Question

The impedance of a material is Z=rhoc, the density times the speed of sound. Consider the following information:Zlung=0.18×106kg s/m2.Zfat=1.34×106kg s/m2.What fraction of the incident intensity of a sound pulse will be reflected when sound travels from fat into lung?

Answer 1

The acoustic impedance of fat tissue is 1.34 × 10^6 kg/(m²·s). Approximately 0.81 of the incident intensity of the sound pulse will be reflected when sound travels from fat into lung.

Step-by-step explanation:

The acoustic impedance of a material is given by Z = p × u, where p is the density of the material and u is the speed of sound. Using the values for density and speed of ultrasound given in Table 17.5, we can calculate the acoustic impedance of fat tissue as follows:

Zfat = pfat × ufat = (0.92 × 103 kg/m³) × (1.46 × 103 m/s) = 1.3432 × 106 kg/(m²·s)

Thus, the acoustic impedance of fat tissue is indeed 1.34 × 106 kg/(m²·s).

To calculate the intensity reflection coefficient when sound travels from fat to lung, we need to consider the acoustic impedances of the two materials. The intensity reflection coefficient, R, can be calculated using the formula:

R = (Zlung - Zfat) / (Zlung + Zfat)

Substituting the given values, we have:

R = (0.18 × 106 kg/(m²·s) - 1.34 × 106 kg/(m²·s)) / (0.18 × 106 kg/(m²·s) + 1.34 × 106 kg/(m²·s)) = -0.81

Therefore, when sound travels from fat into lung, a fraction of approximately 0.81 of the incident intensity of the sound pulse will be reflected.

Answer 2

58.24%

Step-by-step explanation:

`\rho` = Density of material

c = Acoustic velocity

Impedance is given by

`Z=\rho c`

Acoustic impedance of lung

`Z_l=\rho_lc_l=0.18* 10^6\ kgs/m^2`

Acoustic impedance of fat

`Z_f=\rho_fc_f=1.34* 10^6\ kgs/m^2`

Reflection coefficient is given by

`R=\left((Z_(l)-Z_(f))/(Z_(l)+Z_(f)) \right)^2\\\Rightarrow R=\left((0.18* 10^6-1.34* 10^6)/(0.18* 10^6+1.34* 10^6) \right)^2\\\Rightarrow R=0.5824`

The fraction of incident intensity of a sound pulse will be reflected when sound travels from fat into lung is 58.24%