1 1 Physics 198 Final Exam-Solution December 16, 2008 The exam ...

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Physics 198 Final Exam-Solution December 16, 2008
The exam is closed book, although you may use four sheets of paper to help memory.
For multiple choice questions, mark the answer you believe is MOST CORRECT on
the exam. Please turn in only the exam sheets.
You may assume that the speed of sound in air is 343 m/s for this exam.
1. In the English language, vowel sounds are determined by:
a. the pitch of the vowel as it is spoken.
b. the change in pitch of a vowel as it is spoken.
c. ⇒formant frequencies.
d. changes in the intensity of the sound as the vowel is spoken.
e. changes in the phase of the sound wave as the vowel is spoken.
2. How would the frequency of the fundamental bending mode of a bar change if
the length of the bar were doubled?
a. It would go up by a factor of 4.
b. It would go up by a factor of 2.
c. It would not change.
d. I would go down by a factor of 2.
e. ⇒It would go down by a factor of 4.
The bending mode frequencies are proportional to 1/L 2 so if L goes up
by 2 then the frequency goes down by 4.
3. Which statement most accurately describes the shape of bars and the harmonic
structure of a marimba.
a. The bars are shaped so they are thicker in the center giving all
harmonics with the even ones emphasized.
b. The bars are the same thickness in the center and at the edges. They
have partials that are not harmonic, and the pitch is an average value of
the strong partials.
c. The bars are the same thickness and at the edges. The partials consist
only of odd harmonics with the third harmonic emphasized.
d. The bars are shaped so they are thicker in the center giving only odd
harmonics.
e. ⇒ The bars are shaped so that they are thinner in the center giving
partials approximately aligned with the first, fourth and ninth
harmonics and a lower overall pitch.
4. There are resonators under the bars of a marimba causing:
a. the notes to last longer when a bar is struck.
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b. the fundamental would be diminished in relation to the higher
harmonics giving a more brilliant tone
c. ⇒the fundamental to be coupled more effectively to the air
emphasizing the fundamental and decreasing the length of a note.
d. Marimbas do not have resonators under the bars.
5. The normal primary vibration mode of a tympani is
a. The (0,1) mode that has no nodal lines through the center and a nodal
circle only at the rim. The drum is struck at its center to strongly
excite this mode.
b. ⇒The (1,1) mode that has a single nodal line through the center and a
nodal circle only at the rim. The drum is struck a point somewhere
between the center and the rim to strongly excite this mode.
c. The (0,2) “sombrero” mode that has no nodal lines through the center,
a nodal circle at the rim and another nodal circle between the rim and
the center. The drum is struck at its center to strongly excite this
mode.
6. What instrument has “stretched octaves”? ⇒ piano
The strings have some resistance to bending that implies that the overtones
are not exactly harmonic: the first overtone slightly exceeds a factor of
two. Since the octaves are tuned to be without beats, this implies that the
octaves are very slightly stretched.
7. Which of the following would have the smallest reverberation time?
a. ⇒A room the size of the Physics 198 lab room with walls and ceiling
covered with felt.
b. A room the size of the Physics 198 lab room with walls and ceiling
covered with standard (hard) painted sheetrock.
c. A room the size of the Physics 198 lecture hall with walls and ceiling
covered with felt.
d. A room the size of the Physics 198 lecture hall with walls and ceiling
covered with standard (hard) painted sheetrock.
8. Give one advantage and one disadvantage of increasing absorption to obtain a
short reverberation time. ⇒ A short reverberation time implies that very fast
music is possible and speech should be distinct. However people have less of
a feeling of being “bathed in reverberant sound,” and the sound level will be
lower perhaps requiring amplification.
9. What is the criterion in acoustics for an “intimate” concert hall? ⇒ That the
time between the direct sound and 1 st reflected sound is short, about 20 ms or
less.
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10. A talented flute player blows over the top of a long tube with both ends open
that emits sound. Sketch the standing wave pattern for the first overtone (not
the fundamental!) inside the picture of the tube below. Label the nodes and
antinodes.
11. Assuming that the tube in problem 10 is 0.343 meters long, and that the
diameter is small enough to neglect end effects, what are the frequencies of the
fundamental and first overtone?
⇒ All harmonics are present for an open/open pipe.
f 1 = v/(2L) = 343/(2*.343) = 1000/2 = 500 Hz; f 2 = 1000 Hz.
12. The flutist in problem 10 then covers the far end of the 0.343-meter tube with
her hand and again blows over the tube again, and the open/closed tube emits
sound. Sketch the standing wave pattern for the first overtone (not
fundamental!) inside the picture of the tube below. Label the nodes and
antinodes.
13. Assuming that the open/closed tube in Problem 12 is 0.343 meters long, and
that the diameter is small enough to neglect end effects, what are the
frequencies of the fundamental and first overtone? ? ⇒ Only odd harmonics
are present for an open/closed pipe.
f 1 = v/(4L) = 343/(4*.343) = 1000/4 = 250 Hz; f 3 = 750 Hz.
14. What is the wavelength in meters in air of a wave with frequency of 1029 Hz?
a. 1/4 = 0.25
b. 1/3 = 0.33 ⇒ λ = v/f = 343/1029 = 1/3 = 0.33
c. 1/2 = 0.50
d. 2
e. 3
f. 4
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15. A mass m=0.25 kg is attached to an ideal spring and the mass-spring system
undergoes simple harmonic motion with no friction. If the mass is increased
by a factor of four to 1.0 kg, how will the frequency of oscillation change?
a. It does not change.
b. It goes up (faster oscillations) by a factor of four.
c. It does up (faster oscillations) by a factor of two.
d. ⇒It goes down (slower oscillations) by a factor of two
f = (1/(2π))*√(k/m) so multiplying m by 4 divides f by 2.
