11.1.1. Derivation of the One-Dimensional Wave Equation

Let us derive the wave equation for a vibrating string under tension.

Assumptions: • The string is perfectly flexible and homogeneous. • The motion is only in the vertical plane (transverse). • No damping or external forces act on the string.

To derive the one-dimensional wave equation, we start by making some assumptions about the system we are studying, which is a vibrating string. These assumptions help simplify the model:

1. The string is flexible; it can bend easily without breaking.
2. It is homogeneous, meaning its mass is evenly distributed along its length.
3. The string only moves up and down (vertically), so we ignore any sideways motion.
4. There are no external forces acting on the string, such as air resistance or other damping forces that could slow it down.

Think of a guitar string. When you pluck it, the string vibrates, moving only up and down while remaining firmly stretched between two points. The assumptions help us understand the standard vibrations of a simple guitar string without the influence of external factors like wind or bad tuning.

Let: • 𝑇 be the constant tension in the string, • 𝜇 be the mass per unit length, • 𝑢(𝑥,𝑡) be the transverse displacement at point 𝑥 and time 𝑡.

In deriving the wave equation, we define three key variables:

1. Tension (T): This is the force that's keeping the string tight. Think of it like pulling a tight rope. A higher tension means the string will vibrate faster.
2. Mass per Unit Length (μ): This represents how heavy the string is over a certain length. A heavier string requires more force to move, affecting how the vibration travels through it.
3. Transverse Displacement (u(x, t)): This refers to how far the string moves up or down from its rest position at a particular point (x) and time (t). In simpler terms, it shows how much the string is vibrating at any given time and location.

Imagine plucking a rubber band stretched between two fingers. The tension (how tight you hold it) affects how it vibrates. If you have a thicker rubber band (more mass), it won’t vibrate as fast as a thinner one. The height to which you pull the band while plucking it indicates how far it will displace (u).

Consider an infinitesimal element of the string between 𝑥 and 𝑥 + Δ𝑥.

By Newton's second law: ∂²𝑢 𝜇Δ𝑥⋅ = 𝑇sin(𝜃₂)−𝑇sin(𝜃₁) ∂𝑡²

We analyze a small piece of the string, which we’ll call an infinitesimal element between two points, x and x + Δx. According to Newton's second law, the acceleration of this piece can be determined by the tension forces acting on it.

• The left side (𝜇Δ𝑥 ⋅ ∂²u/∂t²) represents the inertial force, which is the mass of the string segment times its acceleration.
• The right side, 𝑇sin(𝜃₂) − 𝑇sin(𝜃₁), represents the net force acting on the piece due to tension pulling at angles θ₂ and θ₁.

Think of holding a slinky stretched between two hands. If you pull one side down, the slinky moves in response. Newton's second law is saying that the movement (acceleration) of a small part of that slinky is due to the forces (how hard you're pulling) acting on it. The slinky's tension keeps it tight, similar to our tension in the string.

Using the small-angle approximation: sin(𝜃) ≈ tan(𝜃) = ∂u/∂x, we get:

∂²u ∂u ∂u 𝜇Δ𝑥⋅ = 𝑇[ (𝑥+Δ𝑥)−(𝑥)] ∂𝑡² ∂𝑥 ∂𝑥

In physics, for small angles, we can make the approximation that sin(θ) is very close to tan(θ), which simplifies calculations. We replace sin(θ) with the derivative of displacement with respect to position, ∂u/∂x. By substituting this into our earlier equation, we express the forces acting on the string segment in terms of its displacement, leading us to a simplified form where the tension forces are easier to manage mathematically.

When swinging a swing at a small angle, the angle's height doesn’t vary much from how far you pull down; this makes the math easier because you can treat the swing's movement as a straight line. This simplification allows us to calculate the behavior more easily.

Dividing by Δ𝑥 and taking the limit as Δ𝑥 →0: ∂²𝑢 𝜇 = 𝑇 ∂𝑡² ∂²𝑢 Thus: = ⋅ ∂𝑡² 𝜇 ∂𝑥² Let 𝑐² = 𝑇, then we obtain the standard wave equation: ∂²𝑢 ∂²𝑢 = 𝑐² ∂𝑡² ∂𝑥²

To finalize our wave equation, we divide through by Δx and take the limit as Δx approaches zero. This gives us a relationship that links the acceleration of the segment of string, the tension, and the mass density. We define c² as T/μ, where c represents the speed of the wave. The relationship leads us to the canonical form of the wave equation, which describes how waves propagate through the medium.

Picture a ripple moving through a pond. The speed of that ripple (like c) depends on how tightly the water’s surface is pulled. The derived wave equation helps predict how fast waves travel across the surface and how they’ll behave over time.

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