Chapter 17
Sound
Sound waves are longitudinal. Our ears can hear the range of frequencies from 20Hz to 20,000Hz. This range is called the audible range. Frequencies above this range is called Ultrasonic that cats and dogs can hear. Ultrasound has medical application. By sending ultra sound wave to internal parts of body and analyzing the reflected sound waves, it is possible to create the image of a desired part by some imaging conversion process. Very high frequency (10^{9}Hz) ultrasound generated by electronic stimulation of quartz crystals is used in acoustic microscopy for generating sharp images.
The motion of sound waves in a gas, such as air, depends on the gas pressure and density. When a person speaks, the vibrations of his or her vocal cords cause air molecules to vibrate as well. Vibrations, at their best, propagate in air on spheres that expand and contract repeatedly transmitting their energy to bigger and bigger spheres thereby transmitting the sound energy via longitudinal waves. The shape of transmission in its ideal case is spherical and the sound energy propagates isotropically throughout the medium. Isotropic means same properties in all directions. We know that this is ideal and if you are standing in front of a speaker you hear it louder than standing at other places around it at the same distance. This means that the sound propagation from a speaker is directional and therefore, anisotropic.
The mechanism is that when a sphere of molecules expands, it creates a slight increase in pressure, ΔP, outward while leaving a slight relative vacuum, ΔP, inward (or behind). This ripple in pressure causes a ripple in density that gets transmitted layer by layer, and the result is the transmission of sound waves.
Speed of Sound
The same way speed of waves in a string depends on the square roots of tensile force in the string and its mass per unit length, the speed of sound waves in a fluid depends on the square roots of the bulk modulus, B, of the fluid as well as its mass density, ρ. The bulk modulus of elasticity of a fluid is a measure of how compressible that fluid is when under a certain pressure. B is defined as the ratio of ΔP over (ΔV / V). This simply calculates the change in pressure as a result of a relative change in volume. The Metric unit of B is N/m^{2}. We may mathematically write:
Example 1:
The bulk modulus of elasticity of air is 1.41x10^{5 }N/m^{2} and its mass density is 1.29kg/m^{3}. Calculate the speed of sound waves in air.
Solution: Using the formula v = (B / ρ )^{0.5}, we get: v = [1.41x10^{5} / 1.29] m/s = 331 m/s. (Speed of sound in air at STP)
Example 2:
The bulk modulus of elasticity of water is 2.1x10^{9 }N/m^{2} and its mass density is 1000 kg/m^{3}. Calculate the speed of sound waves in water.
Solution: Using the formula v = (B / ρ )^{0.5}, we get: v = [2.1x10^{9} / 1000]^{0.5} m/s = 1400 m/s.
Example 3: A tuning fork is vibrating at 400.0 Hz. If you are 32.8ft away from it, how many wavelengths do fit between you and the tuning fork? If a different fork plays a 1200.Hz note, how many wavelengths do fit in the same distance? The speed of sound at Room Temperature is 342 m/s.
Solution: 32.8ft is equivalent to 10.0m. The wavelength corresponding to 400.0Hz is λ = V/f = ( 342 / 400.0)m = 0.855m = 85.5cm
# of λ 's = 10.0m / 0.855m = 11.7
The wavelength corresponding to 1200.0Hz is λ = V/f = ( 342 / 1200.)m = 0.285m = 28.5cm (Nearly 1ft)
# of λ 's = 10.0m / 0.285m = 35.1
A Good Link to Try: http://surendranath.tripod.com/Applets.html .
Motion of Sound Waves in pipes, Resonance:
Here, we will discuss how an organ pipe, for example, plays a certain note.
From music point of view, a pipe, open at both ends, is called an "open pipe". A pipe, closed at one end only, is called a "closed pipe."
Sound waves are longitudinal, i.e., the disturbance is parallel to propagation for them. In pipes sound waves or disturbances travel parallel to the pipes' length. The following is reasonable to think or state:
At a closed end, only a node can form. At an open end, an antinode can form if the pipe's length allows that.
This is true, because the closed end of a pipe is a point at which a wave reflects and wave motion has to momentarily come to stop; therefore, a closed end is at zero state of oscillation and consequently a node.
