The Doppler effect is something we commonly experience when we and/or the source of sound are moving relative to each other, causing the sound to change pitch. For instance, an ambulance siren is heard to change pitch as the ambulance approaches, then moves away from us. The pitch (frequency) is higher as it approaches and drops as it moves away.
Although a common effect, it was not "discovered" until sufficiently fast vehicles (i.e. trains) came into use.
Of course, we want a quantitative answer to go along with our qualitative understanding. We proceed by imagining what happens in various cases where the observer or the source of sound is moving. Since frequency is basically the number of wave crests that pass a given location each second, we want to think about "counting" the number of waves that pass our observer's ears each second.
First, imagine a stationary source and an observer moving (directly) towards the source at speed vo. Each second the source emits a certain number of waves which propagate outward radially in all directions. Such waves are called spherical waves, as described in section 14.5. We will represent these waves by drawing equally spaced circles around the source. The circles can be thought of as representing the location of the crest of a wave propagating outward.
If the emitted frequency is f, then in one second, f waves emerge from the source and pass a stationary observer.
Our observer has moved toward the source a distance vot (t = 1 sec, but let's not insert that just yet!).
Consequently our observer passes some additional wave fronts.
The number of additional wave fronts equals the distance moved divided by the wavelength, vot/l, or just vo/l in 1 second.
Thus the total number of waves passing the observer in 1 second is
f' = f + vo/l.
Finally, since l = f/v, where v is the speed of the wave, we can write:
f' = f + fvo/v = f(1 + vo/v) = f(v+vo)/v.
If the observer is moving away from the source, then fewer wave fronts pass the observer.
The number is reduced by the same amount it is increased when the observer moves towards the source:
f' = f - vo/l = f(v-vo)/v.
We can incorporate both results into one equation by using the "±" notation, meaning plus or minus, and remembering to choose "+" when the observer is moving towards the source and "-" when the observer is moving away from the source.
f' = f(v ± vo)/v.
Now let's consider what happens when the source is moving and the observer is stationary.
Recall that the period, T = 1/f, is the time required for a complete cycle of a wave, so it is the time between the emission of one crest and the following crest.
If the source is moving with speed vs, then it moves a distance vsT = vs/f during this time.
If the source is moving towards the observer, then the distance between wave crests (the wavelength) is shortened by this distance:
l' = l - vs/f.
And if the source is moving away from the observer, then the wavelength is increased by this amount:
l' = l + vs/f.
The frequency heard by the observer is f' = v/l' = v/(l -/+ vs/f) = v/((v/f) -/+ (vs/f)) = fv/(v -/+ vs)
where the "-" is used when the source moves toward the observer and "+" when the source moves away.
(Think about the change in frequency -- when the source moves toward you, the frequency is greater, so the denominator must be smaller; when the source moves away, the frequency is lower, so the denominator must be larger.)
We can combine both cases of moving observer and moving source into a single equation:
f' = f(v ± vo)/(v -/+ vs).
A bat flying at 5.0 m/s emits a chirp at 40 kHz. If this sound pulse is reflected by a wall, what is the frequency of the echo received by the bat?
The tricky part here is the intervention of the wall. This makes you think about whether the source or the observer is moving, and whether they move towards or away from each other. So let's break the problem into two parts: (i) the moving bat emits a chirp which is received at a stationary wall, then (ii) the stationary wall re-emits the chirp which is received by the moving bat. In both instances, the bat is moving towards the wall.
(i) Use v = 331 m/s for the velocity of sound in air.
The frequency received at the stationary wall is
f' = f(v +vo)/v
(ii) The wall re-emits the frequency f', and the bat receives frequency:
f" = f'v/(v - vs) = f(v + vo)/(v - vs) = (40 kHz)(331 + 5.0)/(331 - 5.0) = 41 kHz.
Notice that this is the same as if the observer moves towards the source at 5.0 m/s and the source moves towards the observer at 5.0 m/s.
When deriving the doppler shift result for a source moving towards the observer, I slyly skipped an important detail.
