PHY2464 Homework Solutions
Chapter 1
1. v = d/t => d = tv = 60 s x 344 m/s = 20,640 m or 20.6 km
For t = 1 hr, d = 20.6 km/min (from above) x 60 m.hr = 1240 km
For t = 0.002 s,
d = 344 m/s x 2 x 10 ` 3 s = 688 x 10 `3 m = 68.8 cm
4. Given: v = 344 + 0.6 (T – 20) m/s .
This tells us that for every degree Celsius rise in air temperature above 20
degrees, the speed of sound increases by 0.6 m/s.
Thus v = 0.6 m/s per degree Celsius.
Therefore (1.7 m/s)/ [0.6 m/s per degree] = 2.8333 degrees, or about 3 degrees
Celsius. Looking at the formula [or “functional relationship” as Hall calls it], this
says that a temperature change of about 3 degrees Celsius either way (up or down)
will result in a velocity change, and hence a frequency or pitch change of around
½ %.
6. For an estimated chest area S = 1’ x 1’ , this is about 30 cm x 30 cm = 0.09 m2 .
Since PATM = 105 N/m2 = F/S => F = 105 N/m2 x 0.09 m2 = 9000N ;
9000N/500N/person = 18 persons (!)
8. 60 lb/in2 / 14.7 lb/in2 per atmosphere = 4.08 atmospheres or about 4 ATM.
4 ATM x 105 N/m2 per ATM = 4 x 105 N/m2 per tire.
To support a total weight of 800N, we have 400N per tire.
Since p = F/S => S = F/p = 400N/ 4 x 105 N/m2
= 0.001 m2 or about 10 cm2
12. Assuming sinusoidal waves, 0.997 ATM for the rarefactions implies a
pressure of 1.003 ATM for the compressions [1.000 – 0.997 = 0.003]. The
wave amplitude then must be 0.003 x 105 N/m2 = 300 N/m
This is a very loud sound! See the last paragraph above the Summary on p. 13 .
Chapter 2
2. P = 1.f => f = 1/P = 1/ 5 x 10-3 s = 200 Hz
5. v = f = 175 Hz x 2 m = 350 m/s
This differs from 344 m/s; perhaps due to a higher temperature than 20 Celsius, or
maybe due to different composition of the air.
6. = v/fm/s / (1000s) = 0.344 m or 34.4 cm
= 4 m => f = v/ = 344 m/s/ 4m = 86 Hz
13. Amplitude = (3.8 m - 3.2 m)/ 2 = 0.3 m or 30 cm
15. w = mg = 490 N => m = 490 N/ 9.8 m/s2 = 50 kg
Chapter 4.
2. Assume the pillar is 1 foot square, or about 30 cm wide.
Thus for wavelength 30 cm, f = (344 m/s)/ 0.3 m = 1147 Hz (close to ‘treble’C).
Sounds of lower frequency (longer wavelength) will diffract around the pillar and be
heard without difficulty. The shorter wavelengths (frequencies appreciably above 1200
Hz), however, will be ‘bloked’ by the pillar and not heard clearly except perhaps by
reflections.
6. f1~V/v = 0.12 => V = 0.12 v = 012 x 344 m/s = 41.28 m/s…this is about 148
km/hr!
9. If f1 = 172 Hz, then = 344/172 = 2 m. Since path length for both direct and
reflected sound is a multiple of 2 m, we have constructive interference at L.
For f2 = 86 Hz,  = 344/86 = 4m. The direct sound path of 2 m will be oiut of phase
with the 4m reflected path so destructive interference will result.
10. f1 – f2 = 523 Hz – 520 Hz = 2.4 Hz
11. There will be three beat frequencies. If f1 = 440 Hz; f2 = 438 hz, and f3 = 443 Hz,
we get f3 – f1 = 3 Hz; f3 – f2 = 5 Hz; and f1 – f2 = 2 Hz
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Chapter 5.