16. If the spring-mass simple harmonic system in Problem 13 has a spring
constant k = 10N/m and the mass is pulled 0.1 meters from equilibrium and
released at rest, what is the potential energy in Joules in the system just before
the mass is released?
a. ⇒ 0.05 PE = ½ k x 2 = ½*10*0.1*0.1 = 0.05
b. 0.10
c. 0.50
d. 1.00
e. 5.00
17. A sound pressure pulse is reflected from the closed end of a tube. The
reflected sound pulse is:
a. inverted (positive becomes negative)
b. ⇒ non-inverted (positive remains positive)
c. non-existent (pulse disappears at the tube end)
d. can be either inverted or non-inverted depending on the tube length.
18. The tube between the pinna and the eardrum is:
a. the Eustachian tube.
b. ⇒ the auditory canal.
c. the ossicles.
d. the stapes.
e. the cochlea.
19. In one to three clear sentences, explain the “place” theory of pitch.
⇒ The basilar membrane vibrates at different frequencies along its length, and
the brain knows what part is vibrating from sensor cells firing that have hairs
touching the membrane. Hence the “place” refers to the place on the
membrane that vibrates.
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20. A single concertina player on stage in Bandshell park plays to produce a
sound intensity level of 74 dB at a distance of 24 feet from the stage. How
many identical concertina players playing exactly the same note would
produce an intensity level of 86 dB at the same distance? (Careful!)
a. 2
b. 4
c. 8
d. ⇒16 [doubling gives 3 dB and (86-74)/3=4 so must double 4
times⇒ 2, 4, 8, 16]
e. more concertina players than are in the state of Iowa (perhaps true,
but no credit for this!)
21. If the concertina players now produce 86 dB at 24 feet from the stage what
would be the sound intensity level in dB at 48 feet from the stage?
a. 84 b. 82 c. ⇒80 d. 78 e. 76 f. 74
L I drops 6 dB when distance is doubled: 86 – 6 = 80
22. The most sensitive part of one’s hearing range is nearest which frequency
(Hz).
a. 4 b. 40 c. 400 d. ⇒ 4000 e. 40000
23. The just noticeable difference (jnd) in pitch in the range for a person with a
good ear at an optimal part of the hearing range (1-4 kHz) is approximately:
a. 1/1000 of a half-step
b. ⇒ 1/10 of a half step
c. 1 half step
d. 4 half steps
e.
24. A sawtooth wave at 250 Hz is played together with a squarewave at 101 Hz.
Will there be audible beats in any of the lower harmonics of the two waves?
a. Yes, 1 beat per second at about 202 Hz
b. Yes, 2 beats per second at about 251 Hz
c. Yes, 3 beats per second at about 303 Hz
d. ⇒ Yes, 5 beats per second at about 502 Hz.
e. There will be no audible beats in the harmonics.
Sawtooth has all harmonics: 250, 500, 750, …; Square has odd harmonics: 101,
303, 505,…; so beats of 5 Hz will occur at about (500+505)/2 = 502
25. Two pure (only the fundamental is present) tones seem are dissonant if
a. they differ in frequency by about ¼ of a critical band.
b. they differ in frequency by about 2 critical bands.
c. they differ in frequency by about 4 critical bands.
d. they differ in frequency by more than 4 kHz.
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26. If we choose to tune C 4 to 260 Hz as the first note in a Pythagorean scale, what
would be the frequency of G 4 which is a 5’th higher? The fifth’s are perfect in
Pythagorean scale give a ratio of 3/2. Hence (3/2)*260 = 3*130 =
⇒ 390 Hz.
27. Which tuning/temperament (Pythagorean, mean, just, equal) exactly closes the
circle of 5’ths?
⇒ Equal Temperament
28. Briefly but clearly explain the advantage of multiple pickups on an electric
guitar.
⇒ The anti-nodes for different harmonics occur in different places, and the
anti-node locations also change as frets are used. If several pickups are
used, all harmonic strengths can be sampled and mixed electronically.
29. In lecture, Simon kindly demonstrated a “pedal” tone with his tuba; Nolan
(trombone) and Ross and Chad (trumpets) later played these in labs. Which is
the most accurate description.
a. A pedal tone is the highest peak in the harmonic series of brass
instruments, and is often beyond one’s hearing range in pitch.
b. ⇒ A pedal tone is the lowest note in the harmonic sequence for a
brass instrument. The instrument does not actually have a resonance
there, but, when the instrument is played, the lips can be stabilized at
this frequency by the higher harmonics that do exist in the horn.
c. A pedal tone is obtained by “popping” the mouthpiece giving a
frequency of about 800 Hz.
30. Briefly but clearly explain what Formant Tuning is and where it is used in
music.
⇒ It is possible to align one of the resonant frequencies one of the vocal
cavities (formants) with the frequency of the note being sung. This
can give the sound great power. This is useful at the top end of the
soprano range where the note frequency is above most vowel
formants. I believe that Tom also used it effectively in his
demonstration of “throat singing” in lecture and in lab as well. (The
first sentence is sufficient for full credit.)
End of Exam and End of Course!
Have a Great Winter Break!
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