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15-1
TRAVELING WAVES AND SOUND 15 Q15.1. Reason: (a) In a transverse wave, the thing or quantity that is oscillating, such as the particles in a string, oscillates in a direction that is transverse (perpendicular) to the direction of the propagation of the wave. (b) Vibrations of a bass guitar string are a form of transverse wave. You can see that the oscillation is perpendicular to the string. Assess: The plucking action makes the segment of the string move perpendicular to the string, but the disturbance is propagated along the string.
Q15.2. Reason: For a longitudinal wave, the particles of the medium oscillate in a direction parallel to the direction of the disturbance passing through the medium. It is possible to create a longitudinal mechanical wave in a slinky by compressing and then releasing a number of coils at one end. The individual coils oscillate parallel to the length of the slinky and the disturbance moves along the length of the slinky. Assess: A disturbance in a slinky created as described above is an excellent example of a longitudinal wave because it is possible to see the individual coils oscillate and the disturbance move through the slinky.
Q15.3. Reason: Wave speed is independent of wave amplitude, = =1 2 3.v v v Wave speed for mechanical waves depends on the properties of the medium, not the amplitude of the vibration. Assess: If the wave speed were dependent on the amplitude then it might be the case that a later shout could overtake an earlier whisper.
Q15.4. Reason: Since all sound waves travel in air at the same speed (the speed of sound for that temperature of air), one sound wave cannot overtake and pass another. Assess: Sound can have different speeds in air due to the temperature, but all sound will travel through that air at the same speed.
Q15.5. Reason: Equation 15.2, µ=string s/ ,v T gives the wave speed on a stretched string with tension sT and
linear mass density µ = / .m L We will investigate how sT and µ are changed in each case below and how that affects the wave speed. Use a subscript 1 for the original string, and a subscript 2 for the altered string. We are given that =string 1( ) 200 cm/s.v
(a) µ µ= =s 2 s 1 2 1( ) 2( )T T
µ µ µµ µµ
= = = =
= = =
string 2 s 2 2 s 2 2 s 1 1
string 1 s 1 1 s 1 1s 1 1
string 2 string 1
( ) ( ) / ( ) / 2( ) / 2( ) ( ) / ( ) /( ) /
( ) 2( ) 2(200 cm/s) 280 cm/s
v T T Tv T TT
v v
(b) µ µ= = ⇒ =s 2 s 1 2 1 2 1( ) ( ) 4 4T T m m
µ µ µµ µµ
= = = =
= = =
string 2 s 2 2 s 2 2 s 1 1
string 1 s 1 1 s 1 1s 1 1
string 2 string 1
( ) ( ) / ( ) / ( ) /4 1( ) ( ) / ( ) / 2( ) /
1 1( ) ( ) (200 cm/s) 100 cm/s2 2
v T T Tv T TT
v v
15-2 Chapter 15
(c) µ µ= = ⇒ = 1s 2 s 1 2 1 2 14( ) ( ) 4T T L L
µ µµµ µµ
= = = =
= = =
1string 2 s 2 2 s 1 1s 2 2 4
string 1 s 1 1 s 1 1s 1 1
string 2 string 1
( ) ( ) / ( ) /( ) / 2( ) ( ) / ( ) /( ) /
( ) 2( ) 2(200 cm/s) 400 cm/s
v T TTv T TT
v v
(d) = =s 2 s 1 2 1( ) ( ) 4T T m m and µ µ= ⇒ =2 1 2 14L L
µ µ µµ µµ
= = = =
= =
string 2 s 2 2 s 2 2 s 1 1
string 1 s 1 1 s 1 1s 1 1
string 2 string 1
( ) ( ) / ( ) / ( ) / 1( ) ( ) / ( ) /( ) /
( ) ( ) 200 cm/s
v T T Tv T TT
v v
Assess: Notice as in part (d) that if both the mass and length of the string are increased by the same factor, then µ is not changed, so the speed is the same (with no change in tension).
Q15.6. Reason: The speed of sound in a gas γ=sound( / )v RT M depends on the temperature of the gas. Since air is made up of gases, the speed of sound in air must also be temperature-dependent. Assess: The temperature dependence of the speed of sound in air is small but noticeable. For example, as shown in Table 15.1 when the temperature of air goes from °0 C to °20 C, the speed of sound in air increases by 2 m/s.
Q15.7. Reason: The speed of sound in air depends on the temperature of the air. The distance across the stadium can be measured to give the path length of the sound. Then, you can take the path length and divide by the time between the emission and detection of the pulse to get the speed of the sound. Finally, once the speed of sound is known, you can find the corresponding temperature either by consulting a chart like Table 15.1 or by using Equation 15.3. Assess: The advantage of measuring temperature this way is that it gives you an idea of the average temperature for the whole stadium. It also determines the temperature quickly—the time of measurement is simply the time it takes for sound to travel across the stadium.
Q15.8. Reason: Expounding on Equation 15.2, when the material is less dense the sound waves should travel faster. Similarly, when the tension is higher (molecular bonds are stronger), the sound waves should again travel faster. Both of these factors indicate that sound should travel faster in ice (lower density, stronger bonds) than in liquid water. Assess: Indeed, this is the case. The speed of sound in ice is about 4000 m/s as opposed to the value of 1480 m/s given in Table 15.1 for water.
Q15.9. Reason: Please refer to Figure Q15.9. Since the wave is traveling to the left, the snapshot has the same shape as the history graph. To understand why, consider that the history graph tells us about the displacement at one point in space. As the wave moves to the left, that point witnesses spots on the wave further to the right. Thus increasing t by t∆ on the history graph has the same effect as increasing x by v t∆ on the snapshot graph, where v is the speed of the wave. Consequently the history graph and snapshot graph have the same shape. The only difference is that in going from the history graph to the snapshot graph, the horizontal axis is scaled by a factor of .v
Traveling Waves and Sound 15-3
Assess: A similar argument can be used to show that if a wave is traveling to the right, the snapshot graph can be obtained from the history graph by reflecting the graph about the y-axis and scaling the horizontal axis by a factor of .v
Q15.10. Reason: (a) The wave is traveling to the right. The right edge of the pulse in the snapshot is at −= 2 cmx at = 0.01s.t The disturbance arrives at = 2 cmx later, at = 0.03 s.t (b) It took that leading edge − =0.03 s 0.01s 0.02 s to go from = −2 cmx to = 2 cm,x
= ∆ ∆ = =/ 4 cm/0.02 sxv x t 200 cm/s. Assess: The importance of understanding history graphs and snapshots as views of a function of two variables should not be underestimated. Neither the history graph nor the snapshot would have been sufficient alone to answer this question.
Q15.11. Reason: The relationship between wavelength and frequency is λ = / .v f This tells us that the wavelength is inversely proportional to the frequency—the greater the frequency then, the smaller the wavelength. Since < <1 2 3,f f f we can write λ λ λ> >1 2 3. Assess: Since the speed of sound is a constant for any given situation, the wavelength depends only on the frequency.
Q15.12. Reason: Here is the equation for easy reference: π=( ) cos(2 ).tTy t A
For the oscillation to have a period T means that = +( ) ( ).y t y t T For our function this implies π π π π+= = +cos(2 ) cos(2 ) cos(2 2 ),t t T t
T T T which is true because the period of the cosine function is π2 rad. On the other hand, the previous equality is not true without the π2 . Examine an easy specific case, =(0) ( ).y y T
π = = =
0(0) cos 2 cos(0) ,y A A AT
because =cos(0) 1
The next time that θ =cos( ) 1is whenθ π= 2 . So the argument of the cosine at =t T must be π2 .
π π = = =
( ) cos 2 cos(2 ) ,Ty T A A AT
because π =cos(2 ) 1
That is why the π2 must be there. Assess: It might seem a nuisance to have to remember the π2 , but it really is crucial. And, of course, the whole discussion assumes radians rather than degrees.
Q15.13. Reason: To “see” an object by means of echolocation, the sound used must have a wavelength smaller than the object being viewed. For example, for the dolphin to perceive an object 1 cm in size, it needs to use sound waves with a wavelength of 1 cm or less. The frequency of a sound wave is related to its wavelength by the formula / .f v λ= Thus the frequency increases with the speed of the wave. Bats send their sound waves through air where the speed of sound is around 340 m/s, whereas dolphins send their sound waves through water where the speed is around 1480 m/s. Since the speed of the waves produced by the dolphins is about four times as great, if they are to view the same sized objects using the same wavelength, the dolphins will need a frequency which is four times as great as the bats’. Assess: The factor of four we have predicted seems reasonable in view of Problem 15.76, which suggests a frequency for bats of 26 kHz, which is very nearly four times less than the value of 100 kHz quoted here for dolphins.
Q15.14. Reason: Let’s answer this question first in a very qualitative manner by inspecting the functional relationship between the intensity and the diameter of the beam and then in a very quantitative manner by actually using this functional relationship. Intensity is determined by = /I P A and in this case the area is the cross-sectional area of the beam, π= 2/4.A d Inserting the expression for the area into the expression for the
intensity we have I π π= = =2 2/ /( /4) 4 / .P A P d P d
15-4 Chapter 15
By inspection, we see that the intensity is inversely proportional to the square of the diameter of the beam. When the diameter decrease to one tenth the original diameter =( /10),od d the intensity increases by =100( 100 ).oI I When the diameter increases by a factor of ten =( 10 ),od d the intensity decreases by a factor of =100( /100).oI I We can approach this in a more quantitative manner by writing the expression for the original and new situation as follows:
π= 20 04 /I P d and π= 24 /I P d
Dividing the expression for the new intensity I by the expression for the original intensity ,oI and then inserting the relationship between the original and new diameter we obtain
ππ
= = = =2 220 0
2 2 20 0 0
(4 / ) 100(4 / ) ( /10)
d dI P dI P d d d
or = 0100I I
In like manner for the second case we obtain:
ππ
= = = =2 220 0
2 2 20 0 0
(4 / ) 1(4 / ) (10 ) 100
d dI P dI P d d d
or = 0
100II
Assess: The qualitative method we used is faster than the quantitative method. While you may prefer one method over the other, you need to be comfortable with both methods.
