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 (109Hz) ultrasound generated by electronic stimulation of quartz crystals is used in acoustic microscopy for generating sharp images.
The speed 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 root of tensile force in the string as well as its mass per unit length, the speed of sound waves in a fluid depends on the square root 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 SI unit of B is N/m2. We may mathematically write:
Example 1: The bulk modulus of elasticity of air is 1.41x105 N/m2 and its mass density is 1.29kg/m3. Calculate the speed of sound waves in air.
Solution: Using v = (B/ρ)0.5, we get: v = [1.41x105/1.29]0.5 m/s = 331m/s.
Example 2: The bulk modulus of elasticity of water is 2.1x109 N/m2 and its mass density is 1000 kg/m3. Calculate the speed of sound waves in water.
Solution: Using v = (B/ρ )0.5, we get: v = [2.1x109/1000]0.5 m/s = 1400m/s.
Example 3: A tuning fork is vibrating at 400 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 1200Hz 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 400Hz is
λ = V/f = (342/400.0)m = 0.855m = 85.5cm ; # of λ's = 10.0m/0.855m =11.7
The λ corresponding to 1200Hz is λ = V/f = (342/1200)m = 0.285m = 28.5cm.
# of λ's = 10.0m/0.285m = 35.1
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What is Resonance?
Suppose you are pushing a swing with a kid in it. As it swings at a desired state, it keeps coming to you and going away from you. You normally do your best to give it just a little push exactly at the instant the swing is ready to go away from you. In other words, you try to make your periodic motion (the pushing) not only in phase with the periodic motion of the swing, but also with exactly the same period as the period of the swing. When two oscillatory (periodic) motions have the same period (or frequency) and are also in phase, they add up constructively and create a "resonance state."
The Resonance of Sound Waves in Pipes:
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., their disturbance direction is parallel to their propagation direction. In pipes, sound waves oscillate parallel to the pipe's length. The following is the analysis of what happens to waves at a closed end and at an open end:
1) Closed Pipes:
At the closed end of a closed pipe, only a node can form. Why? The reason is that the air molecules (inside the pipe) that transmit sound waves have to bounce off the closed end after collision. They have to come to stop before bouncing off. Coming to stop means zero state of oscillation at that closed end that results in node formation.
At the open end of a closed pipe; however, both nodes and antinodes are possible to form depending on the wavelength of the sound waves as well as the pipe's length. Why? See Fig. 1 below:
When a pipe has the right length to where an anti-node forms at its open end, it causes the sound waves to amplify (be heard louder) in which case we say: "the oscillations of the air molecules in the pipe are " in resonance " with the oscillations of the blade that is generating them. See Fig. 1 above.
As was mentioned above, a node forms at a closed end, and for resonance there must be an anti-node at the open end. Recall that the distance between any node and the anti-node next to it is λ/4.
On this basis, the first possible length of a closed pipe that can cause 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 pipe lengths for resonance in closed pipes
L1 = 1λ/4
L3 = 3λ/4
L5 = 5λ/4
L7 = 7λ/4
For a closed pipe, resonance occurs when the pipe's length is an odd multiple of λ/4.
2) Open Pipes:
Based on the foregoing discussion, an open pipe is open at both ends; therefore, both ends must form antinodes for the pipe to be "in resonance." Fig. 3 shows such possible waveforms. It is clear that the pipe length must be an even multiple of ( λ/4 ) for resonance.
Possible pipe lengths for resonance in open pipes
L2 = 2λ/4
L4 = 4λ/4
L6 = 6λ/4
L8 = 8λ/4
For an open pipe, resonance occurs when pipe's length is an even multiple of (λ/4 ).
The above discussion allows for an experiment by which the speed of sound can be measured.
The Speed of Sound in a Gas:
The speed of sound waves 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 temperature, pressure, and density. Each dependence has its own formula. We will not discuss this topic any further in this course.
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 40.0cm and two resonances (load sounds) are heard: the first one, at a tube length of 12.5cm, and next, at a tube length of 37.5cm. Calculate the speed of sound at that temperature.
Solution: The smallest length of a closed pipe for resonance is λ/4. The first resonance length of 12.5cm means λ/4 = 12.5cm. The second resonance occurring at a length of 37.5cm means 3λ/4 = 37.5cm. Refer to Fig. 2 for closed pipes. We may write:
λ/4 = 12.5cm ; λ = 4(12.5cm) = 50.0cm. Also,
3λ/4 = 37.5cm ; λ = (4/3)(37.5cm) ; λ = 50.0cm or 0.500m.
Getting the same results should not be surprising because they are both measurements of the same thing.
Knowing λ and the frequency of the waves: f = 686 Hz, the sound speed at that temperature is
v = f λ ; vsound = [686 (1/s)] (0.500m) = 343m/s.
Note that the speed of sound at STP conditions (0oC and 1atm. of pressure) is 331m/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, a greater number of wavelengths per second arrive at our ears than when it is at rest. When the sound source (ambulance) is going away from us, a smaller number of wavelengths arrive at our ears than when the source (ambulance) is at rest. It is possible to calculate the frequency heard in each case.
Five cases may be discussed. Cases (1) and (2) are when the observer is at rest and the source is approaching or receding. Cases (3) and (4) are when the source is at rest and the the observer is approaching or going away. Case (5) is when both observer and source are moving, either approaching or receding. The following formulas apply without proof.
