In deriving the motion of a string under tension we came up with an equation:

∂ 2 y ∂ x 2 = 1 v 2 ∂ 2 y ∂ t 2 ∂ 2 y ∂ x 2 = 1 v 2 ∂ 2 y ∂ t 2 which is known as the wave equation. We will show that this leads to waves below, but first, let us note the fact that solutions of this equation can be added to give additional solutions.

Say you have two waves governed by two equations Since they are traveling in the same medium, v v is the same ∂ 2 f 1 ∂ x 2 = 1 v 2 ∂ 2 f 1 ∂ t 2 ∂ 2 f 1 ∂ x 2 = 1 v 2 ∂ 2 f 1 ∂ t 2 ∂ 2 f 2 ∂ x 2 = 1 v 2 ∂ 2 f 2 ∂ t 2 ∂ 2 f 2 ∂ x 2 = 1 v 2 ∂ 2 f 2 ∂ t 2 add these ∂ 2 f 1 ∂ x 2 + ∂ 2 f 2 ∂ x 2 = 1 v 2 ∂ 2 f 1 ∂ t 2 + 1 v 2 ∂ 2 f 2 ∂ t 2 ∂ 2 f 1 ∂ x 2 + ∂ 2 f 2 ∂ x 2 = 1 v 2 ∂ 2 f 1 ∂ t 2 + 1 v 2 ∂ 2 f 2 ∂ t 2 ∂ 2 ∂ x 2 ( f 1 + f 2 ) = 1 v 2 ∂ 2 ∂ t 2 ( f 1 + f 2 ) ∂ 2 ∂ x 2 ( f 1 + f 2 ) = 1 v 2 ∂ 2 ∂ t 2 ( f 1 + f 2 ) Thus f 1 + f 2 f 1 + f 2 is a solution to the wave equation

Lets say we have two functions, f 1 ( x − v t ) f 1 ( x − v t ) and f 2 ( x + v t ) f 2 ( x + v t ) . Each of these functions individually satisfy the wave equation. note that y = f 1 ( x − v t ) + f 2 ( x + v t ) y = f 1 ( x − v t ) + f 2 ( x + v t ) will also satisfy the wave equation. In fact any number of functions of the form f ( x − v t ) f ( x − v t ) or f ( x + v t ) f ( x + v t ) can be added together and will satisfy the wave equation. This is a very profound property of waves. For example it will allow us to describe a very complex wave form, as the summation of simpler wave forms. The fact that waves add is a consequence of the fact that the wave equation ∂ 2 f ∂ x 2 = 1 v 2 ∂ 2 f ∂ t 2 ∂ 2 f ∂ x 2 = 1 v 2 ∂ 2 f ∂ t 2 is linear, that is f f and its derivatives only appear to first order. Thus any linear combination of solutions of the equation is itself a solution to the equation.

Any well behaved (ie. no discontinuities, differentiable) function of the form y = f ( x − v t ) y = f ( x − v t ) is a solution to the wave equation. Lets define f ′ ( a ) = d f d a f ′ ( a ) = d f d a and f ′ ′ ( a ) = d 2 f d a 2 . f ′ ′ ( a ) = d 2 f d a 2 . Then using the chain rule ∂ y ∂ x = ∂ f ∂ ( x − v t ) ∂ ( x − v t ) ∂ x = ∂ f ∂ ( x − v t ) = f ′ ( x − v t ) , ∂ y ∂ x = ∂ f ∂ ( x − v t ) ∂ ( x − v t ) ∂ x = ∂ f ∂ ( x − v t ) = f ′ ( x − v t ) , and ∂ 2 y ∂ x 2 = f ′ ′ ( x − v t ) . ∂ 2 y ∂ x 2 = f ′ ′ ( x − v t ) . Also ∂ y ∂ t = ∂ f ∂ ( x − v t ) ∂ ( x − v t ) ∂ t = − v ∂ f ∂ ( x − v t ) = − v f ′ ( x − v t ) ∂ y ∂ t = ∂ f ∂ ( x − v t ) ∂ ( x − v t ) ∂ t = − v ∂ f ∂ ( x − v t ) = − v f ′ ( x − v t ) ∂ 2 y ∂ t 2 = v 2 f ′ ′ ( x − v t ) . ∂ 2 y ∂ t 2 = v 2 f ′ ′ ( x − v t ) . We see that this satisfies the wave equation.

