Solutions to Week 12 Assignment

1. v = ÖB/r for a liquid.
   v = ÖB/r= Ö2.8 x 10¹⁰Pa/1.36 x 10⁴kg/m³ = 1.4km/s.
2. (a) Solve for the intensity.
              b = (10db)logI/I₀
   10^b/10dB = I/I₀
               I = I₀10^b/10dB
   Solve for P₀. For T = 20.0°C, v = 343m/s.
     P₀/2rv = I 
                 = I₀10^b/10dB
            P₀ = Ö2rvI₀10^b/10dB
                 = Ö2(1.20kg/m³)(343m/s)(10W/m²)10^120.0dB/10dB
                 = 28.7N/m².
   (b) F = P₀A = (28.7N/m²)(0.550 x 10^-4m²) = 1.58mN.
3. f_n = nv/2L for a pipe open at both ends and v = 343m/s for T = 20.0°C. Find L.
   f₁ = v/2L
   L = v/2f₁
      = (343m/s)/2(130.8Hz)
      = 1.31m.
4. f_beat = Df =3(64.5Hz) - 196.0Hz = 0.2Hz.
5. (a) A source and an observer are traveling toward each other (v_s > 0, v_o < 0)
           f_o = [(1 - v₀/v)/(1 - v_0s/v)] ff_s  = {[1 - (-0.50)]/(1 - 0.5}}(1.0kHz) = 3.0KHz.
   (b) A source and an observer are traveling away from each other (v_s < 0, v_o > 0)
           f_o = [(1- 0.50)/(1 + 0.50)](1.0kHz) = 330Hz.
   (c)  A source and an observer are traveling in the same direction (v_s < 0, v_o < 0)
           f_o = [(1 - 0.50)/(1 - 0.50)](1.0kHz) = 1.0kHz.
6. Use the result of Problem 31 for the frequency of reflected waves during angiodynography
   f_beat = Df = f_r - f = ( 1 + v_cell>/v_sound)/(1 - v_cell/v_sound)f - f = (5.0 x 10Hz)[(1 + 0.1m/s/15700m/s)/(1 - 0.10m/s/1570m/s) - 1} = 640Hz.
   Problem 31 result: