Chapter 16

Waves and Sound

A wave is the motion of a disturbance in a medium.  The medium for ocean waves is water, for example.  When a string, fixed at both ends, is given a vertical hit by a stick, a dent appears in it that travels along the string.  When it reaches an end point, it reflects and inverts and travels toward the other end.  The following figure shows the motion of a single disturbance.

If you hold end A of the above string and try to give it a continuous up-and-down motion, with a little adjustment of the pace of oscillations, you can make at least the following waveforms:

Each  wave travels from A to B and reflects at B.  When each reflected wave reaches point A, it reflects again and the process repeats.  Of course, the hand motion keeps putting energy into the system by constantly generating waves that are in phase with the returned waves creating the above waveforms.  Although the waves appear to be standing as they are called "standing waves,"  they are actually traveling back-and-forth along the string.  The subject of waves is lengthy, complicated, and mathematically very involved.  The above is enough to give you an idea.

Types of Waves:

There are two classifications: one classification of waves is: mechanical versus electromagnetic.

Mechanical waves require matter for their transmission.  Sound waves, ocean waves, and waves on a guitar string are examplesAir, water, and metal string are their media (matter), respectively.

Electromagnetic waves can travel both in vacuum and matter.  If light could not travel in vacuum, we would not see the Sun.  Light is an electromagnetic wave.  Radio waves, Ultraviolet waves, and infrared waves are all electromagnetic waves and travel in vacuum.

Waves can also be classified as transverse and longitudinal. (See Figure).

For a transverse wave the disturbance direction is perpendicular to the propagation direction.  Water waves are transverse.  Waves on guitar strings are also transverse.

For a longitudinal wave the disturbance direction is parallel to the propagation direction.  Waves on a slinky as well as sound waves are longitudinal.

Frequency ( f ):

The frequency ( f ) of a wave is the number of full waveforms generated per second.  This is the same as the number of repetitions per second or the number of oscillations per second.  The SI unit for frequency is (1/s), or (s -1), or (Hz).

Period ( T ):

Period is the number of seconds per waveform, or the number of seconds per oscillation.  It is clear that frequency and period are reciprocals.

T = 1 / f

Also, recall the useful relation between frequency ( f ) and the angular speed ( ω ).    ω = 2π f.    ω is the number of radians per second, but f is the number of turns per second.  Each turn is  radians.

Wavelength ( λ ):

Wavelength ( λ ) is the distance between two successive points on a wave that are in the same state of oscillation.  Points A and B (in the following figure) are the nearest or successive points that are both the same amount passed the maximum and therefore in the same state of oscillation.

Wavelength may also be defined as the distance between a peak to the next, or the distance between a trough to the next, as shown.

Wave Speed ( v ):

The wave speed is the distance a wave travels per second.  Since each wave source generates ( f ) wavelengths per second and each wavelength is ( λ ) units of length long; therefore, the wave speed formula is:

v = f λ.

Example 1:  The speed of sound waves at STP conditions is 331 m/s.  Calculate the wavelength of a sound wave which frequency is 1324 Hz at STP.

Solution:  Using  v = f λ,  solving for λ yields:    λ = v/f    ;    λ = (331m/s) / (1324/s) = 0.250m.

 A Vibrating String      A stretched string fixed at its end points and brought into oscillations forms a vibrating string.  An example is a violin string on which waves keep traveling back-and-forth between its end points.  If a string is observed closely or by a magnifying glass, at times it appears as shown on the right.      The higher the pitch of the note it is playing, the higher the frequency of oscillations and the shorter the wavelength or the sine-waves that appear on it.  The waveforms appear to be stationary, but in reality they are not.  They are called "standing waves."     Nodes are points of zero oscillation and  antinodes are points of maximum oscillation.

Now, look at the following example:

 Example 2: In a 60.0-cm long violin string, three antinodes are observed.  Find the wavelength of the waves on it. Solution:  Each loop has a length of  (60.0cm / 3)  = 20.0cm.    Each wavelength (full sine wave) contains two of such loops; therefore,                               λ = 40.0cm.

Speed of waves in a string depends on the tension in the string, F, as well as the mass per unit length, μ, of it as explained below:

 Speed of Waves in a Stretched String The more a string is stretched, the faster waves travel on it.  The formula that relates the tension ( F ) in the string and the waves speed, v, is:

Example 3: A 120-cm guitar string is under a tension of 400N.  The mass of the string is 0.480 grams.  Calculate (a) the mass per unit length of the string and (b) the speed of waves in it.   (c) In a diagram show the number of (1/2)λ's that appear in this string if it is oscillating at a frequency of 2083Hz.