In the following formulas: fo = the frequency heard by the observer, fs is the frequency of the sound source, vs is the speed of the source, vo is the speed of the observer, and v is the speed of sound in the medium.
1) Denom. < Numerator; fo > fs
2) Denom.> Numerator; fo < fs
3) Denom.< Numerator; fo > fs
4) Denom. > Numerator; fo < fs
5) Choose + & - that make sense.
Possible cases discussed below.
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 least, 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: If an ambulance with its siren on at a frequency of 1350Hz is approaching you at a speed of 33.1m/s at STP conditions, calculate (a) the frequency you hear. (b) If you are driving at a speed of 16.55m/s toward the coming ambulance, what frequency do you hear? (c) If you are driving at a speed of 16.55m/s away from the coming ambulance, what frequency do you hear? (d) What frequency do you hear when the ambulance passes your car and continues in front of you? (d) What frequency do you hear if both of you and the ambulance stop?
Solution: (a) Case 1:
fo = fs [(V+Vo)/(V-Vs)] ; fo = 1350Hz[(331+0)/(331-33.1)] = 1500Hz .
(b) Case 5: Source & obs. move to each other: fo= fs[(V+Vo)/(V-Vs)] = 1575Hz .
(c) Case 5: Source chases observer: fo = fs [(V-Vo)/(V-Vs)] = 1425Hz .
(d) Case 5: Observer chases source: fo = fs [(V+Vo)/(V+Vs)] = 1290Hz .
(e) Case 5: Vo = 0 and Vs = 0. ; fo = fs [(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 y1 = A sin (ω1t) and y2 = A sin (ω2t), we may add the two to get the equation of the combined wave as follows: (For simplicity, let the amplitudes be the same).
Y = y1 + y2 = A sin (ω1t) + A sin (ω2t) or,
The result is a wave function of average frequency favg = (f1 + f2 )/2 whose amplitude is modulated at frequency (f1 - f2 )/2. This modulated amplitude gives a variable loudness at a beat frequency of | f1 - f2| . 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 in which case they add up constructively and result in a greater amplitude and therefore a louder sound. There are also instances at which the two waves are completely 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 in Fig. 4 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 f1 = 340Hz while the building's air conditioning system is producing sound waves of dominant frequency f2. 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 | f1 - f2| , we may write: | f1 - f2| = 2.
Either 340 - f2 = 2 or, 340 - f2 = -2. Finally, f2 = (338 or 342)Hz.
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: I = P/(4πr2), we get: I = 125watts/(4π ∙ 20.02 m2) = 0.0249 w/m2.
Intensity Level: The Decibel Scale:
Human ear can hear an intensity range of 10-12 w/m2 to 1w/m2. 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 /Io )
where Io = 10 -12 w/m2 is the threshold of hearing.
If we set I = Io = 10-12 w/m2, we get β = 0 that is the threshold of hearing.
If we set I = 1w/m2, 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 25.0w 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 Io = 10-12 W/m2.
I = P/(4πr2) or, I = 25.0w/(4π ∙ 16.02 m2) = 0.00778 w/m2.
The intensity level β = 10 log (I /Io) = 10 log (0.00778/10-12) = 98.9dB.
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/m2. The power output may be calculated from I = P/(4πr2).
(b) P = I ∙ (4πr2) = 1w/m2 ∙ [4π ∙ (8.9m)2] = 1000 watts.
(c) β = 10 log (I/Io) = 10 log (1/10-12) = 10 ∙ 12 dB = 120 dB
Test Yourself 1:
1) The 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. The frequency of a buzzer is 440Hz. The wavelength of waves coming out of this buzzer at STP 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)1075s-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 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 (by counting) the sound is heard. Light travels roughly one million times faster than sound. Taking the speed of sound to be 300m/s to one significant figure, how far away has the lightning hit? Of course, sound travels at constant speed. (a) 1800m (b) 2100m (c) 2400m. click here.
7) 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 form a maximum 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 there (b) at an open end maximum deviation from equilibrium or an amplitude can occur (c) both a & b.
11) A node and the antinode next to it are (a) λ/2 (b) λ/4 (c) 1/8 λ apart.
12) The shortest length a closed pipe can have to create resonance for wavelength λ is (a) λ/4 (b) λ/2 (c) λ/16. click here.
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 12.5-cm 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 λ/4 can fit an odd number of times in its length. click here.
16) An open pipe can create resonance for (a) all frequencies (b) for all wavelength (c) wavelengths for which λ/2 can fit any number of times in its length.
17) The Doppler effect (a) occurs when a stationary source changes its frequency to be heard by a stationary observer (b) is 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 at rest.
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 at rest. click here.
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 oC?
3) The density of mercury is 13.6g/cm3. Its bulk modulus is B = 2.8x1010Pa. 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 105 Hz. Find the corresponding wavelength in air at 20oC.
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? click here.
7) A siren has 36 holes on the edge of its rotating disc. Air flow from a nozzle into the rorating holes generates sound. 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 pipe 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 oC. Note: Add 0.61R to each length measurement as a correction because of the diameter of the pipe's opening.
9) In an air column resonance experiment at 15oC, 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 to each measurement. 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 31.7oC.
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 25o.
12) An ambulance moving eastward at 108km/h has its alarm on at a frequency of 500Hz. What change in frequency does the driver of a truck traveling west at 90.0km/h notices 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. click here.
15) To what extent does the intensity level decrease if the distance from an isotropic sound source increases by a factor of 10?