1. [200J/ 4s]/ 5 m2 = 50W/ 5 m2 = 10 W/ m2
3. Since I -= P/S, P = IS = [ 10 -6 W /m2 ] [7 x 10-5 m2 = 7 x 10-11 W or 70
picowatts
Over an elapsed time of 10 s, E = Pt = 70 pJ/s x 10 s = 700 pJ or 0.7 nanoJoules
5. From Table 5.2, 1000 W/ m2 => 150 dB.
Or calculate it: I/Io = 103/10-12 = 1015 ; SIL = 10 log 1015 = 10 (15) = 150 dB
8. Each added violin increases Io by one more, hence 3 violins => 3Io, etc.
So [using Table 5.1 ] 3 violins => I/Io = 3 => 75 dB + 5 dB = 80 dB ;
5 violins => I/Io = 5 => 75 dB + 7 dB = 82 dB ;
Similarly, 10 violins => 10 => 85 dB
SIL = 95 dB => 75 dB + 20 dB, so how many violins are required to produce this
increase? From Table 5.1, 20 dB => 1/Io = 100, or 100 violins!
Similarly, 105 dB is (105 – 70) = 30 dB difference => I/Io = 1000!!!
[If this seems like a ridiculously high number, remember that 105 dB is much louder
than a full orchestra playing fff which includes the loud brasses!]
14. 115 dB – 75 dB = 40 dB difference => intensity ratio of 10 000, or 104 (Table 5.1)
Since (intensity ratio) = (amplitude ratio)2 , amplitude ratio = (10 000) = 100.
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PHY2464 Homework Chapter 6
2. The pinna measures about 5 cm in height. Frequencies having 5 cm wavelength
and less are: 344m/s/ 0.05 m = 6880 Hz and up. So high frequencies will be
more directed by the pinna. Lows will be diffracted and more difficult to place.
7. 2 dB => I/Io = 1.6, from Table 5.1 . The difference is (1 – 1.6) = 0.6 = 60%
8. 300 Hz/2 = 150 Hz ; (500 Hz) x 4 = 2000 Hz
9. (a) 400 Hz/ 180 Hz = 2.22 ; (b) 7000 Hz/ 400 Hz = 1.75
Frequency ratio (a) is more than an octave; (b) is less than an octave, so (a) will
be perceived as the greater change in pitch.
11. Use the curve in Figure 6.13 and a straightedge or ruler to find the loudness-level
corresponding to the given loudness:
(a) 0.5 sone  ~35 phon
(b) 4 sone  ~ 60 phon
(c) 25 sone  ~80 phon
17. From Figure 6.13, 40 phon is ~ 0.8 sone, and is equivalent to 40 dB SIL at 1 kHz.
Increasing the SIL by 40 dB is a factor of 10,000 and put s the SIL at 80 dB
[factory noise level, see Table 5.2 !]. So the neighbors will complain about the
noise. Note that the extreme bass and treble frequencies will seem even louder!
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Chapter 9
5. In the x-direction there are 2 possible modes: (1) both masses moving together in
the same direction; (2) the masses moving toward & away from each other.
There are two similar modes for the y [up & down] direction and for the z [in &
out of the page] direction. But note that the ‘moving together’ and the ‘opposed’
y and z vibrations will have the same frequencies since the springs are similarly
stretched.
So. 6 modes [2=2=2] with 2 pairs “degenerate”.
10. f1 = 110 Hz [A2]; f2 = 2.76 f1 = 303.6 Hz +. ~Eb4
f3 = 5.4 f1 = 594 Hz => ~D5
f4 = 8.93 f1 = 982.3 Hz => ~B5
Displayed on the treble staff, we have a 7th plus an augmented 5th, both out of
tune…..a ‘clang’.
15. [Remember, strike point = antinode; support or touch point = node]. Striking the bar
at the center eliminates mode 2. To get mode 2, strike the bar ¼ the distance
between the support points from one end. Touching the bar at the center
eliminates mode 1 and .leaves mode 2 undisturbed.
