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Dive deep into the world of drum modes, energy transfer, and sound resonance, covering topics like sonograms, tuning forks, sphere energy spread, and decibel measurement. Understand the complex relationship between vibrating objects and resonant frequencies.
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Lecture 17 October 29, 2004
The Drum • Each of these modes are usually excited. • The tension of the drum determines the frequency of each mode. • The modes may NOT be harmonic • Each mode dies out at a different rate. • The player can change the basic “tone” of the drum by changing the tension of the drum head.
Modes • All modes are excited at first strike. • These vibrations may excite others … resonance. • Each mode decays in a different time. Amplitude time
So … back to the tuning forks RESONANCE !
Objects will resonate when • They are in contact with something that vibrates at its resonant frequency. • Buzz in cars is a good example • Sound can cause resonance if it is at a frequency that is the resonant frequency of another object nearby. • It must have enough energy.
The Trip TRAVEL AND ROOM ACOUSTICS CREATION Of Sound
Sound Travels ENERGY
Something is Missing • As time progresses, the amount of energy received by the ear increases. • We need a measure of energy per unit time. ENERGY
Example = The Light Bulb • Consider a 60 Watt light bulb. • It requires 60 Joules of energy each second. • One Joule = 1 Newton Meter • Joule=1( N-m)x (1 lb/4.45N) x (3.28ft/m) • Joule=0.738 ft-lbs
Thinking about light bulbs • 60 Joules = 44.2 ft-lbs • Lift a ~4 pound weight 10 ft. or about one story of a building. • Do this every second for 60 watts. Joule=0.738 ft-lbs A 100 lb woman would have to run up about 2 floors of a building per second to generate this much power!
Energy Spreads Out These AREAS increase with r2. Power per unit area therefore DECREASES with r2.
Let’s go to a concert. 50 Meters 30 watts Ear Canal ~ 0.5 cm = 0.005 m Area = (0.005)2=0.000025 m2
Houston we have another problem 30 watts, 50 meters Ear Area = (0.005)2=0.000025 m2 How much energy gets to the ear?
To the ear …. Area of Sphere =pr2 =3.14 x 50 x 50 = 7850 m2 50m Ear Area = 0.000025 m2 30 watt
In the ear… How do we deal with all of these zeros???
Answer: Scientific NotationChapter 1 in Bolemon, Appendix 2 in Johnston 0.000000095 watts = 9.5 x 10-8 watts
NOTE • 10a/10b=10a-b • Example • 1000/10=103/101=10(3-1)=102=100 • 10000/0.005=104/5 x 10-3=(1/5)x10(4-(-3)) =(1/5)x 107 =(10/5) x 106 = 2,000,000 • You can actually get used to doing it this way! But you probably won’t!
Decibels - dB • The decibel (dB) is used to measure sound level, but it is also widely used in electronics, signals and communication.
Suppose we have two loudspeakers, the first playing a sound with power P1, and another playing a louder version of the same sound with power P2, but everything else (how far away, frequency) kept the same. The difference in decibels between the two is defined to be 10 log (P2/P1) dB where the log is to base 10. Decibel continued (dB) ?
What the **#& is a logarithm? • Bindell’s definition: • Take a big number … like 23094800394 • Round it to one digit: 20000000000 • Count the number of zeros … 10 • The log of this number is about equal to the number of zeros … 10. • Actual answer is 10.3 • Good enough for us!
Back to the definition of dB: 10 log (P2/P1) • The dB is proportional to the LOG10 of a ratio of intensities. • Let’s take P1=Threshold Level of Hearing which is 10-12 watts/m2 • Take P2=P=The power level we are interested in.
An example: • The threshold of pain is 1 w/m2
The sensitivity range for human hearing depends on the loudness and pitch. Noises along each black line would be heard with the same volume.