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

1. Two flagpoles are standing next to each other, andduring a thunderstorm, the two flagpoles begin sway-ing from side to side repeatedly. The shorter flagpoleis observed to sway such that it undergoes 5 cyclesin 8 seconds. You also observe that the taller flag-

pole undergoes 3 cycles in the same time which theshorter flagpole undergoes 4 cycles. What are thefrequencies of the two flagpoles? If the thunderstormbecomes more violent such that the frequency of thetaller flagpole increases to 0.75 Hz, what would thefrequency of the shorter flagpole be, if its frequencyincreases in the same proportion as that of the tallerflagpole?

Answer: The shorter flagpole undergoes 5 cycles in8 seconds, which means that it has a frequency of5

8 Hz i.e. 0.625 Hz. The shorter flagpole would un-

dergo 4 cycles in a duration given by 8 seconds times4

5i.e. 6.4 seconds, which is the same duration taken

by the longer flagpole to undergo 3 cycles. Hencethe frequency of the longer flagpole is given by 3

6.4

Hz i.e. 0.46875 Hz. If the frequency of the longer

flagpole increases to 0.75 Hz, then the frequency ofthe shorter flagpole would be equal to 0.625 Hz times0.75

0.46875 i.e. 1 Hz.

2. A bassoonist plays a musical note on her bassoonwhich has a frequency of 55 Hz. A piccolo player then

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plays a note on his piccolo with a frequency of 3,520

Hz. How many octaves are there between these twonotes? If an oboeist then plays a note on her oboewith a frequency of 440 Hz, how many octaves arethere between the oboe note and the piccolos andbassoons note? If a clarinet plays a note which is twooctaves above the bassoon note, what is its frequencyand how many octaves is this note below the piccolosnote? If the piccolos note had a frequency of 3,600

Hz, and all the other notes still maintained the samerelationship to the piccolos note, what would theirfrequencies be?Answer: Since going up an octave means a doublingof the frequency, if we go up from the bassoons noteby an octave six times, i.e. multiply 55 Hz by 2 sixtimes, we will obtain the piccolos frequency of 3,520Hz. Hence the bassoons note is 6 octaves below thepiccolos note. If the oboes note has a frequency of440 Hz, this is 3 octaves above the bassoons notesince 55 Hz times 2 three times gives a frequency of440 Hz. Thus the oboe note is 3 octaves below thepiccolos note. The clarinets note is 2 octaves abovethe bassoons note and its frequency is thus given by55 Hz times 2 times 2 i.e. 220 Hz, and is 4 octavesbelow the piccolos note. If the piccolo note has afrequency of 3,600 Hz, the bassoons note which is

6 octaves below would have a frequency of 3,600 Hzdivided by 26 i.e. 56.25 Hz. The clarinet note whichis 2 octaves above the bassoon note would have afrequency of 56.25 Hz times 4 i.e. 225 Hz, and theoboe note which is a further octave up would have a

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frequency of 225 Hz times 2 i.e. 450 Hz.

3. During a parade, a marching band is 50 metres awayfrom you, and its sound registers a reading of 57 dBon a sound level meter which you are carrying. Whenthe band is in front of you, the sound power reachingyou from the band is 1,000 times more powerful thanwhen it was 50 metres away. When the band marchesaway from you and is quite a distance away, it reg-isters a reading of 47 dB on the sound level meter.

What is the reading on the sound level meter whenthe band is right in front of you? How much lesspowerful is the sound from the band which reachesyou when its sound registers 47 dB on the sound levelmeter, as compared to when it was in front of you?(Assume that the sound level meter reading is dueonly to the marching band.)Answer: An increase in sound power of 10 times

results in an increase in the sound level meter read-ing of 10 dB, and hence an increase in sound powerof 1,000 times or an increase of 10 times 10 times10 times will result in an increase in the sound levelmeter reading by 10 dB plus 10 dB plus 10 dB i.e. 30dB. Thus when the band is in front of you, the read-ing on the sound level meter should be 30 dB higherthan when it was 50 metres away which means it

should be equal to 57 dB plus 30 dB i.e. 87 dB.When the band registers 47 dB, this is 40 dB lessthan when it was in front of you. Hence the soundreaching you is 10 times 10 times 10 times 10 timesi.e. 10,000 times less powerful than when the band

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was in front of you.

4. A melody for the flute has a musical score with atime signature of 14/16 at its beginning. The addi-tion of a dot to a note or a rest increases the durationof the rest or note by 50%. In one of the bars of thismelody, the bar starts with a dotted quaver and asemiquaver rest, and ends with 4 semiquavers. De-termine the number of semiquaver notes which couldfit exactly into the middle of the bar so that the bar

contains a combination of rests and notes which cor-responds with the time signature of 14/16. If theinitial quaver is not dotted, how many semiquaverswould fit in the middle of the bar? If the time signa-ture was 20/16 instead of 14/16, and the initial qua-ver was not dotted, how many semiquavers would fitinto the middle?Answer: If the score has a time signature of 14/16,

this indicates that each bar of the melody shouldbe filled with the duration equivalent of 14 semiqua-vers. The bar is already filled with a dotted quaverwhich is equivalent to 3 semiquavers, a semiquaverrest which is equivalent to 1 semiquaver, and 4 semi-quavers, making a total of 8 semiquavers. Hence themiddle of the bar needs another 6 semiquavers. Ifthe initial quaver is not dotted, it would be equiva-

lent to 2 semiquavers, so that the bar would alreadyhave 7 instead of 8 semiquavers. The middle of thebar would thus need 7 semiquavers. If the time sig-nature is 20/16, this means that the bar should havea total of 20 semiquavers, and the middle of the bar

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requires 20 minus 7 i.e. 13 semiquavers.

5. On a standard piano keyboard, the interval from anykey to its immediate neighbouring key is always asemitone. An octave is the interval from any key tothe next key above or below it having the same let-ter name (i.e. A, B, C etc), and is always made upof 12 semitones. Starting from the piano key withthe letter name of F which is just below Middle Con a piano keyboard, how many semitones are there

from this F to the key with the letter name of A justabove Middle C? What is the letter name of the keywhich is an identical number of semitones below theF? Determine how many octaves there are from thislower note below the F to the A just above MiddleC.Answer: From the F just below Middle C to theA just above Middle C, you can count a total of 16

semitones between them. Going down by 16 semi-tones from the F, you will reach a note with theletter name of D flat or C sharp. From this noteto the A just above Middle C, there is a total of 32semitones. Since an octave is exactly 12 semitones,this is the same as 2 octaves plus 8 semitones or 2and two-thirds of an octave.

6. We define the ratio of a musical interval from a mu-

sical note to another musical note of a higher pitchas the ratio of the frequency of the higher note to thefrequency of the lower note. Starting from a musicalnote with a frequency of 640 Hz, if we go down byan interval with a ratio of 16

9, what is the frequency

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of the note on which we will arrive? If we now go up

from this second note by an interval with a ratio of19

8, what is the frequency of the third note on which

we will arrive? Calculate the ratio of the intervalbetween the first note and the third note.Answer: Going down from the first note with afrequency of 640 Hz, we have to divide 640 Hz bya ratio of 16

9 , which is the same as multiplying 640

Hz by 916

, which gives us a frequency of 360 Hz. To

go up from 360 Hz with an interval of ratio 19

8, wehave to multiply 360 Hz by 198

which gives 855 Hz.The interval from 640 Hz to 855 Hz is given by 855

640

which can be simplified to 171128

. This ratio can alsobe obtained by dividing 19

8 by 16

9.