Question

Question 1: SOUND

a) A bystander hears a siren vary in frequency from 570 Hz to 398 Hz as a fire truck approaches, passes by, and moves away on a straight street. What is the speed of the truck? (Take the speed of sound in air to be 343 m/s.) By how many decibels do you reduce the sound intensity level due to a source of sound if you triple your distance from it? Assume that the waves expand spherically.

b) If one sound is 18 dB greater than another sound, what is the ratio of their intensities (greater/smaller)?

c)  The sound intensity levels for a machine shop and a quiet library are 85 dB and 49 dB, respectively. What is the difference between the intensity of the sound in the machine shop and that in the library?

d) At a distance of 12.0 m from a point source, the intensity level is measured to be 65 dB. At what distance from the source will the intensity level be 32.5 dB?

e) A gas-powered lawnmower is rated at 94 dB. How many times more intense is the sound of this mower than that of an electric-powered mower rated at 76 dB?

f) The intensity levels of two people's speech are 61.0 dB and 67.0 dB, respectively. What is the intensity level of the combined sounds?

Answer 1

a)

as the train approcahes apparent frequency f1 = f0*v/(v-Vt)

as the train leaves apparent frequency f2 = f0*v/(v+Vt)

f2/f1 = (V-Vt)/(V+Vt)

398/570 = (343-vt)/(343+vt)

v = 61 m/s <<<=========ANSWER

++++++++++++++++++++++

I2 / I1 = (r1/r2)^2

I2/I1 = (r1/3r1)^2 = 1/9

I2 = I1/9

dB1 = 10*log(I1/I0) = 10*(logI1 - logI0)

dB2 = 10*log(I2/I0) = 10*(logI2 - logI0)

dB1 - dB2 = 10*(logI1 - logI2)

dB1 - dB2 = 10*log(I1/I2) = 10*log(9) = 9.542 <<<<========ANSWER

____________________________

(b)

dB1 - dB2 = 10*log(I2/I1)

18 = 10*log(I2/I1)

I2/I1 = 63.1 <<<=====ANSWER

(c)

threshold frequency I0 = 10^-12 W/m^2

dB1 = 10*log(I1/I0) = 85

10*log(I1/10^-12) = 85

I1 = 3.16*10^-4

dB2 = 10*log*(I2/I0)

49 = 10*log(I2/10^-12)

I2 = 7.94*10^-8

I1 - I2 = 0.0003161484

===================================

dB1 = 10*log(I1/I0) = 85

10*log(I1/10^-12) = 65

I1 = 3.16*10^-6

dB2 = 10*log*(I2/I0)

32.5 = 10*log(I2/10^-12)

I2 = 1.78*10^-9

I2/I1 = (r1/r2)^2

(1.78*10^-9)/(3.16*10^-6) = (12/r2)^2

r2 = 505.6 m

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dB1 = 10*log(I1/I0) = 94

10*log(I1/10^-12) = 94

I1 = 0.0025118864

dB2 = 10*log*(I2/I0)

76 = 10*log(I2/10^-12)

I2 = 3.9810717055*10^-5

I1/I2 = (0.0025118864)/(3.9810717055*10^-5) = 63 times

===================

dB1 = 10*log(I1/I0) = 61

10*log(I1/10^-12) = 61

I1 = 1.25*10^-6

dB2 = 10*log*(I2/I0)

67 = 10*log(I2/10^-12)

I2 = 5.01*10^-6

I3 = (I1 + I2 + 2*sqrt(I1*I2)

I3 = (1.25*10^-6) + (5.01*10^-6) + 2*sqrt(1.25*10^-6*5.01*10^-6)

I3 = 1.12*10^-5

dB3 = 10*log(I3/I0)

dB3 = 10*log(1.12*10^-5/10^-12) = 70.5 dB