Physics 198
Exam 2
Fall 2014
21. The predominant frequency of a certain fire engine’s siren is 1550 Hz when at rest. What
frequency do you detect if you move with a speed of 30.0 m s toward the fire engine?
(Speed of sound: 340 m/s)
A)
B)
C)
D)
E)
1690 Hz
1580 Hz
1550 Hz
1520 Hz
1410 Hz
 v  vL 

Solution: Doppler effect: f L  f S 
 v  vS 
For observer moving towards stationary source:
 340m / s  30m / s 
f L  1500 Hz 
  1687 Hz
 340m / s  0 
22. In order to obtain a good single slit diffraction pattern, the slit width could be:
A)
B)
C)
D)
E)
/100
/10

10
100
Solution:
From equation D sin  m  m follows that for D< there are no fringes.
For D< the minima (dark fringes) exist only for m  1 , and   90  , i.e. at infinity.
It is possible to observe several (up to 19) fringes for D=10.
For D=10 the angle between fringes is too small.
23. The Eustachian tube
A)
B)
C)
D)
E)
Connects the outer ear to the oral cavity
Connects the middle ear to the oral cavity
Connects the inner ear to the oral cavity
Connects the outer and middle ear
Connects the middle an inner ear
Solution:
The Eustachian tube connects the middle ear to the oral cavity. (See lecture 10 and/or section
5.2 in the textbook)
1
Physics 198
Exam 2
Fall 2014
24. At low frequencies (below 1000 Hz) localization of sound is possible due to what factor?
A)
B)
C)
D)
E)
Time difference between sound traveling to two ears
The intensity difference at two ears (shadow effect)
Reflection from surrounding objects
Harmonics decomposition
Localization is impossible at these frequencies
Solution:
At high frequencies (above 4000 Hz) localization is due to intensity difference at two ears. (High
frequency – low wavelength – shadow effect)
At low frequencies (below 1000 Hz) localization is due to time difference between sound
traveling to two ears. (Due to diffraction at low frequency there is no shadow.)
25. According to the Fechner law sensation is what function of stimulus?
A)
B)
C)
D)
E)
Linear
Quadratic
Harmonic
Logarithmic
Exponential
Solution:
Fechner law: As stimuli are increased by multiplication sensation increase by addition
(Logarithmic scale)
26. What equation is correct?
A) log( 400)  2  2 log 2
B) log( 400)  3  log 3
C) log( 400)  4  log 3
D) log( 400)  4 log 2
E) log( 400)  6 log 2
Solution:
log( 400)  log( 100  2 2 )  log 100  log 2 2  2  2 log 2
2
Physics 198
Exam 2
Fall 2014
27. A source of a 500 Hz sound is operated at a power of 60 microwatts. The sound radiates
with equal intensity in all directions. What is the intensity level 50 m from the source?
A)
B)
C)
D)
E)
82 dB
70 dB
65 dB
41 dB
33 dB
Solution:
I
W
60  10 6 watt 6 9

 10 watt / m 2  1.9  10 9 watt / m 2
2
2

4r
4 50m 
LI  10dB  log
I
1.9  10 9
 10dB  log
 33db
I0
10 12
28. Suppose that a single violin gives an average sound intensity level of 70 dB at a seat in the
audience. Now suppose that 2 violins play the same piece in the same way. What average
sound intensity level would you expect at the same seat in the audience?
A)
B)
C)
D)
E)
70 dB
73 dB
76 dB
79 dB
82 dB
Solution: If intensity is doubled, then intensity level increases by 3dB.
I  2I1 
LI  10dB  log
 10dB  log


2I
I
I
 10dB  log 1  10dB  log 1  log 2 
I0
I0
I0


I1
 10dB  log 2  70dB  3dB  73dB
I0
29. Suppose that a single violin gives an average sound intensity level of 70 dB at a seat in the
audience that is L meters from the violin. What average sound intensity level would you
expect at a seat that is L/2 meters from the violin?
A)
B)
C)
D)
E)
70 dB
73 dB
76 dB
79 dB
82 dB
3
Physics 198
Exam 2
Fall 2014
Solution:
If distance is halved, then the intensity level increases by 6dB.
I 2 r12
P
P
.
I 