Solution: (a)   μ = M / L    ;   μ =   (0.480x10-3kg) /1.20m    =    4.00x10-4 kg/m.

(b)   v = (F/μ)1/2 ;    v = (400N / 4.00x10-4 kg/m)1/2 = 1000 m/s.

(c)   v = f λ        ;    λ = v / f    ;    λ = (1000m/s) / (2083 /s) = 0.480m.

(1/2)λ = 0.480m / 2 = 0.240m.

The number of (λ/2)'s that fit in the string length of 120cm is

1.20m / 0.240m  =  5.00, as shown:

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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 pipe's length.  The following is reasonable to think or state:

At a closed end, only a node can form.   At an open end, both nodes and antinodes can form.

If the pipe's length allows, when an antinode is formed at an open end, the sound waves amplify (become loud) at which state we say "Resonance" has occurred.

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.

An 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 nonzero 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 nearest antinode on a wave is 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

 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 ).

Possible 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 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.  For more detailed information, refer to a text.

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 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 the figure above for closed pipes.  We may write:

λ/4 = 12.5 cm    ;    λ = 4(12.5 cm) = 50.0 cm.    Also,

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.

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) = 343 m/s.

Note that the speed of sound at STP conditions (0°C and 1atm. of pressure)  is  331 m/s.

The Doppler Effect:

When an ambulance is approaching us, a higher pitch 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 pass by our ears than when it is at rest.  When the sound source is going away from us, less number of wavelengths pass by our ears than when it 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 stationary and the source is approaching or receding.  Cases (3) and (4) are when the source is stationary 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.  For proof refer to your text.

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 ; 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 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 moving 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 & observer moving toward each other: fo = fs[(V + Vo)/(V-Vs)]  = 1575 Hz .

(c) Case 5, source chasing the observer: fo = fs [(V-Vo)/(V-Vs)]  = 1425 Hz .

(d) Case 5, observer chasing the source: fo = fs [(V + Vo)/(V + Vs)] = 1290Hz .

(e) Case 5, Vo = 0  and Vs = 0.     ;    fo = fs [(V+0)/(V+0)]1350Hz .

Chapter 16 Test Yourself 1:

1) A wave is the motion of  (a) a particle along a straight line in a back-and-forth manner  (b) a disturbance in a medium.  (c) a disturbance in vacuum  (d) both b & c.      click here

2) A mechanical wave (a) can travel in vacuum  (b) requires matter for its transmission  (c) both a & b.

3) An electromagnetic wave (a) can travel in vacuum  (b) can travel in matter  (c) both a & b.

4) A longitudinal wave travels (a) perpendicular to the disturbance direction  (b) parallel to the disturbance direction  (c) in one direction only.      click here

5) A transverse wave (a) travels perpendicular to the disturbance direction  (b) can also travel parallel to the disturbance direction  (c) travels sidewise.

6) A stick that gives a downward hit to a horizontally stretched string, generates a trough that travels along the string.  When the moving trough reaches a fixed end point, it returns not as a trough, but a hump.  The reason is that (a) the moving disturbance is not capable of pulling that end point down  (b) conservation of momentum requires the string to be pulled upward by the fixed point and hence the wave reflects  (c) the conservation of gravitational potential energy must be met   (d) both a & b.      click here

7) Wavelength is (a) the distance between two crests  (b) the distance between a crest and the next one  (c) the distance between a trough and the next one  (d) b & c.

8) The frequency of a wave is the number per second of (a) full wavelengths generated by a source  (b) full waves passing by a point  (c) a & b.      click here

9) The formula for wave speed, in general, is (a) V = λ / f    (b) V = f λ     (c) V = f / λ.

10) 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

11) 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.

12) 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

13) 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.

14) 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.

15) If the tension in a stretched string is quadrupled, then a disturbance made in it travels (a) 4 times faster    (b) 1/2 times slower    (c) 2 times faster.      click here

16) If the tension in a stretched string is increased by a factor of 9, then a disturbance made in it travels (a) 3 times faster  (b) 1/3 times slower   (c) 9 times faster.

17) The speed of waves (not necessarily sound waves) in a stretched string is proportional to (a) F, the tension   (b) F1/2, the square root of the tension   (c) F2, the square of the tension.      click here

18) The quantity μ = M/L, mass of a string divided by its length, is called  (a) mass per unit length  (b) mass length  (c) length per unit mass.

19) The Metric unit for μ is (a) kg/m.   (b) slug/ft   (c) gr/cm.

