17.2. Derivation of the One-Dimensional Wave Equation

Let the string extend from x=0 to x=L, and let u(x,t) denote the transverse displacement at position x and time t.

In this initial step, we establish the context of the problem by defining the string's length and how we measure displacement. Here, u(x,t) signifies the displacement of the string from its rest position at a point x along the length of the string and at a time t. Essentially, we are setting up a system to understand how the string moves back and forth due to vibrations.

Think of a guitar string; when you pluck it, you can see the string moving up and down. The displacement u(x,t) represents how far the string is at any point from its original position at any moment.

Consider a small element of the string between x and x+∆x. Let the tension at position x be T, and the angles made by the string with the horizontal at x and x+∆x be θ and θ+∆θ, respectively.

Here, we are examining a small segment of the string. We denote this small piece of string as being between two points, x and x + ∆x. The tension (T) represents the force that is pulling on the string, while θ and ∆θ describe the angles of this segment relative to the horizontal. This particular setup helps us understand how forces act on different parts of the string and are vital for calculating the resulting motion.

Imagine you are holding a slinky stretched between two points. The tension in the slinky pulls it straight, but if you wiggle one end, the forces cause the coils to shift at different angles. This is similar to what we are observing in this portion of the string.

Using Newton’s second law in the vertical direction: Net vertical force=ρ∆x ∂2u ∂t2 (cid:18) (cid:19) T sin(θ+∆θ)−T sin(θ)≈T (x+∆x)− (x) ∂u ∂u

t= T ∆x

t={} ∂x ∂x ∂2u =T ∆x ∂x2

In this chunk, we apply Newton's second law, which states that the net force acting on an object is equal to its mass times acceleration. We focus only on the vertical forces acting on our small piece of the string. The net vertical force acting on this segment relates to the tension difference between the adjacent segments, which can be approximated using small-angle approximations. Ultimately, this leads us to relate the forces to the acceleration of the string segment, helping us to derive the wave equation.

Imagine pulling a rope up with a steady force while someone else pulls it down with a much smaller force. The difference in the forces will cause a change in the acceleration of the rope. This situation parallels how we balance forces in our string segment.

Equating force and acceleration: ∂2u ∂2u T ∆x=ρ∆x ∂x2 ∂t2 Cancelling ∆x, we obtain the one-dimensional wave equation: ∂2u ∂2u =c2 ∂t2 ∂x2 where c=p T/ρ is the wave speed.

After establishing the relationship between acceleration and the force due to tension, we equate these two concepts to derive our core equation. We simplify the equation by canceling out the small segment length (∆x), leading us to the famous wave equation in one dimension: the acceleration of the displacement over time and space relates by the constant c, which signifies the wave speed. This relationship allows us to model wave motion mathematically.

This derivation is similar to how we calculate the speed of a wave in water. If you throw a stone in a pond, the ripples that spread are dictated by the tension of the water (how tightly it is held together) and the density of the water, which determines how quickly the waves travel.

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