = u(x) e^{i \omega t} \tag{6}$$

(remembering that we really mean “real part of…” this complex expression.) Substituting in eq. (5) then gives

$$T \dfrac{d^2u}{dx^2} + m \omega^2 u = 0. \tag{7}$$

This is the simple harmonic equation, so we already know that the general solution is

$$u(x) = A \cos kx + B \sin kx \tag{8}$$

where $A$ and $B$ are arbitrary constants, and

$$k^2 =\dfrac{m \omega^2}{T} = \dfrac{\omega^2}{c^2} \tag{9}$$

where $c=\sqrt{T/m}$.

The quantity $k$ is called *wavenumber *and is a kind of “spatial frequency”. It bears the same relation to wavelength $\lambda$ as a frequency $\omega$ bears to the period of oscillation $\tau$: $\omega=2 \pi/\tau$, and $k=2 \pi/\lambda$.

One interpretation of this solution is that sinusoidal waves can propagate along the string in either direction, and $c$ is the speed of these waves. To see this, it is better to use the complex form of the general solution:

$$u(x) =A’ e^{ikx} + B’ e^{-ikx} \tag{10}$$

where $A’$ and $B’$ are new arbitrary constants. The time-varying solution then looks like

$$w(x,t) = A’ e^{i(\omega t + kx)} + B’ e^{i(\omega t – kx)} \tag{11}$$

$$=A’ e^{i\omega (t + x/c)} + B’ e^{i\omega (t – x/c))}. \tag{12}$$

Now to find the modes we have to satisfy the boundary conditions at the ends of the string — we are assuming that $u = 0$ at both ends, which we can take to be at the positions $x = 0$ and $x = L$. From $u(0) = 0$ we can deduce from eq. (8) that $A = 0$. Now to satisfy $u(L) = 0$ we require

$$B \sin kL = 0 \tag{12}$$

so that the only allowed values of k are

$$k=n \pi/L \mathrm{~~for~~} n = 1, 2, 3, … \tag{13}$$

From eq. (9) these correspond to allowed values of the frequency $\omega$. So we have a sequence of *mode shapes*

$$u_n(x)=\sin (n \pi x/L) \tag{14}$$

with corresponding *natural frequencies*

$$\omega_n = n \pi c/L \mathrm{~~for~~} n = 1, 2, 3, … \tag{15}$$