What happens if vsT > l.
Our calculation gives l' < 0, but this is nonsense!
Since T = 1/f, the above inequality can also be written as:
vs > lf = v.
That is, the speed of the source is greater than the speed of sound!
Such a situation results in a shock wave, and in the case of an aircraft traveling faster than sound, a sonic boom is heard.
If sound travels to your ears by two different paths, the waves from those paths can interfere. An example is shown in the text, where two tubes carry sound from a source to a receiver. By changing the length of one tube relative to the other, the sound can be made to interfere constructively or destructively.
Standing waves are fixed oscillations that can exist in some media. A common example is the vibration of a string. A string secured at both ends, can be made to oscillate such that the incoming and reflected waves interfere in a regular pattern. This regular pattern looks like a wave standing still, hence the name standing wave. There are points where the string never moves; these are called "nodes". Locations where the string moves at its maximum amplitude are called "anti-nodes".
Notice that all points on the string oscillate at the same frequency about the equilibrium position. But they move with different amplitude.
These are some of the qualitative aspects of standing waves. Now we will discuss more quantitative aspects. First, because both ends of the string are fixed, they must be nodes of the standing wave. This means that standing waves can have only certain wavelengths, namely wavelengths which will produce a node at each end of the string.
What are these wavelengths?
If the length of the string is L, then the first (longest wavelength) is l1 = 2L.
This is a "half wave" vibrating on the string. [See Figure 14.18]
The frequency of this wave, called the fundamental frequency or first harmonic is f1 = v/l1 = v/2L.
Using the relation for the speed of a wave on a string, v = Sqrt{F/m}, we can write:
f1 = (1/2L)Sqrt{F/m}.
The next higher frequency on the string is a "full wave", such that l2 = L, and f2 = v/L = 2f1. This is called the second harmonic (twice the frequency).
Likewise, the next higher frequency is the third harmonic, f3 = 3v/2L = 3f1, and above that is the fourth harmonic f4 = 2v/L = 4f1.
Generally, we can write down the frequency for the nth harmonic:
fn = nv/2L = (n/2L)Sqrt{f/m} = nf1.
Standing waves can also be setup in air columns. This is the basis of musical instruments such as pipe organs, and brass instruments (trumpet, trombone, etc.).
A difference between standing waves in air columns and those on strings are the conditions at the "end".
On a string, the ends are fixed, so the standing wave must have a node there.
In an air column, normally at least one end is open to the air, and here the air molecules are free to move.
This results in the open end corresponding to an anti-node of the standing wave.
If the air column has a closed end, then the air molecules are constrained and a node results.
Therefore,:
closed end = node
open end = anti-node.
So first, let's consider a tube of length L with both ends open. We want to find the standing waves that will have anti-nodes at both ends. The first (lowest frequency, longest wavelength) harmonic is a half wave with anti-nodes at each end and a node in the middle. [See Figure 14.22] The wavelength is l1 = 2L, and the frequency is f1 = v/2L.
The second harmonic has a wavelength l2 = L, and a frequency f2 = v/L = 2f1.
The nth harmonic has a wavelength of ln = 2L/n, and a frequency of fn = nv/2L = nf1.
Now consider a tube of length L with one end closed and the other open. The standing waves supported in this column have a node at the closed end and an anti-node at the open end. [See Figure 14.22] The first harmonic is a quarter wave, with l1 = 4L and f1 = v/4L.
The next higher harmonic is a three-quarter wave. It has a wavelength of 4L/3 and a frequency of 3v/4L = 3f1. This is not the second harmonic! Since the frequency is 3 times the first harmonic, this is the third harmonic. Further thought will show that a tube with one end closed will not support even harmonics, i.e. second, fourth, sixth, ... Only the odd harmonics exist. Therefore, l3 = 4L/3, and f3 = 3v/4L = 3f1.
And the nth odd harmonic has wavelength ln = 4L/n and frequency fn = nv/4L = nf1.