Problem Set 10 solutions:
1.
We know that the closer the microphone is to a speaker, the louder the volume it detects because
the intensity I of the sound waves is high. However, the question says unusually loud volume,
which sounds like we are being asked about the effects of constructive and destructive
interference. We are asked for where the microphone records very loud and very low volumes,
and so we had better draw a diagram showing what we know and specifying a suitable location
marker (Figure W17.1). Let us choose this marker to be the distance from the speaker on the
left, which we call speaker 1.
Constructive interference occurs at point P in the Figure W17.1 if the sound waves from the two
speakers arrive at P in phase. To achieve this, the phase difference at P must be one cycle: 0 or
2π or –2π or 4π . . . . Because there is one wavelength in each cycle, the number of wavelengths
in distances D1 and D2 must differ by an integral number of wavelengths: 0, ±λ, ±2λ, . . . . So for
constructive interference,
Destructive interference occurs where the two waves arrive at a given point out of phase by
180º= π rad. Using the same argument as above, we can say
We have to make sure that our answers remain between the two speakers, which requires that D1
+ D2 = L and that 0 < D1 < L and 0 < D2 < L.
We know that D1 + D2 = L. Therefore D2 = L – D1. Consequently, for constructive interference,
Now we can plug in the first few values of n to find all the constructive-interference points
between the two speakers:
These last two distances are not between the speakers, which are a distance 10/3 = 20/6 apart,
and so we discard them. Thus the positions between the speakers where the microphone detects
unusually loud volumes are at these distances from speaker 1:
Using the same procedure, we find that the distances from speaker 1 that produce destructive
interference satisfy the expression
:
Again we want only positions that lie between the two speakers, which are 10/3 = 40/12 apart.
The last three values we just calculated are not on this center-to-center line, and so our
destructive-interference points are
We expect constructive interference halfway between the speakers because the waves from the
two speakers have to travel the same distance to reach this point. This distance is
, and we see that this value does appear in our constructive-interference
list.
We expect the spacing of constructive and destructive points to be related to the wavelength
in some simple way. In other words, the same type of interference should reoccur as the
difference between D1 and D2 changes by one wavelength.. This is easiest to see with the
constructive points. The midpoint between the speakers is a place of constructive interference.
As we move from this midpoint toward speaker 2, D1 increases and D2 decreases. If D1
increases by /2, D2 decreases by /2, and the difference changes by one wavelength. Thus
constructive points should be a distance /2 apart, destructive points should be a distance /2
apart, and this is just what we see
We also expect a destructive point halfway between each pair of constructive points, which is
just what we see, as, for example,
).
We made one assumption that can affect the accuracy of our answers. We assumed that the
amplitude of the waves is constant along the line joining the speakers, and this is not true. The
closer you are to one speaker, the higher the amplitude of the wave coming from it and the lower
the amplitude of the wave coming from the other speaker. Thus, places where the phase
difference corresponds to a half-wavelength are not places where the crest of one wave cancels
the trough of the other. This causes the constructive and destructive points to be at locations
slightly different from those we calculated. Our only calculated point that is strictly accurate is
the constructive point halfway between the speakers.
2. You know that if you are moving toward a stationary sound source, the frequency you hear is
higher than the frequency you would hear if you were standing still. The question here is how
fast must you be moving in order for the frequency of the notes you hears to be shifted upward
by a factor of 0.16. The important information you need to know is how much of a frequency
change is required to change a sound from one note to the next note (from do to re, for instance,
or from la to ti). In western musical theory, one octave is a factor of 2 in frequency. That is, to
take just one example, the frequency of A above middle C is twice the frequency of A below
middle C. For a single note (a half step) the factor is about 1.06 for all notes. Thus we need to
find the speed such that fo/fs = 1.06.
For a moving observer, the change in frequency is given by
Looks like all we need to do is solve for vo and put in numbers.
A little algebra yields
Putting in the speed of sound in air and fo/fs=1.06, we get
The negative sign means that you and the wavefronts are traveling in opposite directions, which
is good because that’s what happening. This speed is pretty fast, but perhaps not unreasonable.
3. A wake is a type of shock wave caused because the speedboat travels faster than the waves do
across the water’s surface. We’re asked to find out (a) what this wave speed is and (b) Mach
number.
Equation 17.18 relates the wave speed to the boat speed and to the angle of the shock wave:
(a) Solving for the wave speed gives
(b) Mach number is defined as the ratio of the speed of an object in a given medium to the speed
of waves in that medium. The water-Mach number is therefore
vs/c= (65 km/h)/(12.4 km/h)=5.2.
The answer to part a indicates that water waves spread across the surface at about 10 ft/s, a value
that is not unreasonable.
4. XXXX
sin  
5. The half angle of the shock wave cone is given by
vS 
v light
sin 

