Question

1. An experimenter wishes to generate in air a sound wave that has a displacement amplitude of 5.70 x 10-6 m. The pressure amplitude is to be limited to 0.850 Pa. What is the minimum wavelength the sound wave can have? (Take the equilibrium density of air to be p = 1.20 kg/m3 and assume the speed of sound in air is v = 343 m/s.) m 2. A sound wave in air has a pressure amplitude equal to 3.98 x 10 Pa. Calculate the displacement amplitude of the wave at a frequency of 10.1 kHz. (Note: Use the following values, as needed. The equilibrium density of air is p = 1.20 kg/m3; the speed of sound in air is v = 343 m/s. Pressure variations AP are measured relative to atmospheric pressure, 1.013 x 105 Pa.) m 3. Suppose you hear a clap of thunder 8.00 s after seeing the associated lightning strike. The speed of sound waves in air is 343 m/s and the speed of light in air is 3.00 x 108 m/s. (a) How far are you from the lightning strike? km (b) Do you need to know the value of the speed of light to answer? Yes No

Answer 1

ware -3 1) displacement amplitude of ware = A. = 5.20 X 106m Pressure amplitude of = Po=0-850 Pa density of air = 3= 1020 Kg m speed of sound in air = V = 343 mst! Let minimum ware length of =) Now, relation between pressure amplitude and displacement amplitude is by ware giren Po = vw. Ao @ ware 2 Where w is angular frequency of W = da f where is fre quency of relation between frequency, ware length and speed of sound is by ware giren

2 v=Xf 3) So, from © me have f sk By using @ in ime get w = 27 V in o, we get [apa = kg m 'š By using ③ Po = SV v (274) A. => Po = 2712 A. - P. = ** VREME ㅢ So, From © we hare 2av2SA. Po y = xoo m = Now, using values in , we zet . 2a(343 m5) 2 X2-20 kg m-3 X5.70 Xlo 0.850 Pa : =) 11764.9 m252 x 1.20 kgm? x5.70.810 m = 27 X -2 .850 Rg mi S =) -6 5056199.6 x 10 5.05 62 = =) 1 : m 0.850 0.85o x = 5.9 48 47 m ł 5.95m Sol minimum warelength of sound ware is 5.95 m