Lets take the example of a Gaussian pulse. f ( x − v t ) = A e − ( x − v t ) 2 / 2 σ 2 f ( x − v t ) = A e − ( x − v t ) 2 / 2 σ 2

Then ∂ f ∂ x = − 2 ( x − v t ) 2 σ 2 A e − ( x − v t ) 2 / 2 σ 2 ∂ f ∂ x = − 2 ( x − v t ) 2 σ 2 A e − ( x − v t ) 2 / 2 σ 2

and ∂ f ∂ t = − 2 ( x − v t ) ( − v ) 2 σ 2 A e − ( x − v t ) 2 / 2 σ 2 ∂ f ∂ t = − 2 ( x − v t ) ( − v ) 2 σ 2 A e − ( x − v t ) 2 / 2 σ 2 or ∂ 2 f ( x − v t ) ∂ t 2 = v 2 ∂ 2 f ( x − v t ) ∂ x 2 ∂ 2 f ( x − v t ) ∂ t 2 = v 2 ∂ 2 f ( x − v t ) ∂ x 2 That is it satisfies the wave equation.

Figure 1

To find the velocity of a wave, consider the wave: y ( x , t ) = f ( x − v t ) y ( x , t ) = f ( x − v t ) Then can see that if you increase time and x by Δ t Δ t and Δ x Δ x for a point on the traveling wave of constant amplitude f ( x − v t ) = f ( ( x + Δ x ) − v ( t + Δ t ) ) . f ( x − v t ) = f ( ( x + Δ x ) − v ( t + Δ t ) ) . Which is true if Δ x − v Δ t = 0 Δ x − v Δ t = 0 or v = Δ x Δ t v = Δ x Δ t Thus f ( x − v t ) f ( x − v t ) describes a wave that is moving in the positive x x direction. Likewise f ( x + v t ) f ( x + v t ) describes a wave moving in the negative x direction.

Another way to picture this is to consider a one dimensional wave pulse of arbitrary shape, described by y ′ = f ( x ′ ) y ′ = f ( x ′ ) , fixed to a coordinate system O ′ ( x ′ , y ′ ) O ′ ( x ′ , y ′ )

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We will often use a sinusoidal form for the wave. However we can't use y = A sin ( x − v t ) y = A sin ( x − v t ) since the part in brackets has dimensions of length. Instead we use y = A sin 2 π λ ( x − v t ) . y = A sin 2 π λ ( x − v t ) . Notice that y ( x = 0 , t ) = y ( x = λ , t ) y ( x = 0 , t ) = y ( x = λ , t ) which gives us the definition of the wavelength λ λ .

Also note that the frequency is ν = v λ . ν = v λ . The angular frequency is defined to be ω ≡ 2 π ν = 2 π v λ . ω ≡ 2 π ν = 2 π v λ . Finally the wave number is k ≡ 2 π λ . k ≡ 2 π λ . So we could have written our wave as y = A sin ( k x − ω t ) y = A sin ( k x − ω t ) Note that some books say k = 1 λ k = 1 λ

Lets go back to our solution for normal modes on a string: y n ( x , t ) = A n sin ( 2 π x λ n ) cos ω n t y n ( x , t ) = A n sin ( 2 π x λ n ) cos ω n t y n ( x , t ) = A n sin ( 2 π x λ n ) cos ( 2 π λ n v t ) . y n ( x , t ) = A n sin ( 2 π x λ n ) cos ( 2 π λ n v t ) . Now lets do the following: make use of sin ( θ + φ ) + sin ( θ − φ ) = 2 sin θ cos φ sin ( θ + φ ) + sin ( θ − φ ) = 2 sin θ cos φ Also lets just take the first normal mode and drop the n's Finally, define A ≡ A 1 / 2 A ≡ A 1 / 2 Then y ( x , t ) = 2 A sin ( 2 π x λ ) cos ( 2 π λ v t ) y ( x , t ) = 2 A sin ( 2 π x λ ) cos ( 2 π λ v t ) becomes y ( x , t ) = A sin [ 2 π λ ( x − v t ) ] + A sin [ 2 π λ ( x + v t ) ] y ( x , t ) = A sin [ 2 π λ ( x − v t ) ] + A sin [ 2 π λ ( x + v t ) ] These are two waves of equal amplitude and speed traveling in opposite directions. We can plot what happens when we do this. The following animation was made with Mathematica using the command

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