Physics of Music

Physics 341

Assignment 3

1) In the graph of dB versus frequency,

i) What frequencies correspond to -30dB?

The -30 dB level is given by the dotted line below. The frequencies which correspond to these can be read from the two vertical lines, and correspond to roughly 150 Hz and 1300 Hz. What pitches correspond to these frequencies?

There would be two ways of determining the pitches. The easiest would be to read them off the graph. Remember that the scale along the top is the semitones with the major marks corresponding to the various octaves of A (A440 is the one marked 4). In this case the mark is roughly 4.5 to 5 semitones above A2. 5 semitones correspond a perfect fourth, so this would be just less than a fourth above A2, which would thus lie between C3# and D3, probably closer to D3. (Remember that the naming of the octaves switches at C not A. Thus the notes go A2, B2, C3, D3,...)
The higher note at 1300Hz is above A5 and is just above 7 semitones above A5. This is just over a perfect fifth higher than A5. It would correspond to the note a bit sharp of E6.

Figure 1: Figure 1

ii)What dB correspond to 1500 Hz? 200Hz? 7000Hz?
1500Hz= -31 dB
200Hz= -27dB
7000Hz= -46.5dB

Note: This curve is the "resonance response curve" of an oscillator whose resonant frequency is 440Hz. Ie, it is a plot of the energy in the oscillator if it is driven by the same amplitude of force at various frequencies. Note that far from the resonance the response falls at 6dB per octave in both directions. This is typical of all resonances.

2) In graph 2, Add the two waves to get the composite wave.

Below are the added waves. Note that the key feature is that the sudden changes in the middle graph are preserved in the sum graph ( the solid lines in the lower graph ) The temptation is to take too few points at which to sample the sum, and then connect them with a smooth curve. However if one of the graphs is not smooth, then the composite graph will in general also not be smooth.

Figure 2

3) In graph 3, estimate which what the highest harmonic which would be neededto make up the complex wave form?

In estimating the highest harmonic, that highest harmonic would have to have at least the same number of peaks that the graph has. This one has four peaks, and thus the highest harmonic would have to be at least four. However, we notice that there is more structure than just four peaks. At time .15 there is the hint of another pek. At time .5 there is again the hint that that broad peak could be resolved into two peaks. This would suggest that perhaps the sixth harmonic is present. Looking at the sharpness of the structures, it would seem to be consistant that the sixth harmonic is present-- trying to fit a sin wave to any of the features, it would seems that a period of just less than .2 sec would be consistant with the period of that shortest wave.

Figure 3

4)In graph 4 of the "spectrum" ( the Fourier components of a sound), which harmonics of the fundamental are present?

In the graph I have labeled all of the interger multiples of the first (low) peak. Note that every one of the peaks in the graphs falls on teh location of one of these multiples. This confirms that this is the fundamental at 100 Hz. The peaks fall at the 1st, the second, the fourth, the seventh and the nineth multiples of this lowest peak, implying that the harmonics 1 2 4 7 and 9 of the 100Hz fundamental are present. Note that if the 100Hz peak had not been there, I would first have tried 200 Hz as the fundamental. However the peak at 700Hz and 900Hz would not have fit as multiples of this 200Hz. They would howver fit as the multiples of 100Hz, implying that 100Hz would still be regarded as the fundamental of this tone, even if no power at 100 Hz were present. The ear would probably also hear this note as having the pich corresponding to this 100Hz.

Figure 4

5) Sound travels at 340 m/sec in air at room temperature. What would be the wavelengths of sounds of i) 26Hz ii)110Hz iii)1760 Hz iv) 10 KHz ?

In each of these cases you can either remember directly that the wavelength is the distance which sound travels in one period, and calculate the period for each of these frequencies, and finally calculate the distance sound would have travelled in that time. for example, a freqency of 26Hz has a period of 1/26 second. Multiplying this time by 340 m/s one would get 9.4 meters.
Alternatively one could remember that the wavelength is equal to the velocity of sound divided by the frequency. Thus the wavelength correspoinding to 110 Hz would be
340/110= 3.1m
1760Hz has a wavelength of 340/1760=.193m=19.3cm
10KHz=10000Hz has a wavelength of 340/10000= .034m=3.4cm

6) How much louder (in dB) is a choir of 80 voices than i)a single singer? ii) two singers iii) 10 singers.

A choir of 80 voices would be 80 times as loud as 1 singer, 40 times as loud as 2, and 8 times as lound as 10. Ro convert these into dB, remember that a factor of 10 corresponds to 10dB and a factor of 2 corresponds to 3 dB.
This 80times is 8 times 10 times as loud. The 10 times corresponds to 10dB, the 8, which is 2 times 2 times 2 correspondes to 3 + 3 + 3 dB= 9dB. Thus 80 would correspond to 10+9=19 dB louder.
40 is 10 times 4 which is 10dB plus (3+3)dB = 16dB louder.
8 times louder is (3+3+3)=9dB louder.

Assume that each singer produces the same energy output in sound while singing. How much closer (outdoors) would you have to be to the 10 voices so that they would be as loud as the 80 voices?

The 80 voice choir is 8 times as loud as the 10 voice choir. For each doubling of distance the sound gets 4 times softer. In this case you would need to be about just under 3 times closer to the 10 voice choir so it would sound as loud as a 10 voice choir. ( 3 squared is 9, so it would need to be slightly less than 3. The more exact answer would be 2.8 times closer.)

7)A string is plucked at a point 3/5 of the way from one end. Estimate the strength of vibration of the first 5 harmonics of the string. How would this compare to plucking the string at a distance 1/2 of the way from an end?

If you pluck a string at a node of a certain mode, that mode will not be excited. Since the modes of the nth mode occur at the multiples of 1/n along the string, the point 3/5 along would be a node of the 5th mode. Thus the 5th mode would not sound at all. For the first mode, which has no nodes, (except at the ends) the 3.5 point is close to the maximum of vibration of that mode. Thus the first mode will be strongly excited. The second mode has a node at the 1/2 way point. The point 3/5 along is reasonably close to this node (closer than to the antinode-- the point of greatest vibration-- at 3/4 along) and thus the second mode will be less strongly excited than the first mode. Similarly, the point 3.5 along will be closer to the node, at 2/3 of the third mode, than to the antinode at .5 of that mode. Again, the excitation will be less than the that of the first mode. The fourth mode has its nearest nodes at .5 and .75, and the plucking point at .6 is quite near the point midway between those nodes at .625, which would be an antinode. Thus the fourth mode would be excited as strongly as the first mode was. Ie, the strongest exctiation would be of the first and fourth, and then the second and third would be less excited, and finally the fifth would not be excited at all.