Q15.15. Reason:
=
= =
= =
= =
=
A
B
C
B C A
energypowertime
2 J 1 W2 s10 J 2 W5 s
2 mJ 2 W2 ms
>
P
P
P
P P P
Assess: We have seen this concept before; it is the ratio of energy to time that matters, not the energy number alone.
Q15.16. Reason: The speed of the disturbance traveling down the towel depends on the linear mass density of the medium as µ= / .v T The linear mass density µ is the critical factor. At your hand (where there is a lot of towel mass) the mass per unit length µ( ) is largest and the disturbance (you giving the towel a shake) travels through the towel very slowly. As the disturbance travels down the length of the towel, µ decreases and hence v increases, resulting in a fine snap at the end. Assess: Good theory frequently leads to experimentation. Do so at you own risk and with some thought for your own safety.
Q15.17. Reason: Since the decibel scale is a logarithmic one, an increase of a fixed number of dB for each click does not correspond to an increase of the output power by a fixed amount. An increase by a fixed number of dB does correspond to a common factor increase in the power output, however. For example, an increase in the level of 10 dB means the power increased by a factor of 10. Assess: While the power output goes up by a given factor, not a given amount, for a given increase in sound level, it does sound to the ear that the stereo is getting louder at approximately the same rate.
Q15.18. Reason: Because the bullet is traveling faster than the speed of sound it creates a shock wave—a little sonic boom. This is what people at a distance from the gun may hear louder. Assess: Even the sharp crack of a bull whip is a sonic boom, as the tip of the whip is traveling faster than the speed of sound in air.
Traveling Waves and Sound 15-5
Q15.19. Reason: The observer will hear the same frequency as the frequency emitted by source 4. The observer will hear a frequency lower than the frequency emitted for all receding sources (1, 2, and 3). The observer will hear a frequency higher than the frequency emitted for all approaching sources (5, 6, and 7). For the receding sources, we will make comparisons to the frequency heard from source 2 since it is receding at a constant rate. Since source 1 is speeding up, the observer will hear a frequency greater than that heard from source 2. Since source 3 is slowing down, the observer will hear a frequency lower than that heard from source 2. For the approaching sources, we will make comparisons to the frequency heard from source 6 since it is approaching at a constant rate. Since source 5 is speeding up, the observer will hear a frequency greater than that heard from source 6. Since source 7 is slowing down, the observer will hear a frequency lower than that heard from source 6. Combining all of the above information, we may write the following: = = > > = =5 6 7 4 1 2 3.f f f f f f f Assess: In general, the frequency is lowered for receding sources and raised for approaching sources.
Q15.20. Reason: Knowing the frequency of the electromagnetic waves traveling from the station, we can determine their wavelength by λ = = × × =8 8/ (3.0 10 m/s)/(1.05 10 Hz) 2.85 m.c f The station would be named KOOL 2.85. The correct choice is B. Assess: The number obtained is in the radio wave range.
Q15.21. Reason: The fundamental relationship for periodic waves is given in Equation 15.10, λ= .v f
We are asked for the frequency of an electromagnetic wave (which travels at a speed of × 83.00 10 m/s) of a given wavelength λ =( 400 nm).
λ −
×= = = ×
×
814
9
3.00 10 m/s 7.50 10 Hz400 10 m
vf
The correct choice is D. Assess: Note the tremendously high frequency of visible light: seven and a half hundred trillion hertz! You may not remember the exact number (which, of course, varies for different colors), but it would be fun to tuck away the tidbit that ≈ 1410 Hzf for visible light in your brain for later recall. This calculation is easily done in one’s head (keep track of the significant figures).
λ − −
×= = = = × = ×
×
8 815 14
9 7
3.00 10 m/s 3 10 Hz 0.75 10 Hz 7.50 10 Hz400 10 m 4 10
vf
Q15.22. Reason: If the depth of penetration is 200 times the wavelength, the depth for a 5.0 MHz signal may be obtained by λ= = = × =6200 200( / ) 200(343 m/s)/5 10 Hz 1.4 cm.d v f The correct choice is B. Assess: Considering the distance from skin surface to bone, this depth would allow the signal to impact most body tissue.
Q15.23. Reason: The fundamental relationship for periodic waves is given in Equation 15.10, λ= .v f We are told that λ = 32 cm. We’ll compute the frequency from = = =1/ 1/0.20 s 5.0 Hz,f T λ= = (32 cm)v f (5.0 Hz) = 160 cm/s. The correct choice is D. Assess: Equation 15.9 is even more directly applicable, but the customary form (that you should remember) is Equation 15.10. The amplitude was unneeded information.
Q15.24. Reason: As the disturbance travels from string 1 to string 2, the frequency remains the same every disturbance in string 1 creates just one disturbance in string 2 and according to the figure, the wavelength decreases. Knowing the relationship between speed of disturbance, wavelength, and frequency λ=( ),v f we can establish what happens to the speed of the disturbance as it travels from string 1 to string 2. If the frequency is constant, as the wavelength decreases, the speed at which the disturbance travels decreases. As a result the disturbance will travel slower in string 2 than in string 1. The correct choice is C. Assess: The answer is obtained by translating the meaning of the expression λ= .v f
15-6 Chapter 15
Q15.25. Reason: When the measured frequency is lower than 0f it is because the source is moving away from you. This is the case from = 0 st to = 2 s.t After that the measured frequency is higher than 0f so the source is moving toward you. The correct choice is D. Assess: In a doppler shift situation the measured frequency is constant as long as the source’s velocity toward you (or away from you) is constant. The measured frequency doesn’t keep increasing as the source gets closer and closer.
Problems
P15.1. Prepare: The wave is a traveling wave on a stretched string. We will use Equation 15.2 to find the wave speed. Solve: The wave speed on a stretched string with linear density µ is µ=string S/ .v T The wave speed if the tension is doubled will be
µ′ = = = =Sstring string
2 2 2(200 m/s) 280 m/sTv v
Assess: Wave speed increases with increasing tension.
P15.2. Prepare: The wave is a traveling wave on a stretched string. We will use Equation 15.2 to find the tension that corresponds to a wave speed of 180 m/s. To be able to obtain the tension from 15.2, we will first obtain µ from the first part of the problem. Solve: The wave speed on a stretched string with linear density µ is
µµ µ
−= ⇒ = ⇒ = × 3Sstring
75.0 N150 m/s 3.333 10 kg/mTv
For a wave speed of 180 m/s, the required tension will be
µ −= = × =3 22S string (3.333 10 kg/m)(180 m/s) 110 NT v
Assess: Increased wave speed must lead to increased tension, as is previously obtained.
P15.3. Prepare: The wave is a traveling wave on a stretched string. The wave speed on a string whose radius is R, length is L, and mass density is ρ is
µ=string S/v T with ρ ρπµ ρπ= = = =2
2m V R L RL L L
that is, =stringv ρπ 2S/ .T R We can now replace R with 2R and get the new wave speed.
Solve: If the string radius doubles, then
ρπ′ = = = =stringSstring 2
280 m/s 140 m/s(2 ) 2 2
vTvR
Assess: Greater linear mass density decreases wave speed, as obtained above.
P15.4. Prepare: Assume the speed of sound is the 20°C value 343 m/s. A pictorial representation of the problem is given.
Traveling Waves and Sound 15-7
Solve: The time elapsed from the stage to you in the middle row is the distance divided by the speed of sound.
=13 m 0.0379 s
343 m/s
The time elapsed for the same sound to get reflected from the back wall and reach your ear is
+=
26 m 13 m 0.114 s343 m/s
The difference in the two times is thus 0.114 s – 0.038 s = 0.076 s. Assess: Due to the relatively large speed of sound, the observed small time difference would be expected.
P15.5. Prepare: Two pulses of sound are detected because one pulse travels through the metal to the microphone while the other travels through the air to the microphone. Sound travels faster through solids than gases, so the pulse traveling through the metal will reach the microphone before the pulse traveling through the air. We will take the speed of sound in air as 343 m/s. Solve: The time interval for the sound pulse traveling through the air is
∆∆ = = = =air
air
4.0 m 0.01166 s 11.66 ms343 m/s
xtv
Because the pulses are separated in time by 11.0 ms, the pulse traveling through the metal takes ∆tmetal = 0.66 ms to travel the 4.0 m to the microphone. Thus, the speed of sound in the metal is
∆= = =
∆metalmetal
4.0 m 6100 m/s0.00066 s
xvt
Assess: We see from Table 15.1 that the speed obtained as shown is quite typical of metals.
P15.6. Prepare: We will assume the wave speed in water to be 1480 m/s, as given in Table 15.1. Solve: The time interval for the sound wave to travel a distance of 3200 km through water is
∆∆ = = = =
water
3200 km 2162 s 36 min1480 m/s
xtv
Assess: A distance of 3200 km is quite large when the speed of sound in water is about 1.5 km/s, so a time of 36 min is reasonable.
