SAT Physics Subject Test
Chapter 13 Waves
Sound waves are produced by the vibration of an object, such as your vocal cords, a plucked string, or a jackhammer. The vibrations cause pressure variations in the conducting medium (which can be gas, liquid, or solid), and if the frequency is between 20 Hz and 20,000 Hz, the vibrations may be detected by human ears. The variations in the conducting medium can be positions at which the molecules of the medium are bunched together (where the pressure is above normal), which are called compressions, and positions where the pressure is below normal, calledrarefactions. In the figure below, a vibrating diaphragm sets up a sound wave in an air-filled tube. Each dot represents a great number of air molecules.
A pipe organ exploits the very phenomenon pictured and described above, to generate different notes.
An important difference between sound waves and the waves we’ve been studying on stretched strings is that the molecules of the medium transmitting a sound wave move parallel to the direction of wave propagation, rather than perpendicular to it. For this reason, sound waves are said to be longitudinal. Despite this difference, all of the basic characteristics of a wave—amplitude, wavelength, period, frequency—apply to sound waves as they did for waves on a string. Furthermore, the all-important equation λf = v also holds true. However, because it’s very difficult to draw a picture of a longitudinal wave, instead graph the pressure as a function of position.
The speed of a sound wave depends on the medium through which it travels. In particular, it depends on the density (ρ) and on the bulk modulus (B), a measure of the medium’s response to compression. A medium that is easily compressible, like a gas, has a low bulk modulus; liquids and solids, which are much less easily compressed, have significantly greater bulk modulus values. For this reason, sound generally travels faster through solids than through liquids and faster through liquids than through gases. The equation that gives a sound wave’s speed in terms of ρ and B is
The speed of sound through air can also be written in terms of air’s mean pressure, which depends on its temperature. At room temperature (approximately 20°C) and normal atmospheric pressure, sound travels at 343 m/s. This value increases as air warms or pressure increases.
A sound wave with a frequency of 343 Hz travels through the air.
10. What is its wavelength?
11. If its frequency increased to 686 Hz, what is its wave speed and the wavelength?
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10. Using v = 343 m/s for the speed of sound through air, we find that
11. Unless the ambient pressure or the temperature of the air changed, the speed of sound would not change. Wave speed depends on the characteristics of the medium, not on the frequency, so v would still be 343 m/s. However, a change in frequency would cause a change in wavelength. Since f increased by a factor of 2, λ would decrease by a factor of 2, to (1 m) = 0.5 m.
12. A sound wave traveling through water has a frequency of 500 Hz and a wavelength of 3 m. How fast does sound travel through water?
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v = λf = (3 m)(500 Hz) = 1,500 m/s.
Intensity and Decibel Level
How loud we perceive a sound depends on both frequency and amplitude. Given a fixed frequency, we can say that loudness is measured by intensity.
Intensity: I =
Where P is the power produced by the source and A is the area over which the power is spread. Consider a point source emitting a sound wave in all directions.
At a distance r, the A = 4πr2 (the surface area of the sphere). Therefore, I ∞ If a listener doubles the distance to the source, the sound will be heard one-fourth as loud. An alternate way of measuring loudness is with the decibel level (sometimes called relative intensity).
Decibel Level: β = 10log()
I0 is the threshold of hearing and is equal to 10–12 watts. Note that, while β is measured in decibels (dB), it is dimensionless.
A useful fact to remember is that if a sound increases by 10 dB, the intensity increases by a factor of 10.
13. How many times more intense is an 82 dB sound than a 52 dB sound?
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82 dB is 30 dB louder than 52 dB. Each change of 10 dB increases the intensity by a factor of 10. 30 dB is 3 • 10 so the intensity would be 10 × 10 × 10 = 1,000 times more intense.
If two sound waves whose frequencies are close but not identical interfere, the resulting sound modulates in amplitude, becoming loud, then soft, then loud, then soft. This is due to the fact that as the individual waves travel, they are in phase, then out of phase, then in phase again, and so on. Therefore, by superposition, the waves interfere constructively, then destructively, then constructively. When the waves interfere constructively, the amplitude increases, and the sound is loud; when the waves interfere destructively, the amplitude decreases, and the sound is soft. Each time the waves interfere constructively, producing an increase in sound level, we say that a beat has occurred. The number of beats per second, known as the beat frequency, is equal to the difference between the frequencies of the two combining sound waves.
fbeat = |f1 – f2|
If frequencies f1 and f2 match, then the combined waveform doesn’t waver in amplitude, and no beats are heard. For example, pianos can be tuned using the phenomenon of beats. A key is struck, and the corresponding tuning fork is struck; if the piano string is in tune, there should be no beats as the two sounds interfere. If beats are heard, then the piano tuner tightens or loosens the string and repeats until no beats are heard.
A piano tuner uses a tuning fork to adjust the key that plays the A note above middle C (whose frequency should be 440 Hz). The tuning fork emits a perfect 440 Hz tone. When the tuning fork and the piano key are struck, beats of frequency 3 Hz are heard.
14. What is the frequency of the piano key?
15. If it’s known that the piano key’s frequency is too high, should the piano tuner tighten or loosen the wire inside the piano to tune it?
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14. Since fbeat = 3 Hz, the tuning fork and the piano string are off by 3 Hz. Since the fork emits a tone of 440 Hz, the piano string must emit a tone of either 437 Hz or 443 Hz. Without more information, we can’t decide which one is correct. On the test, only one answer is correct per question; pick the one that you see in the answers.
15. If we know that the frequency of the tone emitted by the out-of-tune string is too high (that is, it’s 443 Hz), we need to find a way to lower the frequency. Remember that the resonant frequencies for a stretched string fixed at both ends are given by the equation fn = nv/2L, and that v = . Since f is too high, v must be too high. To lower v, we must reduce FT. The piano tuner should loosen the string and listen for beats again, adjusting the string until the beats disappear.