PHYS207 Set 2
Solutions Spring 2012

Chapter 7 RQ 11,12,13,15 Q 1,3 Ex. 1

RQ11: no

RQ12: yes, if the tone is not steady

RQ13: all of these (vibrato of amplitude is often called "tremolo")

RQ15: any of these tricks will do it

Q1: It sounds like a chord because each of the three tones will have its own harmonic structure

Q3: AP is valuable especially for conductors, and it can be a nuisance for listeners.

Recall the concert pianist in V.C. study of medication influencing pitch perception ...

Ex.1: Divide the string at 2/3 of 65 cm = 43.3 cm

Chapter 8 Q 3 Ex. 7 (why is Fig. 8.11 a bad example?)

Q3: 1) set A4(440) so that there are no beats with the tuning fork

2) set E5 to 660 Hz by eliminating beats between 3rd harmonics of 440

and 2nd harmonics of 660 (i.e. at f = 1320 Hz)

3) temper E5 to 659 Hz by RELAXING the tension until you get 2 beats per second

Ex.7: a) Table 9.2 => f(C6) - f(Bb5) =
1046.5
- 932.33 = 114 Hz f(Bb2) = 116

b) The melody
is "God save the King". The example is bad because the solution is
obvious even without the difference tones (just looking at the score
...).

Chapter 11 E1

L2/L1 = f1/f2 = twelveth root of 2 = 1.06 => L2 = L1*f1/f2 =
275*1.06
= 291.5

this means that the u-shaped slide has to be moved by (291.5-275)/2
= 8.25 cm.

Chapter 12 E5

a) f=v/2L = 343/(2*0.42) = 408 Hz

b) F4 is 4 semitones below A4 which is 440 Hz in standard modern
tuning.

So the f(F4) = 440/(twelveth root of two^4) = 440/(cube root of two)
= 349 Hz

**Chapter 23 Q 2, 3 E2,3,4,8**

E2:
(note: data for occupied wooden seats are not given, so we take
occupied
upholstered seats. The difference is not large (e.g. absorption of
persons
depends on whether they have coats or not, etc. ......)

40m=depth 20m=width 15=height

plywood walls: A=2*20*15*0.17 =102

plaster side walls and ceiling A=(2*40*15 + 40*20)*0.06 = 120

wood floor A=40*20*0.10 = 80

1100 wooden seats*0.02 = 22

550 wooden seats*0.02 +550 occupied seats*0.56 = 319

1100 occupied seats *0.56 = 616

RT(empty) = .161*40*20*15/(102+120+80+22) = 6 sec

RT(half full) = .161*40*20*15/(102+120+80+319) = 3.1 sec

RT(full) = .161*40*20*15/(102+120+80+616) = 2.1 sec

E3:
arrival time of direct sound = 20/344 = 58 ms

first reflection from ceiling = 2*sqrt(10^2+15^2)/344 = 105 ms

first reflection from side = 2*sqrt(10^2+10^2)/344 = 82 (24 ms after
direct sound)

E4:
plaster sides and new ceiling: A = 2*40*15*.06 +40*20*.76 = 680 - this
replaces the value of 120 in calculations of RT in E2 above

RT(full) = .161*40*20*15/(102+680+80+319) = 1.6 sec

E8: when absorption of surfaces can be neglected, the RT becomes

RT = .161 V/mV = .161/m independent of V

for air at 20 C 30% m = 0.136 at 8 kHz => RT =
.161/.136
= 1.2 sec

Additional Problems:

f2/f1 = 2^(cents/1200) = 2^(10/1200) = 1.0058

What will be the beat frequency between the two
fundamentals if the
note is

a) A1 (55 Hz) f2 = 55 Hz*1.0058 = 55.3
=> beat frequency will be .3 Hz

b) A6 (1760 Hz) f2 = 1760*1.0058 = 1770 => beats will be at 10 Hz

c) The piano tuner reduced the beat rate in case b)
to 2 Hz.

What is now the mistuning (in cents) ?

cents = 3968log(1762/1760) = 2 cents

2) a) In a church, you notice a pipe about as tall as you are, producing sound of fundamental frequency 100 Hz.

Is it an open-open or open-closed pipe? Explain your reasoning.

For open/open pipe f1 = v/2L => L = v/2f1 = 340 / 2/100 = 1.7 m. That is indeed about as tall as I am (open/closed pipe would have to be 1.7/2 = 0.85 m). So it is an open/open pipe.

b) If the organ builder now changes the end of the pipe from open to closed - or from closed to open - depending on what you found in a), what will now be the new fundamental frequency, and what will be the frequency of the first overtone?

If one end is now closed, the fundamental will drop to 50 Hz, and the first overtone will be 3*50 = 150 Hz.

3) Compare the intervals contained in chords C4/E4/G4 and D4/F4/A4.

