Lateral vibration of a stretched string

Wave equation

A string of length $l$, tension $P$ (constant) and mass per unit length $\rho$ vibrates with small transverse displacement $w(x,t)$.

  P  ^ (theta at x)        ^ P (theta at x+dx)
      \                   /
  ----- \_______________/ -----
        x          x+dx

For an element $dx$, the net transverse force from the tension is

$$P\left[\frac{\partial w}{\partial x}\right]_{x+dx} - P\left[\frac{\partial w}{\partial x}\right]_x = P\frac{\partial^2 w}{\partial x^2}dx$$

By Newton's law, this equals $\rho\,dx\,\dfrac{\partial^2 w}{\partial t^2}$. Hence

$$\frac{\partial^2 w}{\partial t^2} = c^2\frac{\partial^2 w}{\partial x^2}, \qquad c = \sqrt{\frac{P}{\rho}}$$

$c$ is the wave speed.

Solution

Let $w(x,t) = W(x)\,T(t)$. Separating variables:

$$\frac{W''}{W} = \frac{\ddot T}{c^2T} = -\frac{\omega^2}{c^2} \;\Rightarrow\; W = A\cos\frac{\omega x}{c} + B\sin\frac{\omega x}{c}$$

Boundary conditions $W(0) = 0 \Rightarrow A = 0$ and $W(l) = 0 \Rightarrow \sin\dfrac{\omega l}{c} = 0$ (frequency equation):

$$\omega_n = \frac{n\pi c}{l}, \qquad f_n = \frac{n}{2l}\sqrt{\frac{P}{\rho}}, \qquad n = 1, 2, 3,\ldots$$

Mode shape: $W_n(x) = B_n\sin\dfrac{n\pi x}{l}$. The $n$th mode has $n-1$ nodes between the supports.

Numerical values

$$c = \sqrt{\frac{70}{0.0006}} = 341.6\ \text{m/s}$$ $$f_1 = \frac{c}{2l} = \frac{341.6}{2 \times 0.65} = 262.7\ \text{Hz}, \quad f_2 = 2f_1 = 525.5\ \text{Hz}, \quad f_3 = 3f_1 = 788.2\ \text{Hz}$$

Answer: $f_1 = 262.7$ Hz, $f_2 = 525.5$ Hz, $f_3 = 788.2$ Hz.