18.
f1, f4, f9, f2
22. See page 173. Damping time = 10 t1/2 = 10 (0.2 s) = 2 seconds
23. Note that the center of the drumhead is a node for most of the normal modes except
for modes 1, 4 and 9. So touching the drumhead at the center gets rid of f1 but
leaves most of the others. See Box 9.1 to verify that f1 is not a desirable mode!
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Chapter 10
1. L = 0.6 m;  = 2L/n . 1 = 2(0.6 m)/ 1 = 1.2 m ; 2 = 2(0.6 m)/ 2 = 0.6 m
3 = 2(0.6 m)/ 3 = 0.4 m
5. fn = n/2L [T/]1/2 so doubling T increases f by [2] 1/2 = 1.414
Each semitone is an increase of [2] 1/12 so [2] 1/2 = { [2] 1/2}}6 or 6 semitones.
7. fn = n/2L [T/]1/2 = 1/ 2 (0.75m) [ 540/0.006]1/2 = 200 Hz
8. fn = n/2L [T/]1/2 => T/ = 4 L2f2 => T = 4 L2f2  = 4 (2m)2 (30 Hz)2 (0.06kg.m) =
864 N
12. The shorter string is stiffer since it has less length over which it can flex. Thus it's
more like a rod and hence has greater inharmonicity.
20. The absence of harmonics near n = 9, 10 indicates that there is at least one node in
this vicinity. Hence it must have been plucked about L/9 from the end.
21. For mode 3, we want nodesat L/3 for either end; thus the antinodes will be L/6 for
ends. So p[luck the string at L/6. To kill f1 and f2, touch it at the f3 node, L/3 from end.
This harmonic is the "twelfth", or (octave + a fifth)
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Chapter 11
5. G3/E1 = 196/41,2 = about 4.75 . So, a double bass 'scaled up' from a violin would
have:
body width 21 cm x 4.75 = 1 m; total length 60 cm x 4.75 = 2.8 m (!!) . You'd need a
huge 'reach' to be able to finger the notes on a 'fiddle' nearly 3 m tall.
7. f1 = 1/2L 2L [T/]1/2 =>  = T/4 L2f2 = 60/ 4(0.35m)2 (440)2 = 0.63 grams/ m
15. The goblet is more likely to be shatterable when empty. Liquid in it would not only
raise the resonant frequency (decreasing the volume; making the walls stiffer) but
would absorb much of the sound energy that would otherwise go into making the walls
vibrate. [To shatter it, the vibration amplitude would have to exceed the elastic limit of
the glass].
20. The ratio between semitones is 21/12 = 1.05946. For a change of one semitone on a
string instrument, since f is proportional to 1/L, L is shortened to L/1.05946 = 0.94387L
[or lengthened to 1.05946 L].
The change in length is (1 - 0.94387) = 0.05613L .
On the violin, at the lower end of the string you'd move your finger 33 cm x (0.05613) =
1.85 cm per semitone, or about a finger width.
On the double bass, however, a semitone change in pitch would require the string to be
shortened/lengthened by 104 cm x (0.05613) = 5.8 cm
Moving up a fourth (5 semitones) in pitch would be a move of 5 x 5.8 cm = nearly 30 cm
before you can go to the next higher string. If the strings were tuned in fifths you'd have
to move 7 x 5.8 = over 40 cm. The instrument would be difficult to play unless you had
exceptionally long reach.
There seems to be some disagreement on the last problem assigned....probably due to my
writing the assignment incorrectly on the blackboard in class! Here are both solutions:
21. The change in pitch from E1 to C1 is 4 semitones. (1.05946)4 = 1.26 .
104 cm x 1.26 = 131 cm. 131 - 104 = 27 cm change in length.
23. f1 = 1/2L 2L [T/]1/2 . Solving this for T, we get T = 4 L2f2  . Keeping L and 
constant, a change in f requires that T2/T1 = (f2/f1)2 .