What is different? (Note: there is more to this than meets the eye ...).

Each chord contains a major third, minor third and a fifth. The difference is in the order of the thirds. The first chord is call "major", and it sounds more consonant than the "minor" chord because the "major-third-followed-by minor-third" occurs in the harmonic series, but the inversed order does not.

4) Determine the (main) beat rate in an equally tempered Major 3rd if the bottom tone has frequency of a) 220 Hz b) 1760 Hz

In a Major 3rd, the main beats are between the 5th harmonic of the bottom tone, and 4th harmonic of the top tone.

An equally tempered Major 3rd has a frequency ratio of 2^(400/1200) = 1.25992 (instead of 5/4 = 1.25 )

Therefore, the beat rate = 5*f1 - 4*f1*1.25922 = f1*(5-4*1.25992) = f1*0.04

So for f1 = 220 Hz the beat rate will be 220*.04 = 9 Hz

and for f1 = 1760 Hz we get 1760*.04 = 70 Hz!

5) Determine what happens if you go

a) up 12 fifths then down 7 octaves

in Equal: 12*700 - 7*1200 = 0 cents

in just: f(end) = f(start)*(3/2)^12/(2^7) = 1.0136 which means that the end is 3986log1.0136 = 24 cents higher than the start (Equal temperament divides these 24 cents equally among all the 12 fifths ...)

b) up 3 Major thirds then down 4 minor thirds (this is the
basis of "Chaloupka Opus 0.5 we demonstrated in class ...)

in Equal: 3*300-4*300 = 0 cents

un just: f(end) = f(start)*(5/4)^3/(6/5)^4 = f(start) /1.062 => the
difference is

3986log1.062 = 104 cents!

6) Construct the deCaus tuning scheme

(start with perfect major thirds F-A-C#, then build three perfect

fifths upward from each of these three notes) and compare with equal

temperament.

procedure:

- first calculate all frequencies relative to the F. Going up by a major third means factor of 5/4, going up by a fifth means a factor of 3/2.

- calculate the positions of all 12 notes of the
chromatic scale:

cents = 3986 log(freq. ratio).

NOTE: before taking the logarithm, if the frequency ratio is larger
than 2, divide by 2 until it is between (1 and 2) to get back to within
the original octave.

NOTE alternatively, just ADD 702 cents for each 5th,
and ADD 386 cents for each Major 3rd. When you get over 1200, just
subtract 1200.

- now for each of the 12 major chords, plot the (width of the fifth - 702 cents) vs. (width of the major third - 386 cents) to get the Chaloupka Fingerprint (TM) of this particular tuning

notation used in the Table:

note name / frequency ratio relative to F / cents relative to F

F 1 0 | A 5/4 386 | C# 5/4 5/4 773 |

C 3/2 702 | E 5/4 3/2 1088 | G# 5/4 5/4 3/2 275 |

G 3/2 3/2 204 | B 5/4 3/2 3/2 590 | D# 5/4 5/4 3/2 3/2 976 |

D 3/2 3/2 3/2 906 | F# 5/4 3/2 3/2 3/2 92 | A# 5/4 5/4 3/2 3/2 3/2 478 |

and results are: (compared with Equal Temperament,
with chromatic
step
size added ...)

major chord (compare to equal temperament for ANY chord:) |
defect of fifth
(-2) |
defect of 3rd(M)
(14) |
chromatic step to next
(100) |

C | 0 | 0 | 71 |

C# | 0 | 41 | 133 |

D | -22 | 0 | 70 |

D# | 0 | 41 | 112 |

E | 0 | 0 | 112 |

F | 0 | 0 | 92 |

F# | -22 | 0 | 112 |

G | 0 | 0 | 71 |

G# | 0 | 41 | 111 |

A | 0 | 0 | 92 |

A# | 20 | 41 | 112 |

B | 0 | 0 | 112 |

(to calculate the "defects" = (actual - just), you must add 1200 if the top note is below the bottom note ....)

Note that 6 chords are perfect, at the price of the
other 6 being
awful
(especially A#). Notice also that the chromatic scale is limping!

7) A cathedral of volume 24,000 m^3 has RT=5 sec.

Determine the Reverberation Time at 500 Hz after 1,200 m^2

of painted concrete floor is covered with carpet (on pad).

RT = .161 V/A = 5 sec => A = .161 V/RT =
.161*24000/5 = 773

new A = 773 + 1200*(.57 - .06) = 1385 => new RT = .161*24000/1385 =
2.8 sec

(Note: this is much more precise than calculating RT from scratch. It can also be turned around to calculate the absorption coefficient (of the carpet, in this case) from measured values of RT before and after ... Try this: "before painted concrete was covered, RT was 5 sec, after it was covered with carpet (different from above), RT became 3.2 sec. What is the absorption coefficient of that carpet? )