Chapter 6 Waves II: Sound Waves

113518_3v2chap6.pdf Chapter 6Waves II: Sound Waves6.1 The Important Stuff

6.1.1 Sound Waves; The Speed of Sound

A sound wave is a longitudinal wave in an elastic medium (which could be a gas, liquid or solid). In such a wave the particles of the medium oscillate back and forthalong the direction in which the wave travelssuch that regions of high and low density are created. It is these regions of compression and rarefaction which make up the wave fronts which travel through space and carry energy. The waves have a speed which comes from the elastic properties of the medium.

6.1.2 Interference of Sound Waves

6.1.3 Intensity and Sound Level

I=12ρvω2s2m(6.1)

I=P

4πr2(6.2)

β= (10dB)log10I

I0(6.3)

6.1.4 Standing Waves in Pipes

6.1.5 Beats

fbeat=|f1-f2|(6.4) 97

98CHAPTER 6. WAVES II: SOUND WAVES

6.1.6 The Doppler Effect

When a source or sound is in motion - or if an "observer" is in motion - or ifbothare in motion - the observer will detect a different frequency from what the source is producing. If a source of sound moves toward an observer the wavefronts are bunched up closer together than they would be otherwise (though they travel the same speed). This results in a greater frequency at which they strike the observer. IfvSis the speed of the source toward or away from the observer,vis the speed of sound, andfis the frequency at which the source is making the sound, then the observer hears a frequencyf?given by: f ?=f?v v?vS? (6.5) Here the top sign goes with motion of the sourcetowardthe observer, the bottom sign with motionaway. If the source of sound is stationary and the observer is moving toward the source or away from it, he will also hear a different frequency, but for a different reason. Here the observer encounters the wave fronts which have their usual separation distance (i.e. wavelength) but because of his motion, they are (effectively) coming at him with a different speed, and as a result he receives them at a different frequency. IfvOis the speed of the source toward or away from the observer,vis the speed of sound, andfis the frequency at which the source is making the sound, then the observer hears a frequencyf?given by: f ?=f?v±vO v? (6.6) Again, the top sign goes with motion "toward" and the bottom sign with motion "away". If both the source and obeserver are in motion then in fact Eqs.??and??combine as f ?=f?v±vO v?vS? =?1±vO v

1?vSv?

(6.7)

6.2 Worked Examples

Unless otherwise specified, we will take the speed of sound inair to be 343m sand the (mass) density of air as 1.21kg m3.

6.2.1 Sound Waves; The Speed of Sound

1. A stone is dropped into a well. The sound of the splash is heard3.00slater.

What is the depth of the well?[HRW6 18-7]

6.2. WORKED EXAMPLES99

????? ??? ???

Figure 6.1:(a) Rock falls to bottom of well, in a timet1. (b) Sound of splash travels back up the well, in

a timet2. There are two time intervals we need to think about: The timet1it takes for the rock to fall to the bottom of the well and the timet2it it takes for the sound to travel back up the well to the listener, as diagrammed in Fig. 6.1. . Thesumof these two times is 3.00s: t

1+t2= 3.00s (6.8)

If the depth of the well ish, then from our ample experience in the kinematics of freely- falling objects we can relatet1toh. We have: 1

2gt21=12(9.80ms2)t21=h(6.9)

Now, unlike a falling rock, sound travels at aconstantspeed (which we"ll assume isvsound= 343
m s) so that the relation betweent2andhis v soundt2= (343m s)t2=h(6.10)

We can combine 6.9 and 6.10 (eliminateh) to get:

1

2(9.80ms2)t21= (343ms)t2

Since 6.8 gives ust2= 3.00s-t1, we can substitute fort2to get an equation inoneunknown: 1

2(9.80ms2)t21= (343ms)(3.00s-t1)

Dropping the units for simplicity, we can rearrange this to: (4.9)t21+ (343)t1-1029 = 0 Use the quadratic formula to findt1(just get the positive root; the negative one ismeaningless here): t

1=-343 +?

(343)2+ 4(4.9)(1029)

2(4.9)= 2.88s

100CHAPTER 6. WAVES II: SOUND WAVES

so that we can gett2: t

2= 3.00s-2.88s = 0.12s

and then 6.10 givesh; h= (343m s)(0.12s) = 40.7m

The depth of the well is 40.7m.

2. The audible frequency range for normal hearing is from about20Hzto20kHz.

What are the wavelengths of sound waves at these frequencies?[HRW6 18-8] The wavelength and frequency of a wave are related byλf=v, wherevis the speed of the wave. Here we takev= 343m s; for a sound wave of frequency 20Hz the wavelength is

λ=v

f=(343m s) (20.0Hz)=(343m s) (20.0s-1)= 17.2m For the wave of frequency 20kHz, the wavelength is

λ=v

f=(343m s) (20.0kHz)=(343m s) (20.0×103s-1)= 1.72×10-2m = 1.72cm

6.2.2 Intensity and Sound Level

3. A source emits sound waves isotropically. The intensity of the waves2.50m

from the source is1.91×10-4W/m2. Assuming that the energy of the waves is conserved, find the power of the source.[HRW6 18-17] Eq. 6.2 gives the relation between power, intensity and distance for an isotropic (point) source of sound waves:I=P/(4πr2). We are given the intensityIat a given distancer, so the power of the source is

P= 4πr2I= 4π(2.50m)2(1.91×10-4W

m2) = 1.50×10-2W = 15.0mW

The power of the source is 15.0mW.

