Transcript PowerPoint Presentation - 1. 18 Waves in more than one

2.9 Beats
Consider two sound sources that produce harmonic waves
of equal amplitude A but different frequencies f1 and f2.
harmonic waves,
Source 1
Source 2
Detector , x
What is the signal recorded by a detector placed a
distance x from each source?
Need to use the principle of linear superposition to obtain
the answer.
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•
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•
Let the displacement of the sound wave from source 1 be
y1(x,t) = Asin(k1x-w1t) and for source 2
y2(x,t) = Asin(k2x-w2t)
From principle of linear superposition we have
yt(x,t) = y1(x,t) + y2(x,t)
yt(x,t) = Asin(k1x-w1t) + Asin(k2x-w2t)
Which becomes

 
 k  k x _ w  w
1
2
1
2

y t x , t  2A sin

2

 

t 

 
 k  k x _ w  w
2
1
2
cos 1
 
2
 
 
t 


• For the two sound waves that interact the resulting
displacement is

 
 k  k x _ w  w
1
2
1
2

y t x , t  2A sin

2

 
•

t 

 
 k  k x _ w  w
2
1
2
cos 1
 
2
 
 
t 
To simplify the problem let us assume that x = 0
and (w1+w2)/2 = wave and (w1-w2) = wbeat
w t 
beat

y t x , t  2A sin wavet cos


2


 
•


 
Here we have used sin(-a) = -sin(a) and cos(-a) =
cos(a)

• Thus at x we have two two harmonic waves interacting
together one vibrating at the average frequency and the
other vibrating at the beat frequency.
w t 
beat

y t x , t  2A sin wavet cos


2


 

 
Harmonic wave at
ave frequency
Harmonic wave at
beat frequency
When the ear detects this signal the ear responds to
the intensity of the signal.
 
w t 

I x , t  4A sin w t cos 


2


2
I (x, t )  y t x, t
2
2
ave
2
beat
• We can use the fact that
x 
cos x  2cos2 
 
 1
2 

To give
 
 

I x, t  2A2 sin2 wavet 1  cos wbeatt


So when wbeatt = 2nπ the intensity is a maximum

So when wbeatt = (2n+1)π the intensity is zero
Thus there is a periodic modulation of the intensity
and the period is
T beat 
2
wbeat
2.10 How are beats formed?
Need to consider the way in which the phase of the two sound
waves varies at the point x.
Initially the two waves are in
phase. Hence the displacement
caused by each wave is the same.
So the particle experiences a
maximum displacement.
5
4
3
2
1
0
-1
-2
0
1
2
time
3
As time increases the phase of the
two waves evolves at different
rates because the frequency of
each wave is different (w1t ≠ w2t).
T
5
4
3
2
1
0
-1
-2
0
1
2
3
time
After some time T the phase of
wave 1 has evolved such that it is π
out of phase with respect to wave
2. At this point the displacement
caused by each wave is equal but
opposite resulting in zero net
displacement.
5
4
3
Time increases and the two waves
come back into phase resulting in
the maximum displacement.
2
1
0
-1
-2
0
1
2
time
3
2.11 The Doppler effect
So far the wave source has been stationary.
What happens if there is a relative motion between the
source and the observer?
When there is a relative motion between a wave source and
an observer there is an apparent change in frequency.
This is known as the Doppler Effect.
There are two cases to consider
1. Source stationary and the observer moving.
2. Source moving and observer stationary.
2.11(I) The Doppler effect- Source stationary
observer moving
Consider a stationary source generating sound waves at a
frequency fs and wave speed vs.
The observer moves in a straight line towards the source at
a speed vo.
vs
Source
s
vo
Observer
2.11(I) The Doppler effect- Source stationary
observer moving
The source generates wave fronts that are spaced by s.
As the observer moves towards the observer passes the
wave fronts with an apparent speed va = vo + vs.
The spacing between the wave fronts does not change and so
the observer senses an apparent frequency fa = va/s.
So
fa
v


s
v o
s

But s = vs/fs
Hence fa
v


s

 v o fs
vs
If the observer was moving away from
the source the the apparent frequency f  v s  v o fs
a

shift is
vs


2.11(II) The Doppler effect- Source moving observer
stationary
Consider a stationary source generating sound waves at a
frequency fs and wave speed vs.
The source moves in a straight line towards the observer at
a speed vm.
If the source was stationary then spacing between
successive wave fronts would be s and the observer would
sense the wave fronts at their spacing s.The source moves
in a straight line towards the observer at a speed vm.
As the source is moving at a speed vm then the spacing
between successive wave fronts is changed