At an open end, air or gas molecules are free to move back and forth parallel to the pipe's length (similar to a slinky), and therefore can be at some state of oscillation. If a pipe's length allows and the molecules of gas at its open end are at their maximum state of oscillation, an antinode forms. When an antinode forms at an open end, a loud sound can be heard, and the pipe is said to be " in resonance."
Resonance in Closed Pipes:
As was mentioned above, a node forms at a closed end, and for resonance there must be an antinode at the open end. The distance between any node and its neighbor antinode on a wave is always λ/4 as shown below:
On this basis, the first possible length of a closed pipe that can be " in resonance " for a sound wave of wavelength λ is λ/4. The 2nd possible length is not 2λ/4, but 3λ/4, and so forth..., as shown below:
Possible lengths for resonance in closed pipes
L_{1} = 1λ/4 for resonance in a closed pipe. :
L_{3} = 3λ/4 for resonance in a closed pipe.
L_{5} = 5λ/4 for resonance in a closed pipe.
L_{7} = 7λ/4 for resonance in a closed pipe. For a closed pipe, resonance occurs when the pipe's length is an odd multiple of ( λ/4 ). 

Possible lengths for resonance in open pipes
L_{2} = 2λ/4 for resonance in an open pipe.
L_{4} = 4λ/4 for resonance in an open pipe.
L_{6} = 6λ/4 for resonance in an open pipe.
L_{8} = 8λ/4 for resonance in an open pipe.
For an open pipe, resonance occurs when pipe's length is an even multiple of (λ/4 ) or, any multiple of (λ/2 ).


The above discussion allows for an experiment by which the speed of sound can be measured.
The speed of sound in a medium is a function of the physical properties of that medium. The speed of sound in a gas such as air, for example, is a function of the gas pressure and density as was discussed before. Each dependence has its own formula.
Example 4: A tuning fork oscillating at a rate of 686Hz is brought close to the open end of a closed tube in a room at a certain temperature. The tube's length is changed from 0.0 to 40.0 cm. Two resonances (load sounds) are heard, the first one at a tube length of 12.5cm, and the next one at a tube length of 37.5cm. Calculate the speed of sound at that temperature.
Solution: The first resonance occurs where the pipe length is (λ/4). This means that λ/4 = 12.5cm. The second resonance occurs where the pipe length is 3(λ/4). This means that 3λ/4 = 37.5cm.
λ/4 = 12.5 cm ; λ = 4(12.5 cm) = 50.0 cm, or,
3λ/4 = 37.5 cm ; λ = (4/3)(37.5 cm) ; λ = 50.0 cm or 0.500m.
Getting the same results should not be surprising because they are measurements of the same thing. Of course this is ideal. In reality and experiments, we do not get such precise values.
Knowing λ and the frequency of the waves: f = 686 Hz, the sound speed at that temperature is
v = f λ ; v_{sound} = [686 (1/s)] ( 0.500m) = 343 m/s (at that temperature)
Note that the speed of sound at STP conditions (0°C and 1 atm of pressure) is 331 m/s.
The Doppler Effect:
When an ambulance is approaching us, a higher pitch (frequency) sound is heard than when it is going away from us. The reason is that when the sound source is moving toward us, more number of wavelengths per second are received by our ears than when it is stationary. When the sound source is going away from us, lesser number of wavelengths pass by our ears than when it is stationary. It is possible to calculate the frequency heard in each case.
Five cases may be discussed. (1) and (2) are when the observer is stationary and the source is approaching or receding. (3) and (4) are when the source is stationary and the the observer is approaching or going away. (5) is when both observer and source are moving, either approaching or receding. The following formulas apply without proof. For proof refer to your text.
In the following formulas: f_{o} = the frequency heard by the observer, f_{s} is the frequency of the sound source, v_{s} is the speed of the source, v_{o} is the speed of the observer, and v is the speed of sound in the medium.
1) Denom.< Numerator
; therefore, fo > fs
2) Denom.> Numerator ; therefore, fo < fs
3) Denom.< Numerator ; therefore, fo > fs
4) Denom.> Numerator ; therefore, fo < fs
5) Choose (+) and () signs that make sense. Possible cases are discussed below. 
Case 5:
If both the observer and the source are approaching, the highest possible frequency is heard. To make the fraction the greatest, chose the (+) in the numerator and the () in the denominator.
If both the observer and the source are receding, the lowest possible frequency is heard. To make the fraction the smallest possible, chose the () in the numerator and the (+) in the denominator.
