Energy Transport

Figure 1

Lets calculate the energy in a wave of a string:

Consider a fragment of string so small it can be considered straight, as is shown in the figure

The the kinetic energy is K = 1 2 m K = 1 2 m v 2 2 for the string fragment m = μ ⅆ x m = μ ⅆ x

So we have K = 1 2 μ ⅆ x ( ∂ y ∂ t ) 2 K = 1 2 μ ⅆ x ( ∂ y ∂ t ) 2 and using this we can define the energy per unit length, ie. the kinetic energy density: ⅆ K ⅆ x = 1 2 μ ( ∂ y ∂ t ) 2 ⅆ K ⅆ x = 1 2 μ ( ∂ y ∂ t ) 2 When the string segment is stretched from the length ⅆ x ⅆ x to the length ⅆ s ⅆ s an amount of work = T ( ⅆ s − ⅆ x ) = T ( ⅆ s − ⅆ x ) is done. This is equal to the potential energy stored in the stretched string segment. So the potential energy in this case is: U = T ( ⅆ s − ⅆ x ) U = T ( ⅆ s − ⅆ x ) Now ⅆ s = ( ⅆ x 2 + ⅆ y 2 ) 1 / 2 = ⅆ x [ 1 + ( ∂ y ∂ x ) 2 ] 1 / 2 ⅆ s = ( ⅆ x 2 + ⅆ y 2 ) 1 / 2 = ⅆ x [ 1 + ( ∂ y ∂ x ) 2 ] 1 / 2 Recall the binomial expansion ( 1 + A ) n = 1 + n A + n ( n − 1 ) A 2 2 ! + n ( n − 1 ) ( n − 2 ) A 3 3 ! + … ( 1 + A ) n = 1 + n A + n ( n − 1 ) A 2 2 ! + n ( n − 1 ) ( n − 2 ) A 3 3 ! + … so ⅆ s ≈ ⅆ x + 1 2 ( ∂ y ∂ x ) 2 ⅆ x ⅆ s ≈ ⅆ x + 1 2 ( ∂ y ∂ x ) 2 ⅆ x U = T ( ⅆ s − ⅆ x ) ≈ 1 2 T ( ∂ y ∂ x ) 2 ⅆ x U = T ( ⅆ s − ⅆ x ) ≈ 1 2 T ( ∂ y ∂ x ) 2 ⅆ x or the potential energy density ⅆ U ⅆ x = 1 2 T ( ∂ y ∂ x ) 2 ⅆ U ⅆ x = 1 2 T ( ∂ y ∂ x ) 2 To get the kinetic energy in a wavelength, lets start with y = A sin ( 2 π x λ − ω t ) y = A sin ( 2 π x λ − ω t ) ∂ y ∂ t = − ω A cos ( 2 π x λ − ω t ) ∂ y ∂ t = − ω A cos ( 2 π x λ − ω t ) Lets evaluate it at time 0. ∂ y ∂ t | t = 0 = − ω A cos ( 2 π x λ ) ∂ y ∂ t | t = 0 = − ω A cos ( 2 π x λ ) so ⅆ K ⅆ x = 1 2 μ ω 2 A 2 cos 2 ( 2 π x λ ) ⅆ K ⅆ x = 1 2 μ ω 2 A 2 cos 2 ( 2 π x λ ) now integrate K = ∫ 0 λ ⅆ K ⅆ x ⅆ x = 1 2 μ ω 2 A 2 ∫ 0 λ cos 2 ( 2 π x λ ) ⅆ x K = ∫ 0 λ ⅆ K ⅆ x ⅆ x = 1 2 μ ω 2 A 2 ∫ 0 λ cos 2 ( 2 π x λ ) ⅆ x In order to do this integral we use the following trig identity: cos 2 A = cos 2 A + 1 2 cos 2 A = cos 2 A + 1 2 so we get K = 1 2 μ ω 2 A 2 [ x 2 + λ 8 π sin 4 π x λ ] | x = 0 λ = 1 4 μ λ ω 2 A 2 K = 1 2 μ ω 2 A 2 [ x 2 + λ 8 π sin 4 π x λ ] | x = 0 λ = 1 4 μ λ ω 2 A 2

In similar fashion the potential energy can be found to be U = 1 4 μ λ ω 2 A 2 . U = 1 4 μ λ ω 2 A 2 . Deriving this will be assigned as a homework problem

So E = K + U = 1 2 μ λ ω 2 A 2 E = K + U = 1 2 μ λ ω 2 A 2 Power P = Δ E Δ t = 1 2 μ λ ω 2 A 2 τ = 1 2 μ ω 2 A 2 v P = Δ E Δ t = 1 2 μ λ ω 2 A 2 τ = 1 2 μ ω 2 A 2 v Where I have used τ = 1 / ν τ = 1 / ν and λ ν = λ ν = v thus τ = λ / v τ = λ / v