In addition to producing images, ultrasound can be used to heat tissues of the body for therapeutic purposes. An emitter is placed against the surface of the skin; the amplitude of the ultrasound wave at this point is quite large. When a sound wave hits the boundary between soft tissue and bone, most of the energy is reflected. The boundary acts like the closed end of a tube which can lead to standing waves. Suppose 0.7 MHz ultrasound is directed through a layer of tissue at a bone 0.55 cm below the surface. Remember, sound waves in the body travel at 1540 m/s. Explain.

ΔT = 4.058 10²⁰

Explanation:

In this experiment the system can be approximated as a tube with one end open and the other closed.

The open end is where the ultrasound emitter is and the closed end where the limit between the tissue and the bone is, the length of the tube is L = 0.55 cm = 5.5 10⁻³ m

a node is formed at the closed end and a belly at the open end, so the resonance has the form

             λ = 4L                    1st harmonic

             λ = 4/3 L                third harmonic

             λ = 4/5 L                fifth harmonic

             λ = 4L / (2n + 1)     n = 0, 1, 2, (2n + 1)

This wave is a standing wave therefore energy density remains in place

                P = 1/2 ρ v (w S₀)²

angular velocity is related to frequency

               w = 2π f

we substitute

                E = P = 2π² ρ v f² S₀²

if this energy per unit area is transformed into heat

                E = m c_e DT

let's use the concept of density

                ρ = m / V

                m = ρ V

if there are no losses in the system

       ½ π² ρ v f² S₀² = ρ V c_e ΔT

       ΔT =

When analyzing this expression the temperature increase is

* quadratic at the frequency and maximum amplitude of the wave

* proportional to the speed of the wave in the tissue

* inversely proportional to tissue volume

we can approximate the volume of the tissue to the volume of a cylinder tube

            V = π r² L

           ΔT =

we calculate

            ΔT = π 1450 (0.7 10⁶)² S₀² /( r² 5.5 10-3 c_e)

            ΔT = 4.058 10²⁰

absorption lines are where light has been absorbed by the atom thus you see a dip in the spectrum whereas.

emission spectra have spikes in the spectra due to atoms releasing photons at those wavelengths.

explanation: