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Midterm Exam #1: Solutions
P105 - Basic Physics of Sound - Fall 2001
Closed book. You are allowed a calculator.
Wednesay 26 Sept. 2000, 10:10 – 11:00 a.m.
Circle one and only one answer as the “best” answer.
Where applicable, show your work (on back if necessary) – partial credit is possible.
Each question is worth 4 points unless indicated otherwise.
1. If 1mile = 1.61 km , convert the speed of sound, v sound = 343 m/s , into units of miles
per hour , i.e, mi/hr (mph) ( 1 hr. = 3600 s ).
A) 696 mi/hr ;
B) 724 mi/hr
C) 767 mi/hr
D) 796 mi/hr
E) 815 mi/hr
1 km
1 mile 3600 s
343 m
--------------- × ------------------ × -------------------- × ---------------- = 767 mi/hr , (C)
1000 m 1.61 km
1 hr
s
2. (8 pts) On each of the following, sketch on top of the wave shown:
Amplitude
A) A wave that is higher in frequency, but with same amplitude:
2
1
Time
-1
-2
Midterm Exam #1, P105, Basic Physics of Sound
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Amplitude
B) A wave that is smaller in amplitude, but with the same frequency:
2
1
Time
-1
-2
Amplitude
C) A wave that is the same in amplitude and frequency, but with a different tone or timbre:
2
1
Time
-1
-2
Amplitude
D) A wave that is the same in amplitude and frequency, but180 degrees out of phase
with the original:
2
1
Time
-1
-2
Midterm Exam #1, P105, Basic Physics of Sound
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Amplitude
(excess pressure, N/m2
3. At a certain instant in time, the figure below describes a sound wave travelling in air
( v sound = 343 m/s ). What is the frequency of the wave? (Note that the horizontal axis is
not time.)
A)
B)
C)
D)
E)
2
1
1
2
3
4
5
6
Position, x [m]
-1
-2
0.50 Hz;
2.0 Hz;
25 Hz;
86 Hz;
171 Hz.
Careful here! You are given the hint that the horizontal axis does not represent
time, i.e., you cannot get the period directly from the plot, but you can get wavelength:
v
343 m/s
λ = 2 m , v = fλ , f = --- = ------------------- = 1711/s = 171 Hz , (E)
λ
2.0 m
4. A mass of m = 3.0 kg is hanging at the end of a spring of spring constant k = 20 N/m.
What is the period, T, of oscillation of this system?
A) 4.1 s;
B) 3.6 s;
C) 3.2 s;
D) 2.4 s;
E) 1.8 s.
1 k
1 20 N/m
f = ------ ---- = ------ ------------------ = 0.41 Hz , but the question is not asking for fre2π m
2π 3.0 kg
1
1
quency, but rather for period. T = --- = ------------------- = 2.4 s , (D)
0.41 Hz
f
Midterm Exam #1, P105, Basic Physics of Sound
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5. You stand stationary by the railroad tracks as a train approaches, blowing his horn.
The train passes your position. Comparing the frequency and wavelength when the
train is approaching you ( f towards, λ towards ) to the frequency and wavelength when it has
passed and is moving away, ( f away, λ away ), we have due to the Doppler effect:
A) f towards > f away and λ towards > λ away ;
B) f towards > f away and λ towards < λ away ;
C) f towards < f away and λ towards < λ away ; or
D) f towards < f away and λ towards > λ away .:
From either the equation or class, λ towards < λ away (i.e., wavelengths “squashed”
v
together). Since f = --- , they are inversely proportional, and we must have:
λ
f towards > f away , (B).
Position, x [cm]
6. Two waves are shown below, A (dashed line) and B (solid line). Sketch the superposition, i.e., A+B, of the two waves on the plot below.
4
3
2
1
B
time [s]
0
–1
–2
A
–3
–4
Midterm Exam #1, P105, Basic Physics of Sound
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7. How long does it take for a sound wave to travel 15 km (assume that
v sound = 343 m/s )?
A) 0.16 s;
B) 0.78 s;
C) 4.7 s;
D) 143 s;
E) 43.7 s.
15, 000 m
∆x
∆x
v = ------ , ∆t = ------ = ------------------------ = 43.7 s , (E)
343 m/s
v
∆t
8. Two sinusoidal tones of equal amplitude are sounded simultaneously. You hear a tone
with frequency 440 Hz changing in amplitude with a beat frequency of f beat = 4 Hz .
The two original tones would have frequecies of
f A = 438 Hz
,
f B = 442 Hz
fA + fB
- = 440 Hz .
We have to satisfy two conditions: f A – f B = 4 Hz and also f sum = -----------------2
Another way to look at it is we need two frequency values that are different by 4 Hz but
that have an average of 440 Hz. Some people can do this by inspection, i.e., the answer
must be 2 Hz above the central value of 440 Hz and 2 Hz below the central value of 440
Hz. Alternatively, you could also solve it algebraically: f A + f B = 880 Hz , and from the
first equation, f A = f B – 4 Hz . Can substitute into the previous equation,
f B – 4 Hz + f B = 880 Hz so f B = 442 Hz , f A = f B – 4 Hz = 442 Hz – 4 Hz = 438 Hz
(and remember to include units!!)
9. A car is sitting in a hot parking lot with its car alarm wailing. On the figure below,
sketch the resultant path of the sound wave due to refraction of the sound wave in the
variation of the temperature of the air (and hence speed of sound) above the parking lot.
Note that this is exactly the reverse configuation that I drew on the board or that is in
your text where we had cold air by the ground and warm air above. In this situation, the
sound wave fronts will travel faster in warm air and slower in the cold air. It is just like
the example where we had the marching band going from dry ground (fast, e.g., warm
air,) to muddy ground (slow, e.g., cold air). Check out Fig. 2.-19 in your text. You can see
which direction the wavefronts then swing:
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Cold Air
Warm air
10. An oscillation with a frequency of 12.5 kHz has a period of:
A) 0.08 sec;
B) 0.0125 sec;
C) 12.5 msec;
D) 80 msec;
E) 80 µ sec.
It is very important to note that this is 12.5 kHz, and not 12.5 Hz.
3
f = 12.5 kHz = 12.5 ×10 Hz = 12, 500 Hz ,
–5
–6
1
1
T = --- = -------------------------- = 8.0 ×10 s = 80 ×10 s = 80 µsec .
12, 500 Hz
f
11. Rita accelerates a violin bow with mass of m = 0.2 kg from rest to a speed of
v = 9.0 m/s in a time of t = 0.15 s . What average power does she expend to do this?
(Hint: do not calculate acceleration; rather use energy considerations, i.e., what is the
final energy of the bow? as a step along the way).
A) 54 W;
B) 22 W;
C) 2.2 W;
D) 48 W;
E) 128 W.
The initial kinetic energy is zero. The change in kinetic energy is therefore the final
1
1 2
2
kinetic energy: KE = --- mv = --- ( 0.2 kg ) ( 9.0 m/s ) = 8.1 J . (those units are Joules; I saw
2
2
answers with many different units....). Power is the rate of energy expenditure, i.e.,
8.1 J
KE
E
P = ----- = -------- = -------------- = 54 J/s = 54 W , (A)
0.15 s
∆t
∆t
Midterm Exam #1, P105, Basic Physics of Sound
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Position (m)
Name:
4
3
2
1
0
–1
–2
–3
–4
Time (s)
2
4
6
8
10
12
14
12. The graph above shows the position versus time plot of a student walking down a
hallway. During a certain time interval, the student has positive velocity. The velocity
during this time interval is:
A) 0.5 m/s;
B) 1.0 m/s;
C) 1.3 m/s;
D) 2.0 m/s;
E) 4.0 m/s.
The velocity on a position versus time plot is the slope of the graph at that point.
Positive velocity is where that slope is positive, i.e., the whole segment of the
4m–0m
∆x
rise
graph between 0 and 4 sec. v = slope = -------- = ------ = ------------------------- = 1.0 m/s , (B).
4s–0s
∆t
run
13. What property of waves allow you to hear things around corners?
A) diffraction;
B) refraction;
C) polarization;
D) interference;
E) Huygen’s principle.
Diffraction is the bending of waves as they pass through openings or pass around
obstacles (just like a corner).
and all other things being equal, which can we hear more readily around corners, flutes
or tubas? Why?
We learned that diffraction is most important when the wavelength of the wave
Midterm Exam #1, P105, Basic Physics of Sound
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is comparable to the dimensions of the object. For a given object then (e.g.,
corner of a building), the longer the wavelength, the more diffraction we will have.
Tubas typically produce sound waves of lower frequencies than flutes, and hence
produce waves typically with longer wavelengths than flutes. As a result, the sound
from a tuba will tend to be diffracted more than from a flute, so one can hear this
instrument around corners more.
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