test1a14a

8/12/2019 test1a14a

1/9

THE SCIENCE OF MUSIC (PC1327/GEK1519)

Mid-Term Class Test, Semester 1, 2013/14

This is an open book test. The test is one hour long.Give your answers to ALL 25 questions on the computer-readable sheet provided, using a soft (2B)

pencil to shade the appropriate choice for each question.

1. Which of the following is the least appropriate example of a scientific activity?

(a) A clarinet player investigating how the thickness of her clarinet reed affects the loudnessof a certain musical note for the same blowing effort.

(b) A ukulele player finding out how the material out of which the top of the ukulele is madeaffects the quality or timbre of the musical notes produced.

(c) A mathematician playing a Chopin waltz on a new type of electronic piano which uses anovel method of sound generation.

(d) A biomedical scientist studying the behaviour of the muscles on a pianists arm when thepianist is playing a note on a piano keyboard.

Answer: (c) The clarinet player, ukulele player and biomedical scientist are all performingessentially scientific activities. The mathematician is the only one performing an essentiallymusical activity.

2. Which of the following is the most appropriate example of a technological activity?

(a) An engineer doing a jazz improvisation on her electronic organ.

(b) A percussion player designing and constructing a new type of bass drum which can befolded into a small package for portability.

(c) A guitarist performing in a heavy metal band on an electric guitar using a new kind ofspecial effects pedal to alter the sound.

(d) A trombone player rehearsing on a new kind of trombone which has both a slide andvalves.

Answer: (b) The engineer, guitarist and trombone player are all performing essentially musicalactivities. The percussion player is the only one performing an essentially technological activity.

3. Which of the following objects is the most appropriate example of a vibrating object?

(a) A womans hand patting her babys back just once.

(b) A coconut which falls from a coconut tree onto a patch of thick grass and lies still withoutbouncing or rolling.

(c) A students leg as the student shakes it from side to side repeatedly during a boring lecture.

(d) A child sliding down a slide in a playground just once.

Answer: (c) The womans hand, coconut and child are all undergoing a single motion withoutrepetition and hence cannot be said to be undergoing a vibration. Only the students leg isundergoing a repetitive motion and hence can be said to be undergoing a vibration.

4. Two flagpoles which are side by side are observed to be swaying from left to right repeatedlyduring a strong wind. The shorter flagpole undergoes 6 complete left to right movements orcycles in the same duration during which the taller flagpole undergoes 5 complete cycles. Ifthe taller flagpole undergoes 4 complete cycles in 7 seconds, which of the following is closestto the time the shorter flagpole would take to undergo 8 cycles?

1

8/12/2019 test1a14a

2/9

(a) 23.34 seconds.

(b) 11.67 seconds.

(c) 8.75 seconds.

(d) 6.56 seconds.

Answer: (b) The taller flagpole undergoes 4 complete cycles in 7 seconds, and hence the timewhich it takes to undergo 5 cycles is equal to 7 seconds times 5

4 i.e. 35

4seconds. This is the same

duration during which the shorter flagpole undergoes 6 cycles, and hence the shorter flagpolewould take 354

seconds divided by 6 i.e. 3524

seconds to complete one cycle. The time in whichthe shorter flagpole undergoes 8 cycles is thus equal to 8 times 35

24seconds i.e. 35

3 seconds i.e.

approximately 11.67 seconds.

5. An architect designs a house, and the plans for the house prepared by the architect are given toa contractor who builds the actual house. A businessman sees the house when it is completedand buys it for his family, who then live in the house. A composer composes a symphony bywriting the musical score of the symphony, and an orchestra performs the symphony from themusical score for an appreciative audience at a concert. What has the same relationship to thearchitect as the audience has to the composer?

(a) The plans for the house.(b) The contractor.

(c) The house itself.

(d) The family living in the house.

Answer: (d) The audience listens to and enjoys the symphony which is the product of thecomposers creativity and hence the family is analogous to the audience, since the family livesin the house and experiences the product of the architects creativity.

6. A group of friends at a birthday party is singing Happy Birthday as the birthday girl isblowing the candles on her birthday cake. Just then, a latecomer rushes into the room, and

hearing the singing, starts to sing too as she enters the room in which the party is being held.In order to produce counterpoint with the rest of her friends at the party, what song shouldshe not sing as she enters the room?