v light
vS
.
2.25  108 m s
 2.82  108 m s
sin 53.0
6. On Planet X, the surface pressure is
Psurface(X) = Pgauge(X) + 1 earth-atmosphere
=
+
=
At a depth d on Plane X, the total pressure is
Pd(X) = Pgauge(X) + 1 earth-atmosphere
=
+
Applying Eq. (1) on Planet X gives us
=
Pd(X) = Psurface(X) +
Solving for gX and substituting the pressures we just calculated gives
= 5.99 m/s2
=
7. The rate of flow of the water is
. Using the given information, we have

The radius R of the pipe is
Solving Poiseuille’s law for the pressure difference gives
Using the numerical values we just calculated gives us
,
which we round to 470 Pa.
To find the maximum speed, we use
with r = 0. This gives
=
8. Consider spherical balloons of radius 12.5 cm containing helium at STP and immersed in air at
0°C and 1 atm. If the rubber envelope has mass 5.00 g, the upward force on each is
B  Fg,He  Fg,env   air Vg   He Vg  m env g
 4 
Fup   air   He  r3 g  m env g
 3 


4
3 
Fup  1.29  0.179 kg m 3   0.125 m   9.80 m s2  5.00  10 3 kg 9.80 m s 2  0.040 1 N
3




If your weight (including harness, strings, and submarine sandwich) is


70.0 kg 9.80 m s2  686 N
686 N
 17 000 ~ 104
0.0401
N
you need this many balloons:
.
9.
B  Fg

10.
(a)
We imagine the superhero to produce a perfect vacuum in the straw. Take point 1 at the
water surface in the basin and point 2 at the water surface in the straw:
P1  gy1  P2  gy2



y 2  10.3 m
1.013 105 N m 2  0  0  1 000 kg m 3 9.80 m s2 y 2
(b) No atmosphere can lift the water in the straw through zero
height difference.
11. Take points 1 and 2 in the air just inside and outside the window pane.
P1 
1 2
1
v 1  gy 1  P2  v 22  gy 2
2
2
P0  0  P2 


1
2
1.30 kg m 3 11.2 m s
2
P2  P0  81.5 Pa
The total force exerted by the air is outward,


P1 A  P2 A  P0 A  P0 A  81.5 N m 2 4 m1.5 m  489 N outward
Between sea surface and clogged hole:
P1 
1 2
1
v 1  gy 1  P2  v 22  gy 2
2
2



1 atm  0  1 030 kg m 3 9.8 m s 2 2 m   P2  0  0
P2  1 atm  20.2 kPa
The air on the back of his hand pushes opposite the water, so the net force on
his hand is

3
2  
2
F  PA  20.2  10 N m   1.2  10 m
 4 

(b)
 

2
F  2.28 N
Now, Bernoulli’s theorem is
1 atm  0  20.2 kPa  1 atm 


The volume rate of flow is

1
1 030 kg m 3 v 22  0
2
A 2 v2 

v2  6.26 m s
1.2  10 2 m 6.26 m s  7.08  10 4
4
2
m3 s
One acre–foot is
4 047 m  0.304 8 m  1 234 m
2
3
1234 m3
4
Requiring
7.08 10
3
m s
 1.74 106 s  20.2 days