Thus T2 = T1 (440/415)2 = 1.12T1 . So T was increased by (1.12 - 1) or about 12 % .
Changing the length while maintaining the higher pitch: T2/T1 = [L2ff2/L1f1]2 = 1.195,
which is a change of (1.195 -1) = 20%
If you changed the mass of the string, T2/T1 = f22 2 / f12 1 = f22 (1.11 / f12 1
= 212960/172225 = 1.236. or about 24% .
If all of the changes were made at once, we have to include all of them in the calculation.
The result: the changes are multiplicative, not additive!
So 1.12 x 1.2 x 1.24 = 1.67 which is a net change in tension of 67% !!
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Chapter 12
1. Displacement antinodes correspond to locations where the molecular motion of the air
is greatest, i.e., where the air is least confined. Conversely, pressure antinodes
(maximum pressure!) occur where the air is most confined.
Thus maximum pressure <--> minimum displacement
5. For the open tube, n = 2L/n . So 1 = 17 cm x 2 = 34 cm =>
f1 = (344m/s)/0.34m = 1011 Hz
Similarly, f5 = 344/(34/5) = 5058 Hz
7. Open : fn = nv/2L => L = nv/2f = (344 m/s)/ 2(860) = 0.22m or about 20 cm
closed: fn = (2n - 1)/4L => L = 344/4(860) = 0.1 m or about 10 cm
12. Higher harmonic modes in open pipes weaken rapidly above N = L/D [see text, p.
246, near the top of the page].
So for L= 40 cm and D = 10 cm very few harmonics are likely above N= 40/10 = 4
However, for D = 2 cm , N = L/D = 40/2 = 20....so many more harmonics are likely.
15. We will now have a displacement antinode at the hole, so the pipe acts like an open
pipe with L = 44 cm. Thus f = v/2L = (344m/s)/ 0.88m = 391 Hz [close to G4].
20. Standing wave nodes are /2 apart......review Figures 10.1 and 12.2 if you aren't
sure about this! So an adjoining antinode is halfway between, or /4 away.
22. The 'fat' pipe will have fewer harmonics [see #12, above], hence a 'duller', 'flutier'
timbre. The thin pipe has more harmonics and thus might have some 'stringy' quality.
However, the timbre is affected by several other parameters: height and position of the
pipe mouth;s upper lip, lip thickness, windway opening, wind pressure, etc.
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Chapter 13
3. Clamping the reed with the teeth effectively shortens the vibrating length of the reed,
thus raising its natural vibration frequency.
6. f = (2n -1) v/4L => L = (344 m/s)/4f = 344/(4 x 147) = 0.585 m = 58.5 cm
The actual length is longer: the flaring bell on the end requires additional tube length.
12. fn = (2n - 1) v/4Ln => L2 = 3 (344m/s)/(4 x 116.5m) = 2.215 m = 221.5 cm
For A2, L = (3 x 344)/ (4 x 110) = 2.34 m or 234 cm. So (234 - 222) =12 cm change
in L; 12/2 = 6 cm slide movement.
15. The mute on brass instruments works by limiting air column movement at the bell
(where the sound waves emerge) and hence diminishes the radiation efficiency of the
instrument [see text, pages 286-287]. On the clarinet, except for the very lowest note [all
holes closed] sound waves mostly emerge from the finger holes (displacement
antinodes!). So a mute would only work for the 'bottom' note.
20. The flute has a displacement antinode at the embochure, thus flutes are acoustically
equivlent to tubes open at both ends. On the other hand, the reed at the mouthpiece of the
clarinet creates a pressure antinode, causing this instrument to behave as a closed
pipe....see the discussion in your text, p. 264-265. Furthermore, clarinet reed vibrations
contain more harmonics than the edge tones produced at the flute embochure. Thus the
harmonics of the two instruments will be quite different: the flute sounds a few of the
complete harmonic series (n = 1.2.3) whreeas the clarinet will produce more harmonics,
all of them 'odd' (n = 1.3.5.7.9, etc).
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