(a) Happy Birthday

(b) We are Singapore.

(c) Jingle Bells.

(d) Majulah Singapura.

Answer: (a) To ensure that counterpoint is produced, the latecomer should sing a differentmelody from the one which the rest of her friends are singing. She should thus not sing Happy

Birthday.7. At a student concert, the first item is a rock band which registers a reading of 94 dB on a

sound level meter which you are carrying with you as you sit as a member of the audience.The next item is a choir whose sound power is 100 times less powerful than that produced bythe band. What would you expect the reading on the sound level meter due to the choir tobe? (Assume that the sound level meter readings are due only to the rock band or the choir.)

(a) 114 dB.

(b) 94 dB.

2

8/12/2019 test1a14a

3/9

(c) 84 dB.

(d) None of the above.

Answer: (d) A decrease in sound power of 10 times would result in a decrease in the soundlevel meter reading of 10 dB and hence a decrease in sound power of 100 times, which is thesame as a decrease of 10 times 10 times in sound power, would result in a decrease in thesound level meter reading of 10 dB plus 10 dB i.e. a decrease of 20 dB. The sound level meterreading when the choir is performing is thus 20 dB less than that due to the rock band, which

is equal to 94 dB minus 20 dB i.e. 74 dB.

8. The sopranino recorder is the member of the recorder family with the highest pitch range andthe highest note it can play has a frequency of approximately 6,257 Hz. The sub contrabassrecorder is a member of the recorder and the lowest note it can play has a frequency ofapproximately 66 Hz. How many octaves are there from the lowest note of the sub contrabassrecorder to the highest note of the sopranino recorder, whose frequencies are as given above?

(a) Greater than 4 octaves but less than 5 octaves.

(b) Greater than 5 octaves but less than 6 octaves.

(c) Greater than 6 octaves but less than 7 octaves.

(d) Greater than 7 octaves but less than 8 octaves.

Answer: (c) The octave has a ratio of 21

i.e. 2. Multiplying 66 Hz by 2, we obtain 132Hz. Multiplying by 2 again arrives at 264 Hz and another multiplication by 2 gives 528 Hz.Multiplying by 2 for a 4th time gives us 1,056 Hz, and a fifth multiplication by 2 arrives at2,112 Hz. Multiplying by 2 a sixth time gives 4,224 Hz, and by a seventh time finally arrivesat 8,448 Hz which is higher than 6,257 Hz. Therefore from 66 Hz to 6,257 Hz is an intervalwhich spans more than 6 octaves but less than 7 octaves.

9. A piano which is tuned correctly to Equal-tempered tuning has its A4 note tuned to a frequencyof 442 Hz. If a certain note on this piano has a frequency of approximately 1,984.5 Hz, whichof the following notes on this piano is most likely to be this note? (Take the ratio of anEqual-tempered semitone to be equal to 1.05946 for your calculations.)

(a) A6.

(b) B6.

(c) C7.


Answer: (b) The A4 on the piano has a frequency of 442 Hz, and therefore A5 has a frequencydouble this i.e. 884 Hz, and A6 has a frequency double this again, i.e. 1,768 Hz. If we multiply1,768 Hz by 1.05946, we are going up one Equal-tempered semitone from A6 to arrive atAsharp6 which has a frequency of approximately 1,873.1 Hz. Going up one more semitone andmultiplying by 1.05946 again gives us a frequency of approximately 1,984.5 Hz which is thefrequency of B6. If we were to multiply this frequency by 1.05946 again to go up by anothersemitone to arrive at the note C7, this would have a frequency of approximately 2,102.5 Hz.Hence the note is most likely to be B6.

10. A musical score of a tune for a flute has a time signature at its start of 11/8. A certain bar ofthis tune starts with a quaver rest and ends with a dotted quaver and 3 semiquavers. Of thefollowing combinations of notes, which one would fit exactly into the middle of this particularbar?

3

8/12/2019 test1a14a

4/9

(a) 7 semiquavers and 2 crotchets.

(b) A dotted crotchet and 8 semiquavers.