P15.7. Prepare: Please refer to Figure P15.7. This is a wave traveling at constant speed. The pulse moves 1 m to the right every second. Solve: The snapshot graph shows the wave at all points on the x-axis at t = 0 s. You can see that nothing is happening at x = 6 m at this instant of time because the wave has not yet reached this point. The leading edge of the wave is still 1 m away from x = 6 m. Because the wave is traveling at 1 m/s, it will take 1 s for the leading edge to reach x = 6 m. Thus, the history graph for x = 6 m will be zero until t = 1 s. The first part of the wave causes an upward displacement of the medium. The rising portion of the wave is 2 m wide, so it will take 2 s to pass the x = 6 m point. The constant part of the wave, whose width is 2 m, will take 2 seconds to pass x = 6 m and during this time the displacement of the medium will be a constant (∆y = 1 cm). The trailing edge of the pulse arrives at t = 5 s at x = 6 m. The displacement now becomes zero and stays zero for all later times.
15-8 Chapter 15
P15.8. Prepare: Please refer to Figure P15.8. This is a wave traveling at constant speed. The pulse moves 1 m to the left every second. Solve: This snapshot graph shows the wave at all points on the x-axis at t = 2 s. You can see that the leading edge of the wave at t = 2 s is precisely at x = 0 m. That is, in the first two seconds, the displacement is zero at x = 0 m. The first part of the wave causes a downward displacement of the medium, so immediately after t = 2 s the displacement at x = 0 m will be negative. The negative portion of the wave pulse is 3 m wide and takes 3 s to pass x = 0 m. The positive portion begins to pass through x = 0 m at t = 5 s and until t = 8 s the displacement of the medium is positive. The displacement at x = 0 m returns to zero at t = 8 s and remains zero for all later times.
P15.9. Prepare: Please refer to Figure P15.9. This is a wave traveling at constant speed to the right at 1 m/s. Solve: This is the history graph of the wave at x = 0 m. The graph shows that the x = 0 m point of the medium first sees the negative portion of the pulse wave at t = 1 s. Thus, the snapshot graph of this wave at t = 1 s must have the leading negative portion of the wave at x = 0 m.
P15.10. Prepare: Please refer to Figure P15.10. This is a wave traveling at constant speed to the left at 1 m/s. Solve: This is the history graph of a wave at x = 2 m. Because the wave is moving to the left at 1 m/s, the wave passes the x = 2 m position a distance of 1 m in 1 s. Because the flat part of the history graph takes 2 s to pass the x = 2 m position, its width is 2 m. Similarly, the width of the linearly increasing part of the history graph is 2 m. The center of the flat part of the history graph corresponds to both t = 0 s and x = 2 m.
P15.11. Prepare: We will use Equation 15.9 to find the wave speed. Solve: The wave speed is
λ= = =
2.0 m 10 m/s0.20 s
vT
Traveling Waves and Sound 15-9
P15.12. Prepare: We will use Equation 15.10 to find the frequency of the wave. Solve: The frequency is
λ= = =
200 m/s 50 Hz4.0 m
vf
Assess: The wave covers 200 m in 1 s. With a wavelength or repeat unit of 4.0 m, it thus oscillates 50 times per second, which is its frequency.
P15.13. Prepare: Since the lab temperature is 20°C, we know the speed of sound is 343 m/s, as given in Table 15.1. Solve: (a) = 40 kHzf
λ = = =× 3
343 m/s 8.6 mm40 10 Hz
vf
(b) For the round trip the distance is 5.0 m. ∆
∆ = = = =5.0 m 0.015 s 15 ms
343 m/sx
xtv
Assess: 15 ms is quick by human reaction time standards, but it is easy to have electronics in the detector record time intervals such as this.
P15.14. Prepare: The wave is a traveling wave. A comparison of the given wave equation with Equation 15.7 yields A = 3.5 cm, 2 /π λ = 2.7 rad/m, and 2 / 92 rad/s.Tπ = Solve: (a) The frequency is
1 1 2 92 rad/s 14.6 Hz2 2
fT T
ππ π
= = = =
which is 15 Hz to two significant figures. (b) The wavelength is
2 22.7 rad/m 2.33 m2.7 rad/m
π πλλ
= ⇒ = =
which will be reported as 2.3 m to two significant figures. (c) The wave speed is 34 m/s.v fλ= =
P15.15. Prepare: The wave is a traveling wave. A comparison of the given wave equation with Equation 15.8 yields A = 5.2 cm, π λ2 / = 5.5 rad/m, and π2 /T = 72 rad/s. Solve: (a) The frequency is
ππ π
= = = =
1 1 2 72 rad/s 11 Hz2 2
fT T
to two significant figures. (b) The wavelength is
π πλλ
= ⇒ = =2 25.5 rad/m 1.1 m
5.5 rad/m
to two significant figures. (c) The wave speed λ= = 13 m/sv f to two significant figures.
15-10 Chapter 15
P15.16. Prepare: Please refer to Figure P15.16. This is a sinusoidal wave with amplitude 2.0 cm. If we rewrite the equation as
( , ) (2.0 cm)cos 21 0.5x ty x t π
= −
then we can see that the wavelength is 1 cm and the period is 0.5 s. The minus sign in the formula tells us that the wave is traveling to the right. Solve: (a) At 0 s,t = the wave has a crest at the origin. (b) At 1 / 8 s,t = one fourth of a period has elapsed since 0 s,t = so we redraw the wave one-quarter of a wavelength to the right of its original position. (c) The speed of the wave is the wavelength divided by the period, (1 cm)/(0.5s) 2 m/s.v = =
Assess: The minus sign in the formula for the displacement tells us that the wave is moving to the right, which is evidenced by the fact that the snapshot at 1 / 8 st = is to the right of the snapshot at 0 s.t =
P15.17. Prepare: Please refer to Figure P15.17. Solve: A visual examination of the graph shows that the maximum displacement of the wave, called the amplitude, is 4.0 cm. The wavelength is the distance between two consecutive peaks, which gives λ = − =14 m 2 m 12 m. The frequency of the wave using Equation 15.10 is
λ= = =
24 m/s 2.0 Hz12 m
vf
Assess: It is important to know how to read information from graphs.
P15.18. Prepare: Please refer to Figure P15.18. Solve: The amplitude of the wave is the maximum displacement which is 6.0 cm. The period of the wave is 0.60 s, so the frequency = = =1/ 1/0.60 s 1.667 Hz,f T or 1.7 Hz to two significant figures. The wavelength, using Equation 15.10, is
λ = = =2 m/s 1.2 m
1.667 Hzvf
Assess: It is important to know how to read information from graphs.
P15.19. Prepare: We are given the velocity of the boat and wave relative to the shore. From these we can find the velocity of the boat relative to the wave.
bw bs sw bs ws 4.0 m/s 6.8 m/s 2.8 m/sv v v v v= + = − = − = −
Traveling Waves and Sound 15-11
Solve: The boat is traveling backward relative to the wave with a speed of 2.8 m/s. Since the wavelength is 30 m, the boat must travel 30m relative to the wave to go from one crest to the next one behind it. The time needed for the boat to travel 30 m relative to the wave at a speed of 2.8 m/s is given by the following:
bw
30 m 11 s2.8 m/s
tvλ
∆ = = =
Assess: The boat will encounter a crest every 11 s. Thus the frequency relative to the boat is given by (1 /11) Hz 0.091 Hz.f = = The frequency of the wave relative to the shore is given by
ws / (6.8 m/s)/(30 m) 0.23 Hz.f v λ= = = We see that the frequency has been reduced as predicted by the Doppler effect.
P15.20. Prepare: We are given the wavelength and speed of the wave, so we can find the wave’s period using /f v λ= and 1 / .T f=
Solve: If we plug the formula for the frequency, /f v λ= , into the formula for the period, we get the following: 1 / / (95 m)/(12 m/s) 7.92 s.T f vλ= = = = We can find the displacement of the boat at any time by using the
equation for wave displacement.
(1.2 m)cos[2 ( / (95 m) / (7.92 s)]y x tπ= −
We can assume that the boat is at 0.0 mx = since there is initially a crest at that location. Now if we plug in 0.0 mx = and 5.0 s,t = we obtain 0.81 m.y = − The boat will by 0.81 m below the equilibrium position.
Assess: The period is about eight seconds. So after four seconds, or half a period, we would expect the boat to be on a trough, that is, 1.2 m below the equilibrium position. After six seconds, that is another quarter of a period later, we expect the boat to be at the equilibrium position. Thus, it is reasonable that after five seconds, the boat is somewhere between the equilibrium position and 1.2 m below that position.
P15.21. Prepare: We will use the fundamental relation for periodic waves in Equation 15.10 to solve for the wavelength. We must first look up the speed of sound in water; Table 15.1 says it is 1480 m/s. Solve:
λ = = = =× 3
1480 m/s 15 mm 1.5 cm100 10 Hz
vf
15-12 Chapter 15
Assess: Because the speed of sound in water is over four times the speed of sound in air, dolphins must be quick to process the sonar information.
P15.22. Prepare: In aluminum, the speed of sound from Table 15.1 is 5100 m/s. On the other hand, the speed of an electromagnetic wave is c. We will use Equation 15.10 for the wave speed. Solve: (a) The wavelength is
35100 m/s 2.55 10 m 2.6 mm2.0 10 Hz
vf
λ −6= = = × =
×
(b) The frequency would be 8
113
3.0 10 m/s 1.2 10 Hz2.55 10 m
cfλ −
×= = = ×
×
Assess: Because speed = frequency × wavelength, we expected a much higher frequency for the electromagnetic wave.
P15.23. Prepare: From Table 15.1, the speed of sound in room temperature air is 343 m/s and in water is 1480 m/s. On the other hand, the speed of an electromagnetic wave is c. We will use Equation 15.10 for the wave speed. Solve: (a) The frequency is
λ= = =air 343 m/s 1700 Hz
0.20 mvf
to two significant figures. (b) The frequency is
λ×
= = = × =8
93.0 10 m/s 1.5 10 Hz 1.5 GHz0.20 m
cf
(c) The speed of a sound wave in water is vwater = 1480 m/s. The wavelength of the sound wave would be
λ −= = = × =×
7water9
1480 m/s 9.9 10 m 990 nm1.5 10 Hz
vf
Assess: Because speed = frequency × wavelength, we expected a much higher frequency for the electromagnetic wave.
