Chapter 14 10) Speed of sound in air v = 331 + 0.6 T_C m/s a) at 15 degrees C: v = 331 + 0.6 * 15 m/s = 340 m/s b) at 20 degrees C: v = 331 + 0.6 * 20 m/s = 343 m/s 20) Speed of sound in air at 10 degrees C v = 331 + 0.6 * 10 m/s = 337 m/s The distance traversed in 3.4 s is L= 337 m * 3.4 m/s = 1145.8 m The distance to the rock is half the distance traversed by the sound L = 1145.8 m / 2 = 572.9 m 28) Intensity of the point source I = P/(4*pi*R^2) where P i sthe power of the source For fixed value of the power the ratio of the intensities is 1-1/3 P/(4*pi*R2^2) R1^2 I2/I1 = ------- = -------------- = ---- 1 P/(4*pi*R1^2) R2^2 R2^2 = 3/2 R^1^2 ---> R2 = sqrt(1.5) * R1 = 1.22 * R1 30) Intensity level in dB beta = 10 log(I/I_0) where I_0 = 10^(-12) W/m^2 beta/10 = log (I/I_0) I/I_0 = 10^(beta/10) I = I_0 * 10^(beta/10) a) beta = 60 dB --> I = 10^(-12) W/m^2 * 10^6 = 10^(-6) W/m^2 b) beta = 110 dB --> I = 10^(-12) W/m^2 * 10^11 = 0.1 W/m^2 54) The observer will hear no sound if the oscillation from source A will have the phase shift equal to pi with respect to the wave from source B (half the wavelength distance)and the two waves cancel out. The distance between A and B must be lambda/2 = sound velocity/2*frequency = (331 + 0.6 * 20) / 2 * 600 = 0.29 m 55) Beat frequency is given by the difference of the two frequencies 440 Hz - 436 Hz = 4 Hz 63) The shock wave angle is given by the inverse of the Mach number and is equal to the arc sinus of the ratio of the velocity to the speed of sound. sin(theta) = v/v_sound When the aircraft just breaks the speed of sound its velocity is equal to the speed of sound v= v_sound and the sin(theta)=1. The angle for which sin(theta) = 1 is theta = 90 degrees. 64) sin(theta) = 1/1.5 ---> theta = 41.8 degrees