# Lecture14 - Sound Recall: Speed of wave on a...

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Unformatted text preview: Sound Recall: Speed of wave on a string: Sound Waves Chapter 17 Sound = Longitudinal pressure wave in a medium For sound (3D): Linear density, μ (kg/m) Volume density, ρ (kg/m3) Tension, τ (N) Bulk modulus, B (N/m2) Some examples: Air (20OC) 343 m/s Water (20OC) 1482 m/s Aluminum where 6420 m/s Note: ρwater ≈ 1000 ρair Thus we would expect vwater << vair But: Water is almost incompressible, i.e. if we put pressure on water its volume changes liWle: Compressibility, of water is very small. Thus B is very large. Speed of sound in air depends on temperature. For example, in air it depends on temperature in the following way: Here T and T0 are the absolute temperatures. T0 normally taken as 0o C = 273 K. v = v0 T T0 Traveling sound waves As a sound wave travels, the individual elements of the medium oscillate back & forth, crea[ng traveling areas of compression and expansion. What about the pressure in the wave ? Move all by equal amount s no change in pressure Move by diﬀerent amount s compression & expansion! s How much the pressure changes due to changes in volume is given by B. compression for Then: s s s s s Elements of medium: Intensity of sound Intensity = Rate of energy transfer by wave per unit area, We ﬁnd (for deriva[on see page 478, textbook): When a sound wave originates from a central point, the total mechanical energy is conserved, but is spread over a larger & larger area. If the power of the source is Ps , we have: Example If a Roman gladiator, standing at the center of a circular arena (R=30m), yells ader defea[ng his foe with a power = 10 ­6 W, (a) how intense is his yell as heard by an audience member 5m from the edge of the arena ﬂoor, (b) 20m away from the edge? P P 10−6 W 10−6 W I5 = = = = = 6.49 × 10−11 (W / m2 ) 2 2 2 A 4π r 4π (30 + 5) 4π (30 + 5) 10−6 W I 20 = = 3.183 × 10−11 (W / m2 ) 4π (30 + 20)2 Sound level – the decibel scale The range of displacement amplitudes that humans can hear ranges from 10 ­5 to 10 ­11 m, i.e. a factor of 106. Since intensity is propor[onal to the amplitude squared, the intensity of sound we can hear from faintest to loudest diﬀers by a factor of 1012! To deal with such a large range, it is always wise to use a logarithmic scale. i.e. instead of saying: this is 106 = 1,000,000 [mes as big, we just say it is 6 orders of magnitude bigger. For sound, we use a reference intensity, I0 = 10-12 W/m2, near the lower range of human hearing and deﬁne the decibel scale β as: Example: If we had a sound that is 1,000,000=106 [mes the intensity of the faintest audible sound, the decibel number of the sound would be: Example Calculate the sound level (in decibel) of sounds having the following intensi6es: (a)  Loud music = 2.0x10 ­2 W/m2 (b) Noisy classroom = 5.0x10^ ­6 W/m2 (c) Threshold of hearing = 1x10 ­12 W/m2. The Doppler eﬀect You may have no[ced how the siren from a police car or a ﬁre truck changes when it passes you. Technically, that’s the diﬀerence in the frequency you hear – you observe frequency higher when it is coming toward you and lower when it is going away from you than the frequency you would observe if the police car or the ﬁre truck were sta[onary. This apparent change in frequency when the source (or the observer) is moving is called the Doppler eﬀect. Doppler eﬀect is observed for light as well. The Doppler eﬀect contd … : Moving observer and sta7onary source If the observer is moving toward a sta[onary source of sound, he will encounter more cycles of sound waves than he would if he was sta[onary. More cycles per second means for frequency and he thus observes higher frequency than the actual frequency from the source. Let’s ﬁnd the observed frequency in this case. Speed of sound, v’, rela[ve to the observer: v ' = v + v0 (Here, v = the speed of sound rela[ve to a sta[onary observer, i.e., the ‘actual’ speed of sound and v0 = speed of the observer) Now from v = λf, or f = v/λ, the frequency observed by the observer is: v' v + v f '= λ' = λ 0 v + v0 v + v0 = ⇒ f '= f v/ f v We see that the observed frequency increases by a factor of (v+v0)/v. If the observer is moving away from the source, we can similarly calculate the observed frequency to be Note that in this case the observed frequency is less v − v0 f '= f than the actual frequency. v Both of these formulas can be combined in one (where plus sign for the source moving toward and minus for it moving away): f '= v ± v0 f v The Doppler eﬀect contd… : Moving source and sta7onary observer If the source is moving toward a sta[onary observer, he/she will encounter more wavefronts than he/she would with sta[onary source. This means that the observer (A in the ﬁgure below), measures smaller wavelength. In one [me period T, the source moves a distance of vST. This distance is the decrease in the wavelength. Hence the observed wavelength is v ' v vS λ ' = λ − Δλ = λ − vS T ⇒ =− f' f f v ⇒ f '= f v − vS We note that the observed frequency increases by a factor of v/(v ­vS). If the source is moving away from the observer, the observed frequency is given by: f '= v f v + vS Thus in this case the observed frequency is less than the source frequency. As before, the two formulas can be combined together: f '= v f v vS Here, the upper sign for the source moving toward the observer. The Doppler eﬀect contd… : Observer and source both moving What if observer and source both move? Answer: Just combine the formulas. Observer moving Source moving Both moving v ± vO f'= f v + v f'= f = v ± vS v ± vO f '= f v ± vS A convenient rule for choosing the sign (from your textbook): ...
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## This note was uploaded on 11/16/2010 for the course PHY 2170 taught by Professor Blank during the Spring '08 term at Wayne State University.

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