P15.24. Prepare: Light is an electromagnetic wave that travels with a speed of 3 × 108 m/s. The frequencies of the blue and red light are 450 nm and 650 nm, respectively. Solve: (a) The frequency of the blue light is
λ −
×= = = ×
×
814
blue 9
3.0 10 m/s 6.7 10 Hz450 10 m
cf
(b) The frequency of the red light is
−
×= = ×
×
814
red 9
3.0 10 m/s 4.6 10 Hz650 10 m
f
Assess: A higher wavelength for the red compared to the blue light means that the frequency for the red light is smaller than the blue light.
P15.25. Prepare: Microwaves are electromagnetic waves that travel with a speed of 3 × 108 m/s. Solve: (a) The frequency of the microwave is
λ −
×= = = × =
×
810
microwaves 2
3.0 10 m/s 1.0 10 Hz 10 GHz3.0 10 m
cf
(b) The time for the microwave signal to travel is
Traveling Waves and Sound 15-13
×= = =
×
3
8air
50 km 50 10 m 0.17 ms3.0 10 m
tv
Assess: A small time of 0.17 ms for the microwaves to cover a distance of 50 km shows that the electromagnetic waves travel very fast.
P15.26. Prepare: Radio waves are electromagnetic waves that travel with speed c. On the other hand, the speed of sound in air is 343 m/s. Solve: (a) The wavelength is
λ ×= = =
83.0 10 m/s 2.96 m101.3 MHz
cf
or 3.0 m to two significant figures. (b) The speed of sound in air at 20°C is 343 m/s. The frequency is
λ= = =sound 343 m/s 120 Hz
2.96 mvf
to two significant figures. Assess: A difference in the speed of sound and electromagnetic waves means that the sound frequency will be much smaller than the electromagnetic frequency.
P15.27. Prepare: The power (or energy/time) is the intensity multiplied by the area. The intensity is 6 21.0 10 W/m .I −= × We can deduce from the information given that the area of the eardrum
is 2 2 5 2(4.2 mm) 5.54 10 m .A Rπ π −= = = × Solve:
6 2 5 2 11(1.0 10 W/m )(5.54 10 m ) 5.5 10 WP Ia − − −= = × × = × The energy delivered to your eardrum each second is 55 pJ. Assess: This is an incredibly tiny amount of energy per second; you should be impressed that your ear can detect such small signals! Your eardrum moves back and forth about the width of 100 atoms in such cases!
P15.28. Prepare: Follow Example 15.12. The intensity at the stage is 12 (110 dB/10 dB)2 2
stage (1.0 10 W/m )10 0.10 W/mI −= × =
We do not know the power of the source unless we assume a distance for the given 110 dB data; so let’s assume 1.0 m for the distance from the speaker to the edge of the stage. Equation 15.13 shows that the intensity at 30 m will be 2 2(30 m) /(1 m) 900= times less if we assume that the wave is spherical. Therefore the new intensity is
24 20.10 W/m 1.11 10 W/m
900−= ×
The power (or energy/time) is the intensity multiplied by the area. We can deduce from the information given that the area of the eardrum is 2 2 5 2(4.2 mm) 5.54 10 m .a Rπ π −= = = × Solve:
4 5 92 2(1.11 10 W/m )(5.54 10 m ) 6.2 10 WP Ia − − −= = × × = ×
So the eardrum receives 96.2 10 J−× each second during the concert. Assess: This doesn’t seem like much energy, but prolonged exposure to loud sounds like this can damage your hearing.
P15.29. Prepare: We are asked to find the energy received by your back of area 30 cm × 50 cm in 1.0 h if the electromagnetic wave intensity is 1.4 × 103 W/m2. The energy delivered to your back in time t is E = Pt, where P is the power of the electromagnetic wave. The intensity of the wave is /I P a= where a is the area of your back. Solve: The energy received by your back is
15-14 Chapter 15
E = Pt = Iat = (0.80)(1400 W/m2)(0.30 × 0.50 m2)(3600 s) = 6.1 × 105 J
Assess: This is equivalent to receiving approximately 170 J of energy per second by your back. This energy is relatively large and will certainly lead to tanning.
P15.30. Prepare: We are asked to determine the intensity of electromagnetic waves from the sun just outside the atmospheres of Venus, Mars, and Saturn. The intensity of the wave is / ,I P a= where a is the surface area of the spherical shell a distance r from the sun. This is because the sun radiates waves uniformly in all directions. The intensity I of the sun’s rays at a distance r is 2
son /4 .I P rπ= Psun = 4.0 × 1026 W. The values of r needed for this problem are rsun-Saturn = 1.43 × 1012, rsun-Mars = 2.28 × 1011 m, and rsun-Venus = 1.08 × 1011 m. Solve: (a) At Venus, the intensity of electromagnetic waves is 2
Venus 2700 W/m .I =
(b) At Mars, the intensity of electromagnetic waves is 2Mars 610 W/m .I =
(c) At Saturn, the intensity of electromagnetic waves is 2Saturn 16 W/m .I =
Assess: As expected, the intensity falls off from Venus to Mars to Saturn.
P15.31. Prepare: If the solar panel produces electrical power with efficiency ,ε then the power received from the sun is E (1.0 kW)/ .P ε= This is because if we multiply the power from the sun by ,ε we should get
E 1.0 kW.Pε = If the area of the solar panel is ,a then the intensity of radiation from the sun is (1.0 kW) / .aε
To find the intensity at some other distance from the sun, we use Equation 15.13, 2 21 2 2 1/ / .I I r r=
Solve: Using the previous formula, we can find the intensity of sunlight at the location of Saturn’s orbit.
2 2E
S ES
1.0 kW 1 11 W9.5
rI Ir a aε ε
= = =
Then using ,P Ia= we find that S (11 W)/ ,P ε= that is the solar panel would receive (11 W)/ε of power from the sun. Finally, since the panels work with efficiency ,ε the power which would be produced if the spacecraft were in orbit around Saturn is S 11 W.Pε = Assess: The reduction from 1000 W to 11 W corresponds to a factor of about 100. This is reasonable since intensity is inversely proportional to distance squared. At Saturn’s orbit, the satellite is about 10 times farther away from the sun so the intensity of light it receives is reduced by a factor of about 100.
P15.32. Prepare: Since the solar cells only work at 15% efficiency, if we need 11 kW of power, then the amount of light which must fall on the cells is (11 kW)/(0.15) 73.3 kW.= Now we can use /I P a= to solve for the needed area. Solve: Solving for ,a we obtain the following:
2 2/ (73.3 kW)/(200 W/m ) 367 ma P I= = =
which is 2370 m to two significant figures. Assess: This is a very large area and helps to explain why solar cells, at their present efficiency are not practical for providing us with all our energy needs. Consider a small city with a population of 100,000. It would need a
Traveling Waves and Sound 15-15
total solar cell area of 7 23.67 10 m .× But if this area were in the shape of a square, its length would be 6060 m 3.8 mi.L a= = = In other words, the entire city would need to be covered with solar cells.
P15.33. Prepare: To find the power of a laser pulse, we need the energy it contains, ,U and the time duration of the pulse, .t∆ Then to find the intensity, we need the area of the pulse. Its radius is 0.50 mm. Solve: (a) Using / ,P U t= ∆ we find the following:
3 9 4(1.0 10 J)/(15 10 s) 6.67 10 WP − −= × × = ×
(b) Then from / ,I P a= we obtain 4
10 24 2
(6.67 10 W) 8.5 10 W/m(5.0 10 m)
Iπ −
×= = ×
×
Assess: This is very intense light. Using the data from Problem 15.32, the laser light is about 400 million times as intense as the energy from the sun.
P15.34. Prepare: Knowing that the relationship between the sound intensity level and the intensity 10[ 10dBLog ( / )],oI Iβ = we can determine the sound intensity level β for a source of sound intensity .I
Solve: The sound intensity level may be determined by the following: 6 12 6
10 0 10 10(10 dB)log ( / ) (10 dB)log (3 10 /1 10 ) (10 dB)log (3 10 ) (10 dB)(6.5) 65 dBI Iβ − −= = × × = × = =
Assess: This sound is a loud whisper.
P15.35. Prepare: Table 15.3 tells us that the intensity of a whisper at one meter is 10 21 1.0 10 W/m .I −= × We’ll
use Equation 15.12 and ratios to find what it would be twice as far away, source2
.4PI
rπ=
Solve: source
22 22 12
22source1 2
21
10 10 112 2 22 1
(1.0 m)4 0.25(2.0 m)
4
(0.25) (0.25)(1.0 10 W/m ) 0.25 10 W/m 2.5 10 W/m
PI rr
PI rr
I I
π
π− − −
= = = =
= = × = × = ×
The sound intensity level is given by Equation 15.14 where 12 20 1.0 10 W/m .I −= ×
11 2
10 10 12 20
2.5 10 W/m(10 dB)log (10 dB)log 14 dB1.0 10 W/m
II
β−
−
×= = = ×
Assess: Table 15.3 gives the sound intensity level for a whisper at 1 m as 20 dB; we expect it to be less at 2 m, and 14 dB is just about what we expect.
P15.36. Prepare: We use Equation 15.14, 100
(10 dB)log ,II
β
=
which relates the intensity to the intensity level.
Solve: If we double I, we can use a property of logarithms to get the new value of the intensity level, 'β :
10 10 100 0
10
2' (10 dB)log (10 dB) log (2) log
(10 dB)log 2 3 dB 65 dB 68 dB
I II I
β
β
= = +
= + = + =
Assess: It may seem strange that doubling the intensity only increases the intensity level from 65 dB to 68 dB but remember that multiplying the intensity by a factor of 10 increases the intensity level by only 10 dB.