(c) 10 semiquavers and a dotted quaver.


Answer: (b) The tune has a time signature of 11/8, indicating that each bar of the tune has11 beats, with each beat being a quaver duration. Each bar of the tune must therefore have the

duration equivalent of 11 quavers or 22 semiquavers. At the start of the bar there is already aquaver rest equivalent to 2 semiquavers, and at the end of the bar are a dotted quaver equivalentto 3 semiquavers, and 3 semiquavers. Hence the bar already has a total of 8 semiquavers, andthe middle of the bar needs the equivalent of 14 semiquavers. A dotted crotchet is equivalentto 6 semiquavers, and with 8 semiquavers makes up a total of 14 semiquavers.

11. You start from the note E2 on the keyboard of a normal grand piano and then move upwardsby an interval of an Equal-tempered seventh to reach a second note. Starting from this secondnote, you then move upwards by an interval of one-third of an octave to arrive at a thirdnote. Starting yet again from this third note, you move downwards by the interval of anEqual-tempered fourth to arrive at the fourth and last note. Which of the following is thecorrect letter name of this fourth and last note? (Assume that any interval named is an Equal-tempered interval which occurs on the Equal-tempered piano keyboard between Middle C andanother note of the C scale.)

(a) B2.

(b) Csharp3.

(c) Dsharp3.


Answer: (d) An Equal-tempered seventh consists of 11 Equal-tempered semitones, and hencethe second note is 11 semitones above E2 i.e. Dsharp3. 12 semitones make up one octave andthus one-third of an octave is 4 semitones. From Dsharp3, a move upwards by 4 semitoneswould arrive at the third note G3. Finally, moving downwards by an Equal-tempered fourthor 5 semitones from G3 brings you to the fourth and last note which is the note D3.

12. A solo oboist is playing from the musical score of a melody which consists of just a singlemusical staff with a treble clef at its start. A particular bar of this melody has its first notewritten on the lowest of the five lines of the staff, its second note on the second lowest of thefour spaces of the staff, its third note on the second highest line of the five lines of the staff,and its fourth note on the highest of the four spaces of the staff. What are the letter names,in the correct sequence from first to last, of these four notes starting with the first note?

(a) E, A, C and E.

(b) E, B, D and E.(c) E, A, D and F.


Answer: (d) The treble clef shows that the note G is the second lowest line of the staff.Therefore the lowest of the five lines is the note E, the second lowest of the four spaces is thenote A, the second highest line is the note D, and the highest space is the note E.

4

8/12/2019 test1a14a

5/9

13. A guitar is tuned in accordance with the Equal-tempered scale with its A2 string at a frequencyof 110 Hz. A violin has its A4 note tuned exactly two octaves above the guitars A2 string,and all its strings tuned correctly in Just fifths. Which of the following is closest to the ratioof the interval between the note from the lowest open string on the violin and the note fromthe highest open string on the guitar? Open string means that the notes are not played witha finger on the violins fingerboard or the guitars fretboard, i.e. with the full lengths of therespective strings vibrating. (You may take the ratio of an Equal-tempered semitone to beequal to 1.05946 for your calculations.)

(a) 1.500.

(b) 1.591.

(c) 1.686.

(d) 1.786.

Answer: (c) The lowest string on the violin is its G3 string which is exactly two Just fifthsbelow its A string. Hence the frequency of this G string is 440 Hz divided by 3

2twice, which is

the same as 440 Hz times 49

whose value equals approximately 195.56 Hz. The highest stringon the guitar is its E4 string, which is 5 Equal-tempered semitones below the guitars A4note whose frequency is 440 Hz (since A2 is 110 Hz). Hence the frequency of the E string isgiven by 440 Hz divided by 1.05946 5 times, i.e approximately 329.63 Hz. The ratio is thusapproximately equal to 329.63 Hz divided by 195.56 Hz i.e. approximately 1.686.

14. You start from a first note and then go down by a Just third to reach a second note. From thesecond note you now go up, but this time by an interval of a Pythagorean sixth, arriving at athird note. What is the ratio of the interval between the first note and the third note?

(a) 43

.

(b) 13564

.

(c) 2720

.