If the source is chasing the observer, choose () in the numerator and () in the denominator.
If the observer is chasing the source, choose (+) in the numerator and (+) in the denominator.
Example 5: An ambulance producing sound at a frequency of 1350Hz is approaching a person at a speed of 33.1m/s at STP conditions. (a) What frequency is heard by the person? (b) If the person drives toward the ambulance at a speed of 16.55m/s, what frequency will he hear? (c) If he drives away from it at that speed, what frequency will he hear? (d) What frequency is heard when the ambulance passes the persons moving car in (c)? (d) What frequency is heard if both stop?
Solution: (a) Case 1: f_{o} = f_{s} [ V / (V  V_{s}) ] ; f_{o} = 1350Hz [ 331/( 331  33.1 )] = 1500Hz.
(b) Case 5, source & observer moving toward each other: f_{o} = f_{s}[ (V+Vo)/(VV_{s}) ] =1575 Hz.
(c) Case 5, source chasing the observer: f_{o} = f_{s} [ (VVo)/(VV_{s}) ] = 1425 Hz.
(d) Case 5, observer chasing the source: f_{o} = f_{s} [ (V+Vo)/(V+V_{s}) ] = 1290Hz
(e) Case 5, V_{o} = 0 and V_{s} = 0. ; f_{o} = f_{s} [ (V+0)/(V+0) ] = 1350Hz
Interference in Time: Beats
When two sound waves of close frequencies combine, the result is a periodic variation in the loudness of the sound called " Beats." If the equations of the two interfering sound waves are y_{1} = A sin (ω_{1}t) and y_{2} = A sin (ω_{2}t), we may add the two to get the equation of the combined wave as follows: (For simplicity, let the amplitudes be the same.)
Y_{ } = y_{1 } + y_{2} = A sin (ω_{1}t) + A sin (ω_{2}t), or
The result is a wave function of average frequency f _{avg} = (f_{1} + f_{2} ) /2 whose amplitude is modulated at frequency (f_{1}  f_{2} ) /2. This modulated amplitude gives a variable loudness at a beat frequency of  f_{1}  f_{2}  . The following figure shows two individual waves that slightly differ in frequency and the way they combine to form "beats".
As can be seen, there are instances at which the two waves arrive at a point in phase for which case they add constructively and result in a greater amplitude and therefore a louder sound. There are also instances at which the two waves are out of phase and neutralize each other's effects and result in no sound. During the instances between these two extreme cases, the combined wave has a variable amplitude as shown above that causes the "Beats." phenomenon.
Example 6: A truck is parked by a building and is running on idle generating sound waves of dominant frequency f_{1} = 340Hz while the building's air conditioning system is producing sound waves of dominant frequency f_{2}. The driver of the truck is standing at a point where he hears a variable loudness and realizes that the beats phenomenon is happening. He counts 2 beats per second. Calculate the frequency produced by the air conditioner.
Solution: Since the beats frequency is  f_{1}  f_{2}  , we may write:  f_{1}  f_{2}  = 2.
 340  f_{2}  = 2. or, 340  f_{2} = ±2. Either f_{2} = 338 Hz, or f_{2} =342 Hz.
Sound Intensity:
The intensity I of a sound wave at a point is defined as the energy per second per unit of area arriving at that point normal to the propagation direction. Energy per second means power. We may also say that sound intensity at a point is the arriving power per unit area at that point. Mathematically,
Example 7: Calculate the sound intensity at 20.0m from a sound point source that is rated at 125 watts.
Solution: Using I = P/(4πr^{2}), we get: I = 125watts / (4π ∙ 20.0^{2 }m^{2}) = 0.0249 watts/m^{2}.
Intensity Level: The Decibel Scale:
Human ear can hear an intensity range of 10^{12} W/m^{2} to 1 W/m^{2}. When the intensity I doubles, the loudness sensation does not double. Experiments done by A.G. Bell showed that when the intensity is increased by a factor of almost 10, the loudness sensation doubles. He therefore defined intensity level β as a measure of loudness as:
β = 10 log (I / I_{0}) where I_{0} = 10^{12} W/m^{2 } is the threshold of hearing.
In this formula, if we set I = I_{0 }= 10^{12} W/m^{2}, we get β = 0 that is the threshold of hearing.
If we set I = 1 W/m^{2}, we get β = 120 that is the highest limit for loudness without pain.