15-16 Chapter 15
P15.37. Solve: (a) If a source of spherical waves radiates uniformly in all directions, the ratio of the intensities at distances r1 and r2 is
22
3 3 3 31 2 50 m 2 250 m 2 m2
2 1 2 m
2 m 1.6 10 (1.6 10 ) (2.0 W/m )(1.6 10 ) 3.2 10 W/m50 m
I r I I II r I
− − − − = ⇒ = = × ⇒ = × = × = ×
(b) The sound intensity level is given from Equation 15.14. 3 2
10 10 12 20
3.2 10 W/m(10 dB)log (10 dB)log 95 dB1.0 10 W/m
II
β−
−
×= = = ×
Assess: The power generated by the sound source is P = I2m [4π (2 m)2] = (2.0 W/m2)(50.27) = 101 W. This is a significant amount of power.
P15.38. Solve: (a) The intensity of a uniform spherical source of power sourceP a distance r away is 2
source / 4 .I P rπ= Thus, the intensity at the position of the microphone is
( )3 2
50 m 235 W 1.11 10 W/m
4 50 mI
π−= = ×
(b) The sound intensity may be determined from Equation 15.14 as follows:
( ) 50 m10
0
10 dB log II
β
=
Using 12 20 1.0 10 W/m , we get 90 dB.I β−= × =
Assess: This is quite loud, as you may expect (compare with values in Table 15.3).
P15.39. Prepare: Knowing the sound intensity level 1 120 dBβ = at the point 1 5 m,r = we can determine the intensity 1I at this point. Since the power output of the speaker is a constant, knowing the intensity 1I and distance 1r from the speaker at one point, we can determine the intensity 2I at a second point 2.r Knowing the intensity of the sound 2I at this second point 2,r we can determine the sound intensity level 2β at the second point. Solve: The sound intensity may be determined from the sound intensity level as follows:
1 10 1(10 dB)log ( / ).oI Iβ =
Inserting numbers: 12 210 1120 dB (10 dB)log [ /(10 W/m )]I −= or 12 2
10 112 log [ /(10 / )]I W m−=
Taking the antilog of both sides obtains 12 12 2110 /(10 W/m )I −= which may be solved for 1I to obtain 12 2
1 (10 W/m )I −= 12 2(10 ) 1 W/m .=
Knowing how the sound intensity is related to the power and that the power does not change, we can determine the sound intensity at any other point. Inserting 24 ,a rπ= into 1 1 2 2P I a I a= = obtain 2 2 22 2
2 1 1 2( / ) (1 W/m )(5/35) 2.04 10 W/m .I I r r −= = = × Finally, knowing the sound intensity at the second point 2,r we can determine the sound intensity level at this point by
2 12 102 10 2 10 10(10 dB)log ( / ) (10 dB)log (2.04 10 /1 10 ) (10 dB)log (2.04 10 )oI Iβ − −= = × × = ×
continuing 10
2 10 1010 dB[log (2.04) log (10 )] 10 dB[0.31 10] 103 dBβ = + = + =
Assess: This is still as loud as a pneumatic hammer. Either take ear protection or move farther away from the speaker.
P15.40. Prepare: We will use Equation 15.14 to relate the intensity level to the intensity. Solve: If we solve Equation 15.14 for I, we have:
( /10 dB)0 10I I β= ×
Traveling Waves and Sound 15-17
Now plugging in 60 dB for ,β we get 6 210 W/mI −= and plugging in 61 dB for ,β we get 6 21.3 10 W/m .I −= × The ratio of the latter to the former is 1.3.
Assess: This intensity ratio of 1.3, which represents the smallest increase perceptible to our ears, is much greater than the minimum frequency ratio we can hear. For example, the ratio of the frequency of the musical note middle C# to the frequency of middle C is only 1.06 and this is not even the smallest frequency ratio which can be heard.
P15.41. Prepare: The frequency of the opera singer’s note is altered by the Doppler effect. The frequency is f+ as the car approaches and f– as it moves away. fo is the frequency of the source. The speed of sound in air is 343 m/s. Solve: (a) Using 90 km/hr = 25 m/s, the frequency as her convertible approaches the stationary person is
0
S
600 Hz 650 Hz25 m/s1 / 1343 m/s
ffv v+ = = =
− −
(b) The frequency as her convertible recedes from the stationary person is
0
S
600 Hz 560 Hz25 m/s1 / 1343 m/s
ffv v− = = =
+ +
Assess: As would have been expected, the pitch is higher in front of the source than it is behind the source.
P15.42. Prepare: This is a Doppler effect problem, so we refer to Equation 15.16.
0
S1 /ffv v+ =
−
We are given f0 = 2200 Hz, f+ = 2300 Hz, and we use v = 343 m/s. We want to know vs. Solve: First solve for vs and then plug in the numbers.
0 0 0
S
0S
1 11 /
2200 Hz1 (343 m/s) 1 15 m/s2300 Hz
S Sf v f f vfv v v f f v
fv vf
++ +
+
= ⇒ − = ⇒ − =−
= − = − =
Assess: An examination of our final equation shows that the closer f0 and f+ are to each other, the smaller vs is, as we would expect.
P15.43. Prepare: Sound frequency is altered by the Doppler effect. The frequency increases for an observer approaching the source and decreases for an observer receding from the source. You need to ride your bicycle away from your friend to lower the frequency of the whistle. We will use Equation 15.17. Solve: The minimum speed you need to travel is calculated as follows:
0 00 01 20 kHz 1 (21 kHz) 16 m/s
343 m/sv vf f vv−
= − ⇒ = − ⇒ =
Assess: A speed of 16 m/s corresponds to approximately 35 mph. This is a possible but very fast speed on a bicycle.
P15.44. Prepare: Your friend’s frequency is altered by the Doppler effect. The frequency of your friend’s note increases as he races toward you (moving source and a stationary observer). The frequency of your note for your approaching friend is also higher (stationary source and a moving observer). We will use Equations 15.16. Solve: (a) The frequency of your friend’s note as heard by you is
S
025.0 m/s343 m/s
400 Hz 431 Hz11 v
v
ff+ = = =−−
(b) The frequency heard by your friend of your note is
15-18 Chapter 15
00
25.0 m/s1 (400 Hz) 1 429 Hz343 m/s
vf fv+
= + = + =
Assess: As would have been expected, the pitch is higher in front of the source than it is behind the source.
P15.45. Prepare: First compute 0 3.0/10 s 0.30 Hz.f = = When the detector (you in the boat) is moving, the measured frequency is 2.0/10 s 0.20 Hz.f− = = The problem gives o 1.5 m/s.v = Use Equation 15.17, o 0(1 / ) ,f v v f− = − and solve for ,v the speed of the waves. Solve:
0
o
0
o
0
o
0o
0
1
1
11
0.30 Hz(1.5 m/s) 4.5 m/s0.30 Hz 0.20 Hz
ff
f vf vv fv f
v v
fv vf f
v
−
−
−
−
= −
= −
= −
= −
= = −
Assess: These are pretty fast water waves, but within reason. The speed of surface water waves depends on the depth of the water, but is typically in the range of 2–3 m/s. However, tsunamis can travel much faster than this. Carefully note the subscripts: The o stands for observer while the 0 stands for the initial or original (frequency). They look similar but are different.
P15.46. Prepare: The frequency shift of the ultrasound reflected from blood moving in the artery may be determined by o o2 ( / ).f f v v∆ = ± Here of is the frequency of the ultra sound, v is the speed of the ultrasound in human tissue (1540 m/s) and ov is the speed of the blood cell reflecting the ultrasound. Solve:
6 3o o2 ( / ) 2(5.0 10 Hz)[(0.20 m/s)/(1540 m/s)] 1.30 10 Hzf f v v∆ = ± = ± × = ×
Assess: Compared to examples in the text, this is a reasonable number.
P15.47. Prepare: If the carpenter hits the nail twice per second then the frequency is 2 Hz and the period is 0.50 s. However, it takes half a period (0.25 s) for the hammer to go from the nail to the upraised position—and this same half a period for the sound to get to you at 343 m/s, the speed of sound in 20°C air. Solve:
(343 m/s)(0.25 s) 86 mxx v t∆ = ∆ = =
Assess: This calculation was for the minimum distance; since the carpenter continues regular pounding, it could also have taken 3/2 or 5/2, etc., of a period for the sound to reach you. However the larger distances might make it hard to see the carpenter’s swings.
P15.48. Prepare: Knowing the frequency of the sound and the speed of sound in aluminum, we can determine the wavelength of the sound in aluminum. Next, we know that the frequency will be the same in air as it was in aluminum since for every disturbance leaving the aluminum there will be a corresponding disturbance in air. Finally, knowing the frequency of the sound in air and the speed of sound in air, we can determine the wavelength in air. Solve: (a) 3 3
alum alum alum/ (5.1 10 m/s)/(2.5 10 Hz) 2.04 m.v fλ = = × × = (b) alum air.f f= For every disturbance leaving the aluminum there will be a corresponding disturbance in air.
Traveling Waves and Sound 15-19
(c) 3air air air/ (343 m/s)/(2.5 10 Hz) 0.14 m.v fλ = = × =
Assess: Note that for a constant frequency the wavelength is directly proportional to the speed. As the sound exits the aluminum it is slowed down by a factor of 343/5100 = 0.067. Notice that the wavelength is decreased by a factor of 0.14/2.04 = 0.069. The slight difference is round-off error.
P15.49. Prepare: The explosive’s sound travels down the lake and into the granite, and then it is reflected by the oil surface. A pictorial representation of the problem along with the speed of sound in water and granite is shown.