4. A sound wave of frequency300Hzhas an intensity of1.00μW/m2. What is the

amplitude of the air oscillations caused by this wave?[HRW6 18-19] Eq. 6.1 gives the intensity of a sound wave in terms of the amplitude of the air oscillations (sm) and other things. The formula requires the mass density of air (1.21kg m3), the speed of sound (343 m s) and the angular frequency of oscillation of the air, namely

ω= 2πf= 2π(300Hz) = 1.88×103s-1

6.2. WORKED EXAMPLES101

Then we find:

s 2 m=2I

ρvω2

=

2(1.00×10-6W

m2) (1.21kgm3)(343ms)(1.88×103s-1)= 1.36×10-15m2

So then the amplitude of the air oscillations is

s m= 3.7×10-8m = 37nm. This is averysmall length, but it should be remembered that it representsanaveragein the shifts of the positions of the air molecules.

5. Two sounds differ in sound level by1.00dB. What is the ratio of the greater

intensity to the smaller intensity?[HRW6 18-20] Letβ2be the sound level of the louder sound andI2its intensity. Likewise,β1andI1are the sound level and intensity of the weaker sound. We are given thatβ2-β1= 1.00. Use the definition of the sound level to get some information on the intensities: β

2-β1= 1.00

= (10dB)log 10?I2 I0? -(10dB)log10?I1I0? = (10dB) ? log 10?I2 I0? -log10?I1I0?? (6.11) (In the last step, we factored out (10dB) from each term.) Nowwe observe the the logs in the last step can combine (a minus signinvertsthe argument of a logarithm): log 10?I2 I0? -log10?I1I0? = log

10?I2I0?

+ log

10?I0I1?

= log 10?I2

I0·I0I1?

= log 10?I2 I1?

Put this into 6.11 and find:

1.00dB = (10dB)log10?I2

I1?

This gives us:

log 10?I2 I1? =1.0010= 0.100

102CHAPTER 6. WAVES II: SOUND WAVES

and raising 10 to the power of both sides gives: I 2

I1= 100.100= 1.26

Theratioof the two sound intensities is 1.26.

6. A point source emits30.0Wof sound isotropically. A small microphone inter-

cepts the sound in an area of0.7500cm2,200mfrom the source. Calculate (a) the sound intensity there and (b) the power intercepted by the microphone.[HRW6

18-28]

(a)Eq. 6.2 gives the intensity of an isotropic sound wave from the power of the source and the distance: I=P

4πr2=(30.0W)4π(200m)2= 5.97×10-5Wm2

(b)The power intercepted by the microphone is the intensity at its location multiplied by the area of its sound-sensitive element (assuming it is heldface-on to the direction of the incoming wave):Pmicr=IAmicr. Using 1m2= 10-4cm2, this gives: P micr=IAmicr= (5.97×10-5W m2)(0.750×10-4m2) = 4.48×10-9W.

7. A car horn emits a380Hzsound. If the car moves at17mswith its horn

blasting, what frequency will a person standing in front of the car hear?[Wolf 14-42] Here the source is in motion so that eq.??is applicable. Use the top sign for motion "toward"; usingv= 343m sas the speed of sound, the listener hears a frequency f ?=f?v v-vS? = (380Hz)?343m s

343-17ms?

= (380Hz)(1.05) = 400Hz

8. A fire truck"s siren at rest wails at1400Hz; standing by the roadside as the

truck approaches, you hear it at1600Hz. How fast is the truck going?[Wolf 14-44] Again use the formula for the frequency heard when the sourceis moving; this timnevS is unknown. Use 343 m sfor the speed of sound: f ?=f?v v-vS? =?v-vS=ff?v=?14001600? (343 ms) = 300ms

Rearrange and getvS:

v

S=v-300m

s= 343ms-300ms= 43ms

Appendix A: Conversion Factors

Lengthcmmeterkminftmi

1cm =110-210-50.39373.281×10-26.214×10-6

1m =100110-339.373.2816.214×10-4

1km =105100013.937×104328106214

1in =2.5402.540×10-22.540×10-518.333×10-21.578×10-5

1ft =30.480.30483.048×10-41211.894×10-4

1mi =1.609×10516091.6096.336×10452801

Massgkgslugu

1g =10.0016.852×10-26.022×1026

1kg =100016.852×10-56.022×1023

1slug =1.459×10414.5918.786×1027

1u =1.661×10-241.661×10-271.138×10-281

An object with aweightof 1lb has amassof 0.4536kg. 103