A 4r 2
I 1 r22
LI 2  10dB  log
r12
If r2  r1 / 2 , then I 2  I 1 2  4I 1 .
r2


I2
4I
I
I
 10dB  log 1  10dB  log 1  log 4   10dB  log 1  10dB  log 2 2
I0
I0
I0
I0


 70dB  6dB  76dB
30. If two sounds differ by 20 dB, what is the ratio of their intensities?
A)
B)
C)
D)
E)
2
10
20
log20
100
Solution:
I
 10 LI
I0
10 dB

I2
 10 LI 2  LI 1  10dB  1020 / 10  102  100
I1
31. What part of the ear is mainly responsible for critical band and for jnd (just-noticeable
difference)
A)
B)
C)
D)
E)
Outer ear
Middle ear
Eustachian tube
Basilar membrane
Semi-circular canals
Solution: See lecture 14 and/or section 7.2 in the textbook.
32. What statement is wrong?
A)
B)
C)
D)
E)
Pitch increase with sound level for high frequencies
Pitch decrease with sound level for low frequencies
Pitch shows little changes for middle frequencies (~2000Hz)
Pitch decrease with sound level for high frequencies
Pitch depends on duration and on envelope of sound
Solution: See lecture 14 and/or figure 7.3 in the textbook.
4
Physics 198
Exam 2
Fall 2014
33. What is virtual pitch?
A)
B)
C)
D)
E)
Second harmonic
Missing fundamental
The lowest frequency in the spectrum
The highest frequency in the spectrum
None of the above
Solution: See lecture 14 and/or section 7.4 in the textbook.
34. Pure tones of 440 Hz and 600 Hz are sounded together. What is frequency of beats?
A)
B)
C)
D)
E)
600 Hz
440 Hz
160 Hz
80 Hz
None of the above
Solution:
If frequency difference is more then 15 Hz, we do not hear beats between these two frequencies.
35. What frequency ratio gives an interval that is exactly 1/12 of an octave?
A)
B)
C)
D)
E)
1/6
1/7
1/12
21 12
21 6
Solution: Semitone of equal temperament gives an interval that is exactly 1/12 of an octave. The
frequency ratio is 21 12  1.05946 .
36. What ratio has minor third in equal temperament?




4
A) 21 / 12  21 / 3  1.26
B) 5 : 4  1.25
C) 21 / 12
3
 21 / 4  1.19
D) 6 : 5  1.2
E) 3:2=1.5
Solution: Minor third has 3 half steps.
5
Physics 198
Exam 2
Fall 2014
37. How many octaves have 12 perfect fifth?
A)
B)
C)
D)
E)
5
6
7
8
9
Solution: Perfect fifth has 7 half steps and octave has 12 half steps. (7x12=12x7)
38. In Pythagorean scale what is the ratio of F and C frequencies ( f F f C )?
A)
B)
C)
D)
E)
4/3
3/2
9/8
256/243
None of the above
Solution: The ratio f F f C corresponds to perfect fourth. In Pythagorean scale it is exactly 4/3.
39. Just intonation is based on major triads with frequency ratios:
A)
B)
C)
D)
E)
1:2:3
4:5:6
7:8:9
1:3/2:2
1:5/4:2
Solution: See lecture 17 and/or section 9.4 in the textbook.
40. Which scale has a circle of fifths that closes exactly?
A)
B)
C)
D)
E)
The equal temperament scale
The Pythagorean scale
The just intonation scale
The mean tone intonation scale
None of the above
Solution: See lecture 17 and/or section 9.6 in the textbook.
6
Physics 198
Exam 2
21
A) 1690 Hz
31
D) Basilar membrane
22
D) 10
32
D) Pitch decrease with
sound level for high
frequencies
33
B) Missing fundamental
23
B) Connects the middle
ear to the oral cavity
24
A) Time difference
between sound traveling
to two ears
25
D) Logarithmic
34
E) None of the above
26
A) log( 400)  2  2 log 2
36
27
E) 33 dB
37
C) 7
28
B) 73 dB
38
A) 4/3
29
C) 76 dB
39
B) 4:5:6
30
E) 100
40
A) The equal temperament
scale
Fall 2014
35
D) 21 12

C) 21 / 12

3
 21 / 4  1.19
7