Solve: The echo time is equal to
echo water down granite down granite up water up
granite granitegranite
500 m 500 m0.94 s 793 m1480 m/s 6000 m/s 6000 m/s 1480 m/s
t t t t td d
d
= + + +
= + + + ⇒ =
Assess: Drilling into granite for 3/4 km is not unreasonable.
P15.50. Prepare: The wave pulse is a traveling wave on a stretched string. We will use Equations 15.2 and 15.1 to find the string mass. Solve: The wave speed on a stretched string with linear density µ is
S S Sstring 3
2.0 m (2.0 m)(20 N) 0.025 kg 25 g/ 50 10 s
T T LTv mm L m mµ −= = = ⇒ = ⇒ = =
×
Assess: A mass of 25 g for the 2.0 m long string is realistic.
P15.51. Prepare: We need the formulas from Chapter 14 for maximum speed and acceleration:
max 2v fAπ= and 2max (2 ) .a f Aπ= To use these, we need the frequency which is given from
/ (16 m/s)/(4.0 m) 4.0Hz.f v λ= = = Solve: For the maximum speed, we have max 2 (4.0 Hz)(2.0 cm) 50 cm/sv π= = and for the maximum
acceleration, we have 2 2 2max (2 (4.0Hz)) (2.0 cm) 1300 cm/s 13 m/s .a π= = =
Assess: Notice that the maximum speed of a point on the string, 0.50 m/s, is much less than the speed of the wave, 16 m/s. This is typical with waves having small amplitude compared to their wavelength since the ratio of the maximum speed of a point on the string to the speed of the wave is given by the following:
max 2 2v fA Av f
π πλ λ
= =
If / 2 ,A λ π< then the points on the string will move slower than the wave.
15-20 Chapter 15
P15.52. Prepare: To get the time needed for a wave to travel down a string, we need the length of the string and the speed of the wave. The speed comes from Equation 15.2. We are given the mass and length of the string from which we can find the mass density.
Solve: Since the silk is cylindrical, its volume is given by 2 .V r Lπ= From this equation we can find the linear density as follows:
2 22 6 2 3 9
2 (1.0 10 m) (1300 kg/m ) 4.08 10 kg/mm r m r m rL r L V
π πµ π ρ ππ
− −= = = = = × = ×
The speed of the wave on the string is 3
39
4.9 10 N 1.10 10 m/s.4.08 10 kg/m
Tvµ
−
−
×= = = ×
× The time needed for a
pulse to travel the length of the string is just the length of the string divided by the speed of waves on the string.
3
1.1 m 1.0 ms1.10 10 m/s
Ltv
∆ = = =×
Assess: It is good for the spider that very little time is needed for it to receive this information. Notice that the speed of the wave is greater than the speed of sound in air.
P15.53. Prepare: The wave pulse is a traveling wave on a stretched string. The pulse travels along the rope with speed / 3.0 m/150 s 20 m/s / ,Sv x t T µ= ∆ ∆ = = = where we used the relationship involving the wave’s
speed, the rope’s tension, and the linear density rope/ .m Lµ = We can find the tension in the rope from a
Newtonian analysis. Bob experiences horizontal forces T
and k.f
He is being pulled at constant speed, so he is in
dynamic equilibrium with net 0 N.F =
Thus k k .T f nµ= = From the vertical forces we see that Bob .n w m g= =
Solve: The rope’s tension is
2S k Bob (0.20)(60 kg)(9.8 m/s ) 117.6 NT m gµ= = =
Traveling Waves and Sound 15-21
Using this in the velocity equation, we can find the rope’s mass as follows:
S Srope 22
rope
(3.0 m)(117.6 N) 0.882 kg 880 g/ (20 m/s)
T LTv mm L v
= ⇒ = = = =
to two significant figures. Assess: The mass of 880 g for a 3.0-m-long rope is reasonable.
P15.54. Prepare: We have two expressions for the speed of the disturbance along the cord, / .v L t= L is the length of the cord and t is the time it takes the disturbance to travel the length of the cord, / .v T µ= T is the tension in the cord and µ is the linear mass density of the cord / .m Lµ = Equating these two expressions for the speed, inserting the expression for the linear mass density, and solving for the tension T, we obtain
2 2 ( / )L L m L mLTt t tµ
= = =
Solve: Writing this expression for both cases, we obtain
11
1
mLTt
= and 22
2
mLTt
=
Dividing the second expression by the first and solving for 2,T we obtain
2 1 2 1 1 2( / )( / ) (2.1 N)(3.5 m/2.5 m)(1.6 s/1.6 s) 2.9 NT T L L t t= = =
Assess: Compared to the tension for the first case, this is a reasonable number.
P15.55. Prepare: Please refer to Figure P15.55. The wave pulse is a traveling wave on a stretched string. While the tension TS is the same in both the strings, the wave speeds in the two strings are not. We have
S1
1
Tvµ
= and 2 2S2 1 1 2 2 S
2
Tv v v Tµ µµ
= ⇒ = =
Because 1 1 1 2 2 2/ and / ,v L t v L t= = and because the pulses are to reach the ends of the string simultaneously the previous equation can be simplified to
2 21 1 2 2 1 2
2 22 1
L L Lt t Lµ µ µ
µ= ⇒ =
Solve:
1 21 2
2 1
4.0 g/m 2 22.0 g/m
L L LL
µµ
= = = ⇒ =
Since L1 + L2 = 4 m,
2 2 22 4 m 1.66 mL L L+ = ⇒ = and 1 2(1.66 m) 2.34 mL = =
The two lengths will be reported as 1.7 m and 2.3 m.
P15.56. Prepare: Since the problem mentions maximum acceleration, we will use Equation 14.17, 2
max (2 ) .a f Aπ= We are also given the speed and frequency of the wave from which we can find the wavelength using .v f λ= Solve: (a) To find the time between the earthquake and first detection we simply divide the distance traveled by
the speed of the wave, 200 km 29 s.7.0 km/s
xtv
∆∆ = = =
(b) This was a longitudinal wave since it traveled from east to west and caused the ground to vibrate in an east–west direction.
15-22 Chapter 15
(c) The wavelength was 7.0 km/s 6.4 km.1.1 Hz
vf
λ = = =
(d) The maximum horizontal displacement of the ground, that is, ,A can be obtained by solving Equation 14.17 for A:
2max
2 2
0.25(9.8 m/s ) 0.051 m 5.1 cm.(2 ) (2 (1.1 Hz))
aAfπ π
= = = =
Assess: It is interesting to note that even though the wave traveled incredibly fast— 7.0 km/s is about 16,000 mph, the ground moved much slower. The maximum speed of the ground, using Equation 14.15, is given by the following:
2max
1 mi 3600 s(2 ) 2 (1.1 Hz)(5.5 10 m) 0.38 m/s 0.85 mph,1600 m 1 h
v f Aπ π − = = × = =
which is a snail’s pace.
P15.57. Prepare: Draw a right triangle with the base 3.0 m long and the height 0.15 m long. We will keep some extra significant figures to see the point of this problem. We are given dR = 3.0 m, the distance between the ears is 0.15 m, and we use v = 343 m/s. Solve: (a) Use the Pythagorean theorem to compute the distance from the bird to the coyote’s left ear.
2 2L (3.0 m) (0.15 m) 3.003748 md = + =
To two significant figures this is 3.0 m, but we need the extra digits in this case to get a nonzero answer. (b) Use t = d/v.
5L R L RL R
3.003748 m 3.0 m 1.1 10 s 11 s343 m/s
d d d dt tv v v
µ−− −− = − = = = × =
(c) The period of the sound wave is 31/ 1 10 s.T f −= = × The requested ratio is 5
2L R3
1.1 10 s 1.1 10 0.0111 10 s
t tT
−−
−
− ×= = × =
×
Assess: This is impressive time resolution, even down to a hundredth of a period of the wave.
P15.58. Solve: The difference in the arrival times for the P and S waves is
6S P
S P
1 1120 s 1.23 10 m 1230 km4500 m/s 8000 m/s
d dt t t d dv v
∆ = − = − ⇒ = − ⇒ = × =
This will be reported as 1200 km to two significant figures. Assess: d is approximately one-fifth of the radius of the earth and is reasonable.
P15.59. Solve: The time for the wave to travel from California to the South Pacific is 68.00 10 m 5405.4 s
1480 m/sdtv
×= = =
A time decrease to 5404.4 s implies the speed has changed to 6/ 8.00 10 m/5404.4 s 1480.28 m/s.v d t= = × = Since the 4.0 m/s increase in velocity is due to an increase of 1°C, an increase of 0.28 m/s occurs due to a temperature increase of
1 C (0.28 m/s) 0.07 C4.0 m/s
°= °
Thus, a temperature increase of approximately 0.07°C can be detected by the researchers. Assess: Measuring a temperature increase of 0.07°C is rather trivial with modern scientific instruments.
P15.60. Prepare: Obtaining information from Figure P15.60 and using basic sound related information, we can determine the desired quantities.
Traveling Waves and Sound 15-23
Solve: (a) The wavelength is read directly from the graph as 0.12 m.λ = (b) Referring to Figure P15.60, it looks like the wave has traveled a distance of about 0.012 mx∆ = in a time
0.01 st∆ = to give a wave speed of / 0.012 m/0.01 s 1.2 m/s.v x t= ∆ ∆ = = (c) The frequency of the wave is / (1.2 m/s)/(0.12 m) 10 Hz.f v λ= = = (d) The amplitude appears to be around 4mm. Assess: It is important to know how to obtain information from a pictorial representation.