Example 8: Calculate the intensity level β at a distance of 16.0m from a 25.0watt sound source that can be approximated as a point source.
Solution: The intensity I at the given distance must be calculated first, and then compared with I_{0} = 10^{12} W/m^{2}.
I = P/(4πr^{2}),or I = 25.0watts / (4π ∙ 16.0^{2 }m^{2}) = 0.00778 watts/m^{2}.
The intensity level β = 10 log (I / I_{0}) = 10 log ( 0.00778 / 10^{12} ) = 10 ∙ 9.89 dB = 98.9 dB
Example 9: At a distance of 8.9m from a sound source, a group of people feel uncomfortable from its loudness. Calculate (a) the intensity of the sound at that distance, (b) the power output of the source if it can be treated as a point source, and (c) the corresponding intensity level at that distance..
Solution: If the intensity is very near or at the threshold of pain, we then have (a) I = 1 W/m^{2}. The power output may be calculated from I = P/(4πr^{2}).
(b) P = I ∙ (4πr^{2}) = 1 W/m^{2} ∙ [ 4π ∙ (8.9m)^{2 }] = 1000 watts
(c) β = 10 log (I / I_{0}) = 10 log ( 1 / 10^{12} ) = 10 ∙ 12 dB = 120 dB
Chapter 17 Test Yourself 1:
1) The formula for wave speed, in general, is (a) V = λ / f (b) V = f λ (c) V = f / λ.
2) Sound waves travel at a speed of 331m/s at STP conditions. The frequency of a buzzer is 440Hz. The wavelength of waves coming out of this buzzer at STP conditions is (a) 75cm. (b) 133cm. (c) 1.33m. click here
3) The wavelength of a certain pitch noise is 32.0cm. The speed of sound at the room temperature where the noise is being made is 344m/s. The frequency of the noise is therefore (a) 575Hz . (b) 1075 s^{1}. (c) 480/s.
4) Our ears are sensitive or can hear frequencies ranging from 20/s to 20,000/s. The speed of sound at the normal comfortable temperature of 25ºC is 346m/s. The wavelengths corresponding to these frequencies are (a) 17.3m & 17.3mm. (b) 16.55m & 16.55mm. (c) 14.2m & 14.2mm. click here
5) An empirical formula that calculates the speed of sound as a function of temperature is V(T) = [ 331 + 0.6T ] m/s where T is the ambient temperature in degrees Celsius. The speed of sound at 25ºC is (a) 334m/s. (b) 349 m/s. (c) 346m/s.
6) On a rainy day a lightning is observed and after 7 seconds the sound is heard. Light travels roughly one million times faster than sound. Taking the speed of sound to be 340m/s, how far away has the lightning hit? (a) 1800m (b) 2100m (c) 2400m. click here
7) Musically, a closed pipe is (a) closed at both ends. (b) closed at one end only. (c) neither a nor b.
8) In a closed pipe, sound waves can have a maximum at (a) the open end. (b) the closed end. (c) both a & b.
9) The reason why a closed end cannot have a maximum (an antinode) at is that (a) at a closed end air molecules are not free to oscillate back and forth to form maximum deviation or an amplitude. (b) a closed end is a harder medium than air in the pipe and waves reflect when they reach (or hit) a hard obstacle. (c) both a & b. click here
10) The reason an open end of a pipe can form an antinode is that (a) air molecules are free to oscillate back and forth at an open end. (b) at an open end maximum deviation from equilibrium or an amplitude can occur. (c) both a & b.
11) The distance between a node and the next antinode is (a) 1/2 λ. (b) 1/4 λ. (c) 1/8 λ. click here
12) The shortest length a closed pipe can have to create resonance for wavelength λ is (a) 1/4 λ. (b) 1/2 λ. (c) 1/16 λ.
13) A 12.0cm closed pipe (a) can form resonance for λ = 50.0cm. (b) can't form resonance for λ = 50.0cm.
14) A 12.5cm closed pipe (a) can form resonance for λ = 50.0cm. (b) can't form resonance for λ = 50.0cm. click here
15) A closed pipe, can create resonance for (a) all frequencies. (b) for all wavelength. (c) only a wavelength for which 1/4 λ can fit an odd number of times in the pipe's length. click here
16) An open pipe, can create resonance for (a) all frequencies. (b) for all wavelength. (c) wavelengths for which 1/2 λ can fit any number of times in the pipe's length.
