Figure 1

Consider the forces on a short fragment of string F y = T sin ( θ + Δ θ ) − T sin θ F y = T sin ( θ + Δ θ ) − T sin θ F x = T cos ( θ + Δ θ ) − T cos θ F x = T cos ( θ + Δ θ ) − T cos θ Assume that the displacement in y is small and T T is a constant along the stringthus θ θ and θ + Δ θ θ + Δ θ are smallthen F x ≈ 0 F x ≈ 0 we can see this by expanding the trig functions F x ≈ T [ 1 − ( θ + Δ θ ) 2 2 − 1 + θ 2 2 + … ] F x ≈ T [ 1 − ( θ + Δ θ ) 2 2 − 1 + θ 2 2 + … ] or F x ≈ T θ Δ θ F x ≈ T θ Δ θ which is very small.On the other hand F y ≈ T [ θ + Δ θ − θ + … ] F y ≈ T [ θ + Δ θ − θ + … ] or F y ≈ T Δ θ F y ≈ T Δ θ which is not nearly as small. So we will consider the y y component of motion, but approximate there is no x component F y = T sin ( θ + Δ θ ) − T sin θ ≈ T tan ( θ + Δ θ ) − T tan θ = T ( ∂ y ( x + Δ x ) ∂ x − ∂ y ∂ x ) = T ∂ 2 y ∂ x 2 Δ x F y = T sin ( θ + Δ θ ) − T sin θ ≈ T tan ( θ + Δ θ ) − T tan θ = T ( ∂ y ( x + Δ x ) ∂ x − ∂ y ∂ x ) = T ∂ 2 y ∂ x 2 Δ x Also we can write: F y = m a y F y = m a y m = μ Δ x m = μ Δ x where μ μ is the mass density a y = ∂ 2 y ∂ t 2 a y = ∂ 2 y ∂ t 2 now have T ∂ 2 y ∂ x 2 Δ x = μ Δ x ∂ 2 y ∂ t 2 T ∂ 2 y ∂ x 2 Δ x = μ Δ x ∂ 2 y ∂ t 2 ∂ 2 y ∂ x 2 = μ T ∂ 2 y ∂ t 2 ∂ 2 y ∂ x 2 = μ T ∂ 2 y ∂ t 2 Note dimensions, get a velocity T μ = v 2 T μ = v 2 ∂ 2 y ∂ x 2 = 1 v 2 ∂ 2 y ∂ t 2 ∂ 2 y ∂ x 2 = 1 v 2 ∂ 2 y ∂ t 2 The second space derivative of a function is equal to the second time derivative of a function multiplied by a constant.

Before considering traveling waves, we are going to look at a special case solution to the wave equation. This is the case of stationary vibrations of a string.

For example here, lets consider the case where both ends of the string are fixed at y = 0 y = 0 . Now we vibrate the string. Every point along the string acts like a little driven oscillator. So lets assume that every point on string has a time dependence of the form cos ω t cos ω t and that the amplitude is a function of distance Assume y ( x , t ) = f ( x ) cos ω t y ( x , t ) = f ( x ) cos ω t then ∂ 2 y ∂ t 2 = − ω 2 f ( x ) cos ω t ∂ 2 y ∂ t 2 = − ω 2 f ( x ) cos ω t ∂ 2 y ∂ x 2 = ∂ 2 f ∂ x 2 cos ω t ∂ 2 y ∂ x 2 = ∂ 2 f ∂ x 2 cos ω t Substitute into wave equation ∂ 2 y ∂ x 2 = 1 v 2 ∂ 2 y ∂ t 2 ∂ 2 y ∂ x 2 = 1 v 2 ∂ 2 y ∂ t 2 ∂ 2 f ∂ x 2 cos ω t = − ω 2 v 2 f ( x ) cos ω t ∂ 2 f ∂ x 2 cos ω t = − ω 2 v 2 f ( x ) cos ω t Then every f ( x ) f ( x ) that satisfies: ∂ 2 f ∂ x 2 = − ω 2 v 2 f ∂ 2 f ∂ x 2 = − ω 2 v 2 f is a solution of the wave equation

A solution is (requiring f ( 0 ) = 0 f ( 0 ) = 0 since ends fixed) f ( x ) = A sin ( ω x v ) f ( x ) = A sin ( ω x v ) Another boundary condition is f ( L ) = 0 f ( L ) = 0 so get A sin ( ω L v ) = 0 A sin ( ω L v ) = 0 Thus ω L v = n π ω L v = n π ω = n π v L ω = n π v L

Be careful with the equations above: v v is the letter vee and is for velocity. now we introduce the frequency ν ν which is the Greek letter nu.

recall ν = ω / 2 π ν = ω / 2 π so ν n = n v 2 L = n 2 L ( T μ ) 1 2 ν n = n v 2 L = n 2 L ( T μ ) 1 2 This is a very important feature of wave phenomena. Things can be quantized. This is why a musical instrument will play specific notes. Note, that we must have an integral number of half sine waves λ n = 2 L n λ n = 2 L n end up with f n ( x ) = A n sin ( 2 π x λ n ) f n ( x ) = A n sin ( 2 π x λ n ) leading to 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 where ω n = n π L ( T μ ) 1 2 = n π L v = n ω 1 ω n = n π L ( T μ ) 1 2 = n π L v = n ω 1 ω 1 ω 1 is the fundamental frequency