P15.61. Prepare: We’ll employ the fundamental relation for periodic waves, .v fλ= Solve: (a) We simply want to know the period of the passing waves, 1/ 1/0.3 Hz 3.33 s 3 s.T f= = = ≈ (b) We are given that 30 mλ = and (30 m)(0.3 Hz) 9 m/s.v fλ= = = Assess: We have only one significant figure for the data in our estimation. Horizontal oscillations can also contribute to motion sickness, as can rolling oscillations. In fact, motion sickness can even be produced without physical oscillation of the body, but by moving (including oscillating) visual stimuli such as flight trainers.
P15.62. Prepare: This is a sinusoidal wave. Comparison of the given wave equation with Equation 15.8 gives the following: A = 3.0 cm, λ = 2.4 m, and T = 0.20 s. Solve: (a) The displacement of a wave traveling in the positive x-direction with wave speed v must be of the form y(x, t) = y(x− vt). Since the variables x and t in the given wave equation appear together as x + vt, the wave is traveling toward the left, that is, in the −x direction.
(b) The speed of the wave using Equation 15.9 is 2.4 m 12.0 m/s.0.20 s
vTλ
= = =
The frequency is 1 1 5.0 Hz.0.20 s
fT
= = =
(c) The displacement is
0.20 m 0.50 s(0.20 m, 0.50 s) (3.0 cm)cos 2 2.6 cm2.4 m 0.20 s
y π = + = −
P15.63. Prepare: All quantities, except the period, needed to write the y-equation for a wave traveling in the negative x-direction are given. We will determine the period using Equation 15.9. The y-equation that we are asked to write will look like Equation 15.8.
Solve: The period is calculated as follows: 0.50 m 0.125 s.4.0 m/s
Tvλ
= = =
The displacement equation for the wave is
( , ) (5.0 cm)cos 250 cm 0.125 s
x ty x t π
= +
Assess: The positive sign in the cosine function’s argument indicates motion along the −x direction.
P15.64. Prepare: We need to first find the wavelength and the period to write the y-equation for a wave traveling in the positive x-direction. We will determine the wavelength using Equation 15.10. The y-equation that we are asked to write will look like Equation 15.7. Solve: The wavelength and the period are calculated as follows:
400 m/s 2 m200 Hz
1 1 0.005 s200 Hz
vf
Tf
λ = = =
= = =
The displacement equation for the wave is
15-24 Chapter 15
( , ) (0.010 mm)cos 22 m 0.005 s
x ty x t π
= −
Assess: The negative sign in the cosine function’s argument indicates motion along the +x direction.
P15.65. Prepare: The function of x and t with the minus sign will be a wave traveling to the right. Solve: (a) For simplicity, take the snapshot at 0 st = so the equation becomes ( ) (3.0 cm)cos(1.5 ).y x x=
(b) First compare Equation 15.7 for a wave moving to the right to our wave function to identify λ and .T
( , ) cos 2
( , ) (3.0 cm) cos(1.5 50 )
x ty x t AT
y x t x t
πλ
= −
= −
We see that (2 /1.5) m,λ π= (2 /50) s,T π= and (unneeded) 3.0 cm.A = Now use Equation 15.9 for the wave speed.
(2 /1.5) m 33 m/s(2 /50) s
vTλ π
π= = =
As mentioned, because of the minus sign in Equation 15.7 (and our wave function), the wave is moving to the right. Assess: We do not know what kind of wave this is, but the answer is reasonable for the wave speed on a string, and the amplitude is a reasonable amplitude for a wave on a string. Note that although we had to concern ourselves with the 2 ,π it did cancel out in our last calculation.
P15.66. Prepare: Knowing the speed and wavelength of the wave (periodic disturbance) moving along the string, we can determine the frequency of the disturbance. This frequency tells us the rate at which a point on the string is moving up and down (i.e., its linear frequency). Knowing the linear frequency, we can determine the angular frequency which can be combined with the amplitude information to determine the maximum translational speed of any point on the string. Solve: The frequency of the disturbance and hence the linear frequency of the oscillating particle on the string is determined by / (24 m/s)/(0.3 m) 80 Hz.f v λ= = =
The maximum transverse speed of any point on the string is determined by 2max 2 2 (80 Hz)(10 m)v A fAω π π −= = =
5.0 m/s.= Assess: It is important to distinguish between the speed of the wave along the string and the transverse speed of a point on the string.
P15.67. Prepare: From the diagram we see that 15 cm.λ = We know that the trail of sand will get longer (in the direction of the car’s motion) by one wavelength in one period. It is our old friend, / .v Tλ=
Traveling Waves and Sound 15-25
We do need to calculate the period of the oscillatory motion, and we can do so from the formula for a simple pendulum.
2
1.2 m2 2 2.2 s9.8 m/s
LTg
π π= = =
Solve:
15 cm 6.8 cm/s 0.068 m/s2.2 s
vTλ
= = = =
Assess: Whether the car moves one way or a long strip of paper moves the other way under the stationary pendulum mount doesn’t matter. If the angular amplitude is small the sand trail will be a faithful representation of a sine or cosine function.
P15.68. Prepare: The presence of the wall has nothing to do with the intensity. The wall allows you to see the light, but the light wave has the same intensity at all points 2.0 m from the bulb whether it is striking a surface or moving through empty space. At a distance r from the bulb, the one watt of visible light has spread out to cover the surface of a sphere of radius r. The surface area of a sphere is a = 4πr2. We will now use Equation 15.12 to find the intensity. On the other hand, the full two watts of light is concentrated in a dot of area 22 (0.001 m)a rπ π= = = 3.14 6 210 m .−× Solve: (a) The intensity at a distance of 2 m is
222
4.0 W 0.080 W/m4 4 (2.0m)
P PIa rπ π
= = = =
(b) Unlike the light from a light bulb, a laser beam does not spread out. We ignore the small diffraction spread of the beam. The laser beam creates a dot of light on the wall that is 2 mm in diameter. The full two watts of light is concentrated in this dot of area 2 6 22 (0.001 m) 3.14 10 m .a rπ π −= = = × The intensity is
26 2
2 W 640,000 W/m3.14 10 m
PIa −= = =
×
Assess: Although the power of the laser is half the power of the light bulb, the laser produces light on the wall whose intensity is about eight million times that of the light bulb.
P15.69. Prepare: The radio wave is an electromagnetic wave. At a distance r, the 25 kW power station spreads out waves to cover the surface of a sphere of radius r. The surface area of a sphere is 4πr2. Solve: Thus, the intensity of the radio waves is
35source 2
3 22
25 10 W 2.0 10 W/m4 4 (10 10 m)PI
rπ π−×
= = = ××
Assess: The frequency of the radio waves does not enter the calculations above.
P15.70. Prepare: This problem may be solved by applying the relationship between intensity, power, and area ( / )I P a= to the two different situations. It is also necessary to know the expression for the area of a sphere
2sphere( 4 )a rπ= and circle 2
circle( ).a rπ= Solve: Since the total power emitted by the sun is spread out over a sphere, we can determine it if we know the energy received per unit area each second at every point on the sphere and the area of the sphere. The energy received per unit area each second is just the intensity.
3 11 2 262 2Sun shpere (4 ) (1.38 10 W/m )(4 )(1.5 10 m) 3.9 10 WP Ia I rπ π= = = × × = ×
To determine the power received by the earth, we need to realize that the area the earth presents to the incident radiation is that of a disk, hence
3 6 2 1722earth earthdisk earth( ) (1.38 10 W/m ) (6.37 10 m) 1.8 10 WP Ia I rπ π= = = × × = ×
15-26 Chapter 15
Assess: Notice that this number is greater than total human consumption by a factor of 104.
P15.71. Prepare: We will use Equation 15.12 and the definition of power. Solve: (a) The peak power of the light pulse is
7peak 8
500 mJ 0.500 J 5.0 10 W10 ns 1.0 10 s
EPt −
∆= = = = ×
∆ ×
(b) The average power is
totalavg
10 500 mJ 5 J 5 W1s 1s 1s
EP ×= = = =
The laser delivers pulses of very high power. But the laser spends most of its time “off,” so the average power is very much less than the peak power. (c) The intensity is
7 717 2
laser 2 11 2
5.0 10 W 5.0 10 W 6.4 10 W/m(5.0 m) 7.85 10 m
PIa π µ −
× ×= = = = ×
×
(d) The ratio is 17 2
14laser3 2
sun
6.4 10 W/m 5.8 101.1 10 W/m
II
×= = ×
×
Assess: The laser produces light whose intensity is more than 14 orders of magnitude larger than the sun’s intensity on the target.
P15.72. Prepare: From previous problems, we know that a human eardrum has a diameter of about 8.4 mm, so this seems like a reasonable value for the diameter of the ear canal. From this we can find the area. We are also given the intensity level of the sound, 0 dB.β = We know that this is the threshold of hearing and so
12 20 1.0 10 W/m .I I −= = ×
Solve: The cross sectional area of the ear canal is 2 3 2 5 2(4.2 10 m) 5.54 10 m .a rπ π − −= = × = × We can solve Equation 15.11 to get the power “captured” by the ear,
12 2 5 2 17(1.0 10 W/m )(5.54 10 m ) 5.5 10 W.P Ia − − −= = × × = ×
Assess: It goes to show how incredibly sensitive our ears are that they can detect such small rates of energy flow. Remember, however, that 0 dB does not mean “no sound” but merely the sound that humans can barely hear.
P15.73. Prepare: We have a traveling wave radiated by the tornado siren. To calculate the sound intensity level we will use Equation 15.14. Solve: (a) The sound intensity level of the source is calculated as follows:
2
10 10 12 20
0.10W/m(10 dB)log (10 dB)log 110 dB1.0 10 W/m
II
β −
= = = ×
(b) The maximum distance is calculated as follows:
10 10 12 20
60 dB (10 dB)log 6 log1.0 10 W/m
I II
β −
= = ⇒ = ×
Noting that 10log(x) = x,
60 dB6 6 260 dB12 2
10 1.0 10 W/m1.0 10 W/m
II −
−= ⇒ = ××
Now before we can use Equation 15.12 to find the maximum distance, we must calculate the siren’s power.