17) The Doppler effect occurs (a) when a stationary source changes its frequency to be heard by a stationary observer (b) the change in the frequency observed or heard due to the motion of the source or observer or both. (c) both a & b. click here
18) If an observer moves toward a stationary source, the frequency heard will be (a) lower. (b) higher. (c) the same.
19) If a source moves away from a stationary observer, the frequency heard will be (a) lower. (b) higher. (c) the same. click here
20) If an observer follows a source at the same velocity as the source is moving, the frequency heard will be (a) lower. (b) higher. (c) the same.
21) The maximum Doppler effect (maximum positive change in frequency heard or observed) occurs when the source and observer (a) move away from each other. (b) move toward each other. (c) are both stationary.
22) The minimum Doppler effect (maximum negative change in frequency heard or observed) occurs when the source and observer (a) move away from each other. (b) move toward each other. (c) are both stationary. click here
Problems:
1) Verify that in SI, the unit of v in the formula v = (B / ρ )^{0.5} is m/s.
2) If you are standing 5.80m away from an oscillating tuning fork at a frequency of 385Hz, how many wavelengths do fit between the tuning fork and you (a) at STP and (b) at 21.0 °C?
3) The density of mercury is 13.6g/cm^{3}. Use its bulk modulus of elasticity B = 2.8x10^{10} Pa to calculate (a) the speed of longitudinal waves in mercury. (b) At what wavelength do sound waves that have a wavelength of 66.2cm in air (at STP) travel in mercury? Note: when a wave changes medium, its frequency does not change, but its wavelength does. click here
4) The highest frequency of sound waves emitted by bats is about 10^{5} Hz. Find the corresponding wavelength in air at 20°C.
5) The shortest wavelength of sound waves that dogs can hear is about 1.0cm. Calculate the corresponding frequency using a sound speed of 343m/s.
6) An ultrasound equipment uses a sound frequency of 3.75MHz. The speed of sound in human tissue is 1500m/s. Find the corresponding wavelength (a) in meters, and (b) in mm. (c) Is this wavelength short enough to show the details of an organ? c*400
7) A siren has 36 holes on the edge of its rotating disc that air blows from a nozzle into the holes. Calculate the frequency of the sound generated by this siren when turning at 1320rpm.
8) A tuning fork oscillating at a rate of 486Hz is brought to the open end of a closed tube of radius R = 1.25cm. The tube's length can be changed by lowering the water level in it. Two resonances are heard, the first one at a tube length of 17.0cm, and the next one at a tube length of 52.6cm, both measured from the opening. Calculate (a) the speed of sound at that temperature, and (b) The room's temperature using the empirical formula v(T) = { 0.6T + 331}m/s in which v(T) is the speed of sound at T temperature in °C. Note: Add 0.61R to each length measurement as a correction because of the diameter of the pipe.
9) In an air column resonance experiment at 15°C, a closed pipe is used. The 1st resonance occurs at 9.6cm and the 3rd one at 48.0cm both measured from the open end. Find (a) the speed of sound at that temperature, and (b) the frequency of the sound used. Note: To avoid applying a correction as in problem 8, find the difference of the two measurements. This automatically eliminates the correction that needs to be made. click here
10) A straight pipe is 17.50m long and open at both ends. Calculate the lowest 3 frequencies that the pipe can be in resonance with at a temperature of 31.7°C.
11) A police car has an alarm frequency of 520Hz. What frequency and wavelength does a stationary observer measure if the car is (a) approaching at 108km/h and (b) going away at this speed? Assume an air temperature of 25°C.
12) An ambulance moving eastward at 108km/h has its alarm on at a frequency of 500.0Hz. What change in frequency does the driver of a truck traveling west at 90.0km/h notice as the ambulance passes by? Use a sound speed of 340.0m/s.
13) Calculate (a) the sound intensity from a 4.00w isotropic sound source at a distance of 50.0m from it, and (b) the corresponding intensity level.
14) A 0.250w sound source is placed 20.0m from a 0.400w sound source. If both can be treated as perfect point sources, find (a) the point on the line connecting the two sources at which the intensities are equal. (b) calculate the intensity level from each source at that point.
15) To what extent does the intensity level decrease if the distance from an isotropic sound source increases by a factor of 10?