Energy Transported By A Harmonic Wave

The potential energy of the element is equal to the work done to stretch it from dx to dl. Assuming the tension remains constant

dU = T(dl – dx)

From the Fig. (14.9).

dl = √(dx)² - (dy)² = dx √1 + (δy/δx)²

Using binomial expansion:

dl = dx [1 + 1/2(δy/δx)²]

or dU = 1/2Tdx((δy/δx)²)

Now (δy/δx) = -ωAcos(kx - ωt)

∴ dU =  1/2(Tdx)(k² - A²) cos²(kx - ωt)   (14.18)

The total mechanical energy of the element is

dE = dk + dU

or dE = [(dm)(δy/δt)² +(Tdx)(k² - A²) cos²(kx - ωt) ]

Since ω = vk and for a string v = √T/μ

∴ dE = μω²A²cos²(kx- ωt)dx

The quantity dE / dx is called the linear energy density

dE/dx = μω²A²cos²(kx- ωt)(14.19)

It is obvious from equation (14.19) that the linear energy density is maximum at the mean position (y = 0) and minimum at the extreme position (y = A).

At any point, such as x = 0, the average value of cos₂ωt over one period is 1/2. Thus,

dE_av = (1/2 μω²A²)dx

The average power transmitted by the wave is

Pav = dE_av/dt = (1/2 μω²A²)     (14.20)

where v = dx/dt  is the wave velocity.

The mass per unit length of a wire is given by μ = ρ a

where ρ is the density and a is the cross-sectional area.

The intensity of the wave is given by

I =  P_av/a = (1/2 ρω²A²)   (14.21)

Note that the power and intensity are proportional to the square of the frequency and square of the amplitude.