Energy Transported By A Harmonic Wave

Wave Motion of Class 11

Energy Transported By A Harmonic Wave

As a wave propagates along string, it transports energy. Consider a small element of length dx under the influence of a wave as shown in Fig.14.9. If μ is the linear mass density, then mass dm of the element is

dm = μdx

The kinetic energy of the element is

dK = 1/2(dm)(δy/δt)2

If y = Asin(kx – ωt)

Then δy/δt = -ωAcos(kx - ωt) (14.17)

Energy Transported By A Harmonic Wave

An element of length dx of the string

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,

dEav = (1/2 μω2A2)dx

The average power transmitted by the wave is

Pav = dEav/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 =  Pav/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.

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