(d) None of the above.Answer: (c) The ratio of a Just third is 5

4, and the ratio of a Pythagorean sixth is 27

16. The

ratio of the interval between the first and the third note is obtained by firstly dividing by 54

and then multiplying by 2716

, which is equivalent to multiplying 45

by 2716

. Therefore the ratio ofthe interval between the first and third note is 108

80, which we can reduce to the ratio 27

20.

15. On placing your finger on a vibrating string which has a length of 150 cm at a distance of 30cm from the nearer end of the string, you discover that it vibrates at a frequency of 2,250 Hz.What is the fundamental frequency of the string?

(a) 450 Hz.

(b) 562.5 Hz.(c) 375 Hz.


Answer: (a) Since your finger is 30 cm from the nearer end of the string which is 150 cm long,and 30 cm is one-fifth of 150 cm, the string must be vibrating at its 5th harmonic. Thereforethe fundamental frequency of the string is equal to 2,250 Hz divided by 5 i.e. 450 Hz.

5

8/12/2019 test1a14a

6/9

16. A string which is 120 cm long is vibrating with 4 antinodes between its two ends at a frequencyof 1,100 Hz. A second string which has 7 antinodes between its two ends is vibrating at afrequency of 1,540 Hz. What is the length of the second string which is vibrating with 7antinodes? (Assume that the two strings are identical in all respects except length.)

(a) 140 cm.

(b) 96 cm.

(c) 75 cm.


Answer: (d) The first string has 4 antinodes and is thus vibrating at its 4th harmonic, andhence its fundamental frequency is given by 1,100 Hz divided by 4 i.e. 275 Hz. The secondstring has 7 antinodes and is thus at its 7th harmonic. Therefore its fundamental frequency isgiven by 1,540 Hz divided by 7 i.e. 220 Hz. The length of the second string is thus equal to120 cm times 275

220 i.e. 150 cm.

17. The nodes and antinodes on a vibrating string which is 160 cm long are such that the distancebetween adjacent nodes on this string is 20 cm. When you place your finger on a second stringwhich is 240 cm long at a distance of 40 cm from its nearer end, this string vibrates at a

frequency of 1,920 Hz. What is the frequency of vibration of the first string which is 160 cmlong? (Assume that the two strings are identical in all respects except length.)

(a) 2,880 Hz.

(b) 3,360 Hz.

(c) 3,840 Hz.


Answer: (c) For the second string, since 40 cm is one-sixth of 240 cm, this string is at itssixth harmonic and hence its fundamental frequency is given by 1,920 Hz divided by 6 i.e. 320Hz. The fundamental frequency of the first string which is 160 cm long is thus equal to 320 Hz

times 240

160 i.e 480 Hz. Since the distance between adjacent nodes is 20 cm, which is one-eighthof 160 cm, this string is at its eighth harmonic, and its frequency of vibration is given by 480Hz times 8 i.e. 3,840 Hz.

18. A string labelled A which is vibrating at a frequency of 3,150 Hz has 6 antinodes between itstwo ends. Another string labelled B is 25% longer than the string A, and is cut into 5 piecesof equal length. 3 of these 5 pieces are joined up to make a third string labelled C. Which ofthe following is the frequency of C when it is vibrating with 5 nodes between its two ends (notcounting the nodes at both ends)? (Assume that the strings are identical in all respects exceptlength.)

(a) 1,512 Hz.

(b) 2,100 Hz.(c) 3,500 Hz.


Answer: (d) The string A has 6 antinodes and is thus at its 6th harmonic, and its fundamentalfrequency is equal to 3,150 Hz divided by 6 i.e. 525 Hz. The string B is 25% longer than A,and hence its fundamental frequency is given by 525 Hz times 1

1.25 i.e. 420 Hz. The string C

has a length which is three-fifths that of B, and therefore its fundamental frequency is equalto 420 Hz times 5

3 i.e. 700 Hz. When C is vibrating with 5 nodes, it will have 6 antinodes and

6

8/12/2019 test1a14a

7/9

hence will be at its 6th harmonic. Its frequency will thus be equal to 700 Hz times 6 i.e. 4,200Hz.