22 source source 250 m source22
0.10 W/m (0.10 W/m )4 (50 m) (1000 ) W4 4 (50 m)P PI P
rπ π
π π= = = ⇒ = =
Traveling Waves and Sound 15-27
Using once again Equation 15.12 for the 60-dB sound we obtain the following:
6source 2noise 2 2
(1000 ) W1.0 10 W/m 15.8 km4 4PI r
r rπ
π π−= ⇒ × = ⇒ =
Assess: The sound intensity level (compare with values in Table 15.3) is quite loud, as you would expect. The maximum distance of approximately 10 miles at which the siren could be heard seems reasonable as well.
P15.74. Prepare: We are given the intensity level of the sound but we need the intensity so we solve Equation 15.14, /10 dB) 13 2
0 10 1.0 10 W/m .I I β( −= × = × Let 1r and 1I be the distance to and intensity experienced by you and let 2r and 2I be the distance to and intensity experienced by the mouse. Then we can use Equation 15.13 to get the distance from the leaf to the mouse, 2.r
Solve: Solving Equation 15.13 for 2r , we obtain the following:
12 2
2 212 1 13 2
2
1.0 10 W/m(1.5 m) 4.7 m1.0 10 W/m
Ir rI
−
−
×= = = ×
Assess: To decrease β by 10 dB, the intensity must be decreased by a factor of 10. We know that intensity is inversely proportional to the distance to the source squared, so a reduction in the intensity by a factor of 10 means an increase in r by a factor of 10, which is about 3. So it makes sense that the mouse is about three times farther away from the leaf than you are.
P15.75. Prepare: When the problem says “radiated equally in all directions” we know that it will be a uniform spherical wave, and that we can use Equation 15.12.
source24
PIrπ
=
We are given that source 600 WP = and 5.00 m.r = Solve: (a)
source 222
600 W 1.9 W/m4 4 (5.00 m)PI
rπ π= = =
(b) Now we need Equation 15.14 for the sound intensity level where 12 20 1.0 10 W/m .I −= ×
2
10 10 12 20
1.9 W/m(10 dB)log (10 dB)log 122.8 dB 123 dB1.0 10 W/m
II
β −
= = = ≈ ×
(c) We have to do the previous calculations in reverse. 23 dB less than 123 dB is 100 dB, so we plug 100 dBβ = into Equation 15.15.
12 ( /10 dB) 12 (100 dB/10 dB)2 22(1.0 10 W/m )10 (1.0 10 W/m )10 0.010 W/mI β− −= × = × =
The other part of the reverse calculation (with the new intensity of 20.010 W/m ) is to solve the first equation for r.
15-28 Chapter 15
source2
600 W 69 m4 4 (0.010 W/m )Pr
Iπ π= = =
Phil needs to be 14 times farther away than you. Assess: And at 100 dB he still shouldn’t have any trouble hearing the concert (and neither will you with the earplugs at 5.00 m).
P15.76. Prepare: The bat’s chirping frequency is altered by the Doppler effect. The frequency is increased as the bat approaches and it decreases as the bat recedes. Solve: The bat must fly away from you, so that the chirp frequency observed by you is less than 25 kHz. From Equations 15.16,
( )S
0S
S 343 m/s
26,000 Hz20,000 Hz 100 m/s1 / 1 v
ff vv v− = ⇒ = ⇒ =
+ +
Assess: This is a rather large speed, 100 m/s ≈ 230 mph. This is not possible for a bat.
P15.77. Prepare: The sound generator’s frequency is altered by the Doppler effect. According to Equations 15.16, the frequency increases as the generator approaches the student, and it decreases as the generator recedes from the student. Convert rpm into SI units. Use 343 m/s for the speed of sound. Solve: The generator’s speed is
S100(2 ) (1.0 m)2 rev/s 10.47 m/s60
v r r fω π π = = = =
The frequency of the approaching generator is
010.47 m/s
S 343 m/s
600 Hz 620 Hz1 / 1
ffv v+ = = =
− −
Doppler effect for the receding generator, on the other hand, is
010.47 m/s
S 343 m/s
600 Hz 580 Hz1 / 1
ffv v− = = =
+ +
Thus, the highest and the lowest frequencies heard by the student are 620 Hz and 580 Hz.
P15.78. Prepare: Knowing the wavelength and period of the waves, we can determine the speed of the incoming waves. If we then determine the speed of the boy with respect to the waves it will be as if the boy is running and the wave is standing still. Then knowing the speed of the boy, we can determine how far he can run in one second and how many wavelengths will fit into this distance. There is one crest per wavelength. Solve: First let’s agree that the direction the boy is running is the positive direction. The velocity of the incoming waves relative to land then is / 1.2 m/1.5 s 0.80 m/s.WLv f Tλ λ= − = − = − = −
The velocity of land relative to water is ( 0.80 m/s) 0.80 m/s.LW WLv v= − = − − =
The velocity of the boy relative to water is obtained by adding the velocity of the boy relative to land and the velocity of the land relative to water, 3.5 m/s 0.80 m/s 4.3 m/s.BW BL LWv v v= + = + =
The distance the boy runs relative to water in one second is | | (4.3 m/s)(1.0 s) 4.3 m.LWd v t= = =
Finally, the number of waves that will fit into this distance and hence crests passed by the boy in one second is n = / 4.3 m/1.2 m 3.6.d λ = = Assess: The concept of relative velocity simplifies the problem.
P15.79. Prepare: This is obviously a Doppler effect problem with the source moving, so we look up Equation 15.16
0
s1 /ffv v+ =
−
Traveling Waves and Sound 15-29
for the source approaching, and opposite signs for source receding. We are given 340 m/sv = (instead of the usual 343 m/s). We follow the mathematical derivation given in Example 15.13 and use the result there. Solve: Insert 2f f+ −= everywhere for .f+
s2 (2 1)2 (2 1) 3
f f f f f vv v v vf f f f f
+ − − − −
+ − − − −
− − −= = = =
+ + +
s /3 (340 m/s)/3 113.3 m/s 113 m/sv v= = = ≈ to three significant figures, or110 m/s to two. Assess: We could have done the algebra another equivalent way: Because we are told that 2 ,f f+ −= let’s divide the two equations in Equations 15.16.
s
s
1 /21 /
f v vf v v
+
−
+= =
−
Solving this for sv (the steps are left to the student) also gives s /3.v v= In any event, the answer is independent of 0.f P15.80. Prepare: We can find the speed of the blood using Equation 15.18 for the Doppler effect for sound reflecting off a moving liquid. Then we can find the pressure difference between the heart and the aorta using Bernoulli’s principle, which in this case will have the following form:
2 2b h h b a a
1 12 2
v P v Pρ ρ+ = + ,
where bρ is the density of blood. From chapter 13, 3b 1060 kg/m .ρ =
Solve: (a) Solving Equation 15.18 for the speed of the blood, we get the following:
b0
6000 Hz(1540 m/s) 1.85 m/s2 2(2.5 MHz)
fv vf
∆= = =
(b) Solving Bernoulli’s equation for h a ,P P P∆ = − we get the following:
( )2 2 3 2 2h a b a h
35
1 1 (1060 kg/m )((1.85 m/s) (0 m/s) )2 2
760 mm of Hg1.81 kPa 1.81 10 Pa 14 mm of Hg1.013 10 Pa
P P P v vρ∆ = − = − = − =
= × = ×
Assess: The speed of the blood seems reasonable if you consider that your heart beats once per second and the distance from your heart to your feet is about 1.2 m. If blood travels to your feet and back to your heart once per second, in this simple model, it would have a speed of 2.4 m/s.
15-30 Chapter 15
P15.81. Prepare: Knowing that /v d t= and that the distance d includes out and back, we can determine the time after a pulse is emitted that a bat is ready to detect its echo. Solve:
/ 2 / 2(1.0 m)/343 m/s 6.0 mst d v x v= = ∆ = =
The correct choice is D. Assess: This time will allow the bat to keep track of insects flying at standard insect speeds.
P15.82. Prepare: “Changes in frequency” makes us think of the Doppler effect. Solve: The Doppler effect is able to tell the bat about the speed of its prey. The correct answer is B. Assess: If the bat also measures the time it takes for the reflected pulse to return it could determine distance; and if the bat also measures the intensity of the reflected pulse it might get some idea of size as well. But the changes in frequency indicate speed.
P15.83. Prepare: Some bats emit ultrasonic pulses through their nostrils into a parabolic nose reflector that concentrates the pulse in the forward direction. As a result, the reflected signal from the prey (usually in the forward direction) is much stronger than the signal from other directions. Furthermore, since the pulse is concentrated in a small area, it is more intense, which makes it easier to detect prey. If we could hear the ultrasound produced by the bat, it would be as loud as a jet engine (100 dB), so the bat does need some ear protection from the ultrasound. However this is provided by another mechanism (see Problem P15.81). Solve: Based on the above, answers A and B are correct. Assess: We can learn a lot from bats. They might hold the key to detecting stealth bombers.
P15.84. Prepare: We want to invoke the fundamental relationship for sinusoidal waves, .v fλ= Solve: The speed of sound in air is determined by the properties of the air (including its temperature), but the speed is not altered by a decreasing frequency sound wave. That is, v is constant for given air conditions. So, in ,v fλ= if f is decreasing, then λ must be increasing. The correct answer is C. Assess: Always remember that for mechanical waves and sound waves the speed is determined by the properties of the medium.
