Tutorial #5 Solutions

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PHYS 1240: Sound and Music Friday, July 19, 2019

Instructions: Work in groups of 3 or 4 to answer the following questions. Write your solutions on this copy of the tutorial—each person should have their own copy, but make sure you agree on everything as a group. When you’re ﬁnished, keep this copy of your tutorial for reference—no need to turn it in (grades are based on participation, not accuracy).

1. When telephones were ﬁrst being developed by Bell Labs, they realized it would not be feasible to have them detect all frequencies over the human audible range (20 Hz - 20 kHz). Instead, they deﬁned a frequency response so that the highest-pitched sounds telephones could receive is 3400 Hz and lowest-pitched sounds they could detect are at 340 Hz, which saved on costs considerably.

a) Assume the temperature in air is 15◦C. How large are the biggest and smallest wavelengths of sound in air that telephones are able to receive? We can find the wavelength from the formula v = λf, but first we’ll need the speed: v = 331 + (0.6 × 15◦C) = 340 m/s Now, we find the wavelengths for the two ends of the frequency range given: v 340 m/s λ = = = 0.1 m smallest f 3400 Hz v 340 m/s λ = = = 1 m largest f 340 Hz Therefore, perhaps surprisingly, telephones can pick up sound waves up to a meter long, but they won’t detect waves as small as the ear receiver itself. b) Most adult voices have frequencies below 340 Hz. How is it then that voices are able to be transmitted through telephones with such a poor frequency response? The missing fundamental effect—our voices also have upper harmonics that can be transmitted, and our brains then fill in the gaps to perceive of the lowest, fundamental tone even when it’s not physically present.

1 Tutorial #5 Solutions PHYS 1240: Sound and Music

c) Suppose you hear the following frequencies being transmitted through the telephone: 375 Hz, 625 Hz, 875 Hz. If these are all part of some harmonic series, what is the fundamental frequency? To solve this, we must ﬁnd the greatest common factor. The diﬀerence between each frequency is 250 Hz, but this can’t be the fundamental because none of the frequencies listed are integer multiples of 250 Hz. Instead, it must be half this, 125 Hz . That way, 375 Hz is the third harmonic, 625 Hz is the 5th harmonic, and 875 Hz is the 7th harmonic.

2. The Precise Tone Plan for North American telephones has standardized the dial tone so that it is always composed of two sine waves with frequencies of 350 Hz and 440 Hz.

a) Listening to these two tones, it may appear as if there is a third note, with a frequency caused by periodic changes in the sound’s amplitude. What is this frequency, and what is its cause? When two tones are played together, our brain senses a third frequency called a combination tone, given by the diﬀerence between the two real tones (related to the concepts of beats and the missing fundamental):

440 Hz − 350 Hz = 90 Hz

b) Calculate the frequency ratio between the upper and the lower dial tone frequencies, expressing your answer as a decimal. 440 Hz ≈ 1.257 350 Hz

c) Determine about what musical interval this gives. To do so, start with 350 Hz and multiply by 21/12 to go up a half step. Then, multiply by the same factor again, and repeat until you get close to the upper frequency. How many half steps do you ﬁnd, and which musical interval would that be?

(350 Hz) × 21/12 ≈ 370.81 Hz (350 Hz) × (21/12)2 ≈ 392.86 Hz (350 Hz) × (21/12)3 ≈ 416.22 Hz (350 Hz) × (21/12)4 ≈ 440.97 Hz Thus, we need to go up 4 half steps to reach approximately the right interval. 4 half steps is a major third, so this intervals is a little less than a major third . If it were exactly a major third, then the ratio found in part (b) would exactly equal (21/12)4 ≈ 1.260. Tutorial #5 Solutions PHYS 1240: Sound and Music

d) Is the musical interval you found greater than, exactly equal to, or less than the justly-tuned version of that interval? Do you think this would sound dissonant or consonant? A justly-tuned major third has frequency ratio 5/4 = 1.25, and the frequency ratio found in part (b) is greater than this ratio. It would likely sound slightly dissonant, then, since the upper note would produce beats with the lower note’s harmonics at a frequency of about 2.5 Hz (440 Hz - (350 Hz × 5/4)). However, as with equal-tempered tuning, it may be a small enough diﬀerence that we only hear it as a consonant major third.

3. The tones produced when pressing a key on a telephone are standardized as per the table below—two tones are played, one high-frequency tone and one low-frequency tone. The low-frequency component is determined by the key’s row on the telephone, and the high-frequency component by the key’s column, as shown. None of these are pure, consonant intervals, but calculate the frequency ratios for the 4 key, 8 key, and # key to show that they give roughly the same ratio. Are they close to any musical interval? 1209 Hz 1336 Hz 1477 Hz 697 Hz 1 2 3 770 Hz 4 5 6 852 Hz 7 8 9 941 Hz * 0 #

1209 Hz 4 key: ≈ 1.570 770 Hz 1336 Hz 8 key: ≈ 1.568 852 Hz 1477 Hz # key: ≈ 1.570 941 Hz Thus, all these keys along the same diagonal have roughly the same frequency ratio. These don’t represent any consonant musical interval. It seems relatively close to a perfect ﬁfth (1.5), but they are slightly larger, closest to 8 half steps, since (21/12)8 ≈ 1.587 (a minor sixth, if you’re familiar with the music notation). Tutorial #5 Solutions PHYS 1240: Sound and Music

4. For the ﬁnal problem, we will consider a pentatonic scale, consisting of 5 notes per octave.

a) First, find the frequencies of a pentatonic scale with equal temperament. To do this, we can go through the same process as discussed in class, except instead of multiplying by 21/12 for every interval, we now need to multiply by 21/5, so that we’ll reach an octave (2/1) after five notes instead of twelve. Label your notes 1, 2, 3, 4, 5, and 1’ (the first note but an octave higher). Let the frequency of 1 be 261.5 Hz, and calculate the frequencies of the other five notes. 1: 261.5 Hz 2: 261.5 Hz ×(21/5) ≈ 300.38 Hz 3: 261.5 Hz ×(21/5)2 ≈ 345.05 Hz 4: 261.5 Hz ×(21/5)3 ≈ 396.36 Hz 5: 261.5 Hz ×(21/5)4 ≈ 455.30 Hz 1’: 261.5 Hz ×(21/5)5 = 523 Hz

b) The table below shows frequency measurements for two instruments within a gamelan ensemble, the barung and the demung. Both are xylophone/bell-like instruments, so based on the information given in the table, which instrument is larger? This is a simple harmonic motion stiﬀness/mass problem—the larger instrument will be the one with the lower frequencies. Therefore, the demung is larger, since all its frequencies are about an octave lower than the barung. c) Compare the demung’s frequencies to your equal temperament. Are they the same? In particular, is a gamelan’s octave a pure 2/1 ratio, or is it stretched or squished? The frequencies match pretty closely (within a few Hz for each note, but the octaves for a gamelan are stretched (the frequency of note 1’ is greater than double the frequency of note 1). A tuning system like this will have more diﬃculties staying in tune across several octaves, but since each instrument only plays a span of one octave, it still sounds decent and gives the ensemble a unique, exotic sound. Note Barung Demung 1 524.5 Hz 261.5 Hz 2 600.3 Hz 301.5 Hz 3 688 Hz 343.4 Hz 4 803 Hz 399 Hz 5 918.6 Hz 456.5 Hz 1’ 1055.5 Hz 524.5 Hz