19. An open pipe which has 3 nodes between its two ends is vibrating with the same frequency asa string which has 4 nodes between its two ends (not counting the nodes at both ends). Theopen pipe is then made into a closed pipe of the same length by closing one end of the openpipe. If the fundamental frequency of the string is 390 Hz, what is the frequency of the closedpipe as it vibrates with 3 nodes between its two ends (not counting the node at one end)?

(a) 1,950 Hz.

(b) 2,275 Hz.

(c) 3,250 Hz.


Answer: (b) The string has 4 nodes and hence has 5 antinodes, and is thus at its 5th harmonic.If its fundamental frequency is 390 Hz, its 5th harmonic frequency will be given by 390 Hztimes 5 i.e. 1,950 Hz. This is the same frequency as that of the open pipe with 3 nodes andhence at its 3rd harmonic, so the fundamental frequency of the open pipe is equal to 1,950 Hzdivided by 3 i.e. 650 Hz. The closed pipe with the same length as the open pipe will have a

fundamental frequency half that of the open pipe which is 650 Hz divided by 2 i.e. 325 Hz.When the closed pipe vibrates with 3 nodes, it will be at its 7th harmonic, and its frequencywill be equal to 325 Hz times 7 i.e. 2,275 Hz.

20. The spectrum of a string vibrating at a fundamental frequency of 180 Hz shows all its harmonicsup to the 11th harmonic. The spectrum of a vibrating closed pipe also shows its fundamentalfrequency and all its harmonics up to its 11th harmonic. The 6th line from the left in thespectrum of the string has the same frequency as the 5th line from the left in the spectrumof the closed pipe. What is the frequency of the 6th line from the left in the spectrum of theclosed pipe? (Assume that the frequencies in each spectrum increase from left to right.)

(a) 1,320 Hz.

(b) 1,220 Hz.

(c) 720 Hz.


Answer: (a) The 6th line from the left in the spectrum of the string is its 6th harmonic whichwill have a frequency of 180 Hz times 6 i.e. 1,080 Hz. The 5th line in the spectrum of theclosed pipe is its 9th harmonic, as it only has odd harmonics. If this 9th harmonic also hasa frequency of 1,080 Hz, the fundamental frequency of the closed pipe is given by 1,080 Hzdivided by 9 i.e. 120 Hz. The 6th line in the closed pipes spectrum is its 11th harmonic,whose frequency is thus equal to 120 Hz times 11 i.e. 1,320 Hz.

21. A certain folk music string instrument emits a musical note which has a spectrum showing allits harmonics up to the 12th harmonic. On comparing the spectrum of this instrument withthe spectrum of a square wave, you notice that the 4th line from the left in the square wavespectrum has the same frequency as the 3rd line from the left in the spectrum of the musicalinstruments note. If the fundamental frequency of the musical instruments note is 420 Hz,what is the fundamental frequency of the square wave? (Assume that the frequencies in eachspectrum increase from left to right.)

(a) 120 Hz.

7

8/12/2019 test1a14a

8/9

(b) 180 Hz.

(c) 315 Hz.


Answer: (b) The 3rd line from the left in the musical instruments spectrum is its 3rdharmonic, which will have a frequency of 420 Hz times 3 i.e. 1,260 Hz. The square wavedoes not have even harmonics, and hence the 4th line from the left in its spectrum is its 7thharmonic. If this 7th harmonic has a frequency of 1,260 Hz too, the fundamental frequency of

the square wave will be given by 1,260 Hz divided by 7 i.e. 180 Hz.

22. A closed pipe is vibrating with 5 nodes between its two ends (not counting the node at oneend) at a frequency of 1,650 Hz. This pipe is then cut into 7 pieces to make one short closedpipe and 6 short open pipes of equal length. 4 of the short open pipes are then joined togetherto make a longer open pipe. What is the frequency of this longer open pipe when it vibrateswith 4 nodes between its two ends?

(a) 1,050 Hz.

(b) 2,000 Hz.

(c) 4,725 Hz.