20) Mechanical waves travel faster in a string that is (a) thicker and therefore less flexible   (b) thinner and therefore more flexible   (c) neither a nor b.

21) The speed of waves in a stretched string is (a) directly proportional to μ    (b) inversely proportional to μ    (c) inversely proportional to 1/μ    (d) directly proportional (1/μ)0. 5.      click here

22) The tension in a guitar string is 576N and its 1.00m length has a mass of 0.100gram.  The waves speed in this stretched string is (a) 2400m/s    (b) 3600m/s    (c) 1800m/s.

23) If two full wavelengths can be observed in this string, the wavelength of the waves is (a) 1.00m   (b) 0.500m.   (c) 2.00m.

24) What frequency does the guitar string ( in the previous questions) play, if the wavelength is 0.500m?  (a)1200Hz   (b) 400Hz   (c) 4800Hz      click here

25) Musically, a closed pipe is (a) closed at both ends  (b) closed at one end only  (c) neither a nor b.

26) In a closed pipe, sound waves can have a maximum at (a) the open end  (b) the closed end  (c) both a & b.

27) The reason an antinode cannot form at a closed end is that (a) at a closed end air molecules are not free to oscillate back and forth    (b) a closed end is a harder medium than the air in the pipe and waves reflect when they hit a harder medium   (c) both a & b.      click here

28) 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.

29) The distance between a node and the next antinode is (a) 1/2 λ.    (b) 1/4 λ.    (c) 1/8 λ .      click here

30) 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 .

31) A 12.0-cm closed pipe (a) can form resonance for λ = 50.0cm   (b) can't form resonance for λ = 50.0cm.

32) A 12.5-cm closed pipe (a) can form resonance for λ = 50.0cm   (b) can't form resonance for λ = 50.0cm.

33) 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 time in the pipe's length.      click here

34) An open pipe, can create resonance for (a) all frequencies  (b) for all wavelength   (c) only a wavelength for which 1/2 λ can fit any number of times in the pipe's length.

35) 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.

36) If an observer moves toward a stationary source, the frequency heard will be (a) lower  (b) higher  (c) the same.

37) If a source moves away from a stationary observer, the frequency heard will be (a) lower  (b) higher  (c) the same.

38) If an observer follows a source at the same velocity,  the frequency heard will be (a) lower  (b) higher  (c) the same.

39) 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.

40) 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.

Problems:

1) The speed of sound waves at STP is 331 m/s.  Calculate (a) the frequency of a sound wave that has a wavelength of 75.0cm at STP.  If the temperature increases by 40.0°C, like a hot summer day, (b) what frequency sound will generate the same wavelength?  Use the empirical formula v(T)=[331+0.6T] m/s  to calculate the sound speed at the given temperature.

2) In an 80.0-cm long guitar string, 2 antinodes are observed.  Find (a) the wavelength of the waves in it.  If the frequency at which the string is vibrating is 784Hz, find (b) the speed of waves.  The mass of 81.0m of this string is 12.0grams.  Find (c) its mass per unit length (linear density), and (d) the tension in the string.

3) A 270.0-cm string is under a tension of 81.0N.  The mass of the string is 6.75 grams.   Calculate (a) the speed at which waves travel in it, (b) the wavelength of waves in it when one end of it is fixed and the other end is being oscillated at a frequency of 200Hz,  and (c) draw a figure showing the number of standing waves that can be observed in the string.

4) In an experiment, a 512-Hz oscillating blade generates resonance in a closed tube at three positions.   The first resonance is heard when the tube length is 17.2cm measured from zero (the open end), the second one occurs at 50.1cm from zero, and the third one at 87.1cm from zero.   Calculate (a) the wavelength on the basis of each of the resonance positions, (b) the average value of the wavelength, and (c) the speed of sound on the basis of this average.  Use the empirical formula V=[331+0.6T] m/s to find the air temperature, T (in °C) at the location of the experiment.

5) A police car (Car1) with an alarm frequency of 784Hz is traveling at a speed of 28.0m/s on a straight road where another car (Car2) is moving at a speed of 22.0m/s.  The air temperature is 25°C.   Find the frequency heard by Car 2 for each of the following cases: (a) when they are heading toward each other, (b) they are moving away from each other, (c) Car2 is following Car1, and (d) Car1 is following Car2.  Note: Calculate the speed of sound from the empirical formula v(T)=[331+0.6T] m/s where T is in °C.

6) The speed of certain waves traveling along a string is 244m/s.  Find (a) their wavelength  for a frequency of 610 s-1.   What would be the frequency if the wavelength changes to 0.800m?