Answers to Tutorial No 2,Semester 1, 2013/14

1. When you touch a string which is 120 cm long at acertain distance from its nearer end, it vibrates witha frequency of 1,800 Hz. If the string has a funda-mental frequency of 450 Hz, what is the distance ofyour finger from its nearer end? If the string is nowincreased in length by 20%, what would the vibrat-ing frequency of the string be if it vibrates with yourfinger at a distance of 48 cm from the nearer end ofthe string?Answer: Since 1,800 Hz is 450 Hz times 4, the stringis vibrating at its fourth harmonic. The distance ofyour finger from the nearer end must be one-quarterof the length of the string, which is 120 cm dividedby 4 i.e. 30 cm. If the string is lengthened by 20%,its length will become equal to 120 cm times 1.2 i.e.144 cm. The fundamental frequency of the stringwill now be given by 450 Hz times 120

144 i.e. 375 Hz.If your finger is 48 cm from the string’s nearer end,since 48 cm is one-third of 144 cm, the string willnow be at its third harmonic. Its frequency of vibra-tion will thus be equal to 375 Hz times 3 i.e. 1,125Hz.

2. A string which is vibrating at a frequency of 4,800Hz is observed to have 8 antinodes between its twoends. If a second string which is 120 cm long is ob-served to be vibrating with 5 antinodes between its

two ends at a frequency of 2,400 Hz, what is thelength of the first string vibrating at 4,800 Hz? Ifa vibrating third string is observed to have sevennodes between its two ends and has a length of 150cm, what would be its frequency of vibration? (As-sume that the three strings are similar in all respectsexcept for length.)Answer: The first string has 8 antinodes and isthus at its 8th harmonic, and hence its fundamen-tal frequency is given by 4,800 Hz divided by 8 i.e.600 Hz. The second string has 5 antinodes and isthus at its 5th harmonic, and hence its fundamentalfrequency must be equal to 2,400 Hz divided by 5i.e. 480 Hz. Therefore since the second string has alength of 120 cm, we can deduce that the first stringhas a length given by 120 cm times 480

600 i.e. 96 cm.The third string has a length of 150 cm, and henceits fundamental frequency compared to the secondstring is given by 480 Hz times 120

150 i.e. 384 Hz. Ifthe third string is vibrating with 7 nodes, it must beat its 8th harmonic, and its frequency must be equalto 384 Hz times 8 i.e. 3,072 Hz.

3. Starting from a first musical note, if we go up bythe interval of a Just seventh, we will arrive at asecond note. If we start again from the same firstnote and go up again but now by the interval of aPythagorean seventh, we will arrive at a third note.Which of these two notes arrived at i.e. the secondor third note, has the higher frequency? What isthe ratio of the interval between these two notes? If

the frequency of the first note is 880 Hz, what arethe frequencies of the second and third notes? Ifwe start again from the same first note with a fre-quency of 880 Hz, but go down instead of up by theJust seventh and Pythagorean seventh respectively,what would the frequencies of the second and thirdnotes be?Answer: The Just seventh has a ratio of 15

8 whichis equal to 1.875, while the Pythagorean seventh hasa ratio of 243

128 which is approximately equal to 1.898.Therefore the second note is lower than the thirdnote. If we denote the ratio between the second andthird note as r, we have 15

8 times r equal to 243128 , so

that r is equal to 243128 times 8

15 i.e. 243240 which can be

reduced to 8180 . If the first note has a frequency of

880 Hz, the frequency of the second note is equal to880 Hz times 15

8 i.e. 1,650 Hz, and that of the thirdnote is equal to 880 Hz times 243

128 i.e. 1,670.625 Hz.If we went down by the same ratios, the frequencyof the second would be equal to 880 Hz divided by158 which is the same as 880 Hz times 8

15 i.e. ap-proximately 469.33 Hz. The frequency of the thirdnote would similarly be given by 880 Hz times 128

243

i.e. approximately 463.54 Hz.