(d) None of the above.Answer: (d) The closed pipe has 5 nodes and is thus at its 11th harmonic, and hence itsfundamental frequency is equal to 1,650 Hz divided by 11 i.e. 150 Hz. An open pipe of thesame length would have a fundamental frequency twice this i.e. 300 Hz. The longer open pipecreated from the 4 short open pipes is four-sevenths of the length of the closed pipe, and hencewould have a fundamental frequency given by 300 Hz times 7

4 i.e. 525 Hz. When this open

pipe has 4 nodes, it will be at its 4th harmonic and its frequency will be equal to 525 Hz times4 i.e. 2,100 Hz.

23. Which of the following gives the correct sequence of the following pipes arranged in increasingorder of their frequencies?

(i) The seventh harmonic frequency of a closed pipe of length s cm.(ii) The eleventh harmonic frequency of a closed pipe of length 7s4

cm.(iii) The sixth harmonic frequency of an open pipe of length 5s

3 cm.

(iv) The seventh harmonic frequency of an open pipe of length 9s4

cm.

(a) (ii), (iv), (i), (iii).

(b) (iv), (ii), (i), (iii).

(c) (iv), (ii), (iii), (i).


Answer: (b) Let fHz be the fundamental frequency of a closed pipe of length s cm. Itsseventh harmonic frequency is therefore equal to 7f Hz. A closed pipe of length 7s

4 cm will

have a fundamental frequency given by f Hz times 4s7s

i.e. 4f7

Hz, and its 11th harmonic will

have a frequency given by 4f7

Hz times 11 i.e. 44f7

Hz. A closed pipe of length 5s3

cm will have

a fundamental frequency equal to f Hz times 3s5s

i.e. 3f5 , which means that an open pipe of

length 5s3

cm will have a fundamental frequency double this i.e. 6f5

Hz. Its sixth harmonic

frequency is hence given by 6f5


Hz. Finally an open pipe of length 9s4

cm

will have a fundamental frequency given by 6f5

Hz times 5s3

divided by 9s4

i.e. 120f135

Hz which

can be reduced to 8f9

Hz. Its seventh harmonic frequency is given by 8f9


Hz. The pipes in order of increasing frequency are therefore (iv), (ii), (i) and (iii).

8

8/12/2019 test1a14a

9/9

24. A violinist whose violin strings are properly tuned in Just fifths, with its A string at 440 Hz,plays a musical note which causes a sound wave to travel from her violin to a member of theaudience at a speed of 330 metres per second. If the musical note is one octave above thefrequency of the E string of her violin, what would be the wavelength of this sound wave?

(a) 0.25 metres.

(b) 0.375 metres.

(c) 0.5 metres.


Answer: (a) The E string of the violin is exactly a Just fifth above the A string, and henceits frequency is equal to 440 Hz times 3

2 i.e. 660 Hz. The note one octave above this would

have a frequency equal to 660 Hz times 2 i.e. 1,320 Hz. This is also the frequency of the soundwave which has a speed of 330 metres per second, and hence the wavelength of the sound wavemust be equal to 330 metres per second divided by 1,320 Hz i.e. 0.25 metres.

25. Standing between you and a high vertical cliff is a hunter with a shotgun, such that the linefrom you to the hunter is perpendicular to the cliff. The hunter fires a shot from his shotgun,and 0.2 seconds after you see the flash of the shotgun blast, you hear the sound of the shotgun

which reaches you directly. If the distance between the hunter and the cliff is 74.25 metres,how long after you see the flash will you hear the echo of the shotgun from the cliff? (Assumethat sound travels at a speed of 330 metres per second, and that you see the shotgun flash atthe same time as when the shotgun is fired.)

(a) 0.20 seconds.

(b) 0.45 seconds.

(c) 0.65 seconds.


Answer: (c) Since the direct sound of the shotgun takes 0.2 seconds to reach you, the distance

from the hunter to you is equal to 330 metres per second times 0.2 seconds i.e. 66 metres. Thedistance between the hunter and the cliff is 74.25 metres, so the echo has to travel backwardsby 74.25 metres, then forwards by 74.25 metres plus 66 metres to reach you, a total of 214.5metres. The time taken by the echo is thus equal to 214.5 metres divided by 330 metres persecond i.e. 0.65 seconds.

END OF TEST PAPER

9

test1a14a

Documents