4. One of the most frequently used musical scales infolk music is the common pentatonic scale, whichcan be found on any piano keyboard by playing onlythe black notes on the keyboard. This scale is knownas a pentatonic scale as it consists of five notes (notcounting the note one octave above the beginning

of the scale), which are arranged in the followingsequence of intervals: tone, tone, three semitones,tone, followed by three semitones. This gives a totalof 12 semitones so we arrive at the note one octaveabove the starting note. In Balinese gamelan music,one of the most used scales is also a pentatonic scaleof five notes. However, this scale has a different se-quence of intervals, which is: semitone, tone, 2 tones,semitone, 2 tones, giving a total of 12 semitones. Ifyou start from the D just above Middle C as the firstnote, what are the letter names of the notes on thekeyboard making up the notes of these two differentpentatonic scales? Starting instead from the G justbelow Middle C, what are the names of the notesmaking up these two pentatonic scales?Answer: For the common pentatonic scale, if westart from D, the next note is a tone above givingthe note E, then the next note is also a tone abovegiving F sharp. We then go up three semitones toarrive at A, and then a tone above to give B, andthen up another three semitones to arrive at the noteD exactly one octave above the starting D. For theBalinese pentatonic scale, going up a semitone fromthe starting D, the next note is E flat, then goingup a tone gives F, then 2 tones up gives us the noteA, then up by a semitone gives B flat, and up by 2tones arrives at the D one octave above the startingD. Starting from the note G instead, the notes forthe common pentatonic scale would be A, B, D, E,and G. The notes for the Balinese scale would be A

flat, B flat, D, E flat, and G.

5. A violin has its strings tuned in Just fifths as is nor-mally done, with its A string tuned to a frequencyof 440 Hz. A guitar is tuned to the Equal-temperedscale with its A string tuned to 110 Hz. What is theratio of the interval between the guitar’s A4 note andits E5 note? What would the ratios be between theguitar’s A4 note and the same note E5 as played onthe violin’s E string? Calculate the ratios of the in-terval between the violin’s D string and the guitar’sA4 note, and of the interval between the guitar’s noteequivalent to the note on the violin’s D string, andthe guitar’s A4 note. (Take the ratio of an equal-tempered semitone to be equal to 1.05946 for yourcalculations.)Answer: The interval of a semitone in the Equal-tempered scale on the guitar has the ratio of 1.05946.The guitar’s A4 note has the same frequency as theviolin’s A4 note i.e. 440 Hz, since the guitar’s A2string is tuned to 110 Hz. From the guitar’s A4 noteto its E5 note is an interval of an equal-tempered fifthi.e. 7 equal-tempered semitones, and hence the ratio

is given by ( 12√

2)7

i.e. approximately 1.4983. The Estring of the violin is exactly a Just fifth above its Astring, and hence the ratio of the interval from theguitar’s A4 to the violin’s E5 is equal to 3

2 i.e. 1.5.The violin’s D string is a Just fifth below its A string,and hence the ratio between the D string and the Astring is also a fifth with the ratio of 3

2 i.e. 1.5. Theviolin’s D string is the note D4, and on the guitar

D4 is 7 equal-tempered semitones below the guitar’sA4, and hence the ratio is that of an equal-tempered

fifth whose ratio is ( 12√

2)7

i.e. approximately 1.4983.

6. A musical sound has a spectral diagram or spectrumwhich is a graph representing the fundamental fre-quency and harmonics of the sound by vertical lines,with the lengths of the lines represents the ampli-tudes of the harmonics, and their positions on thehorizontal axis represents their frequencies. A cer-tain musical instrument discovered by archeologistscan emit a musical note producing a spectrum whichshows its fundamental frequency and its harmonicsup to the 12th harmonic. There are no missing har-monics in the spectrum, and the 5th line from theleft in the spectrum has a frequency of 3,500 Hz.What are the frequencies of the 3rd, 4th, 7th and9th lines from the left in the spectrum of this in-strument? The spectrum of a square wave has a 4thline from the left with the same frequency as the 6thline from the left in the spectrum of note from themusical instrument. What are the frequencies of the2nd and 5th lines from the left in the spectrum ofthe square wave?Answer: The 5th line from the left in the instru-ment’s spectrum is its 5th harmonic, and hence thefundamental frequency of the note is equal to 3,500Hz divided by 5 i.e. 700 Hz. The 3rd, 4th, 7th and9th lines from the left in the spectrum are the note’s3rd, 4th, 7th and 9th harmonics respectively. There-fore the frequencies of these harmonics are 2,100 Hz,

2,800 Hz, 4,900 Hz and 6,300 Hz respectively. Thespectrum of a square wave has only odd harmonics,so the 4th line from the left in its spectrum is the7th harmonic. The 6th line from the left of the in-strument’s spectrum is its 6th harmonic, which hasa frequency of 700 Hz times 6 i.e. 4,200 Hz. If thisis equal to the 7th harmonic of the square wave, itsfundamental frequency must be equal to 4,200 Hzdivided by 7 i.e. 600 Hz. Its 2nd and 5th lines fromthe left in its spectrum are its 3rd and 9th harmon-ics, which will have the frequencies 600 Hz times 3i.e. 1,800 Hz, and 600 Hz times 9 i.e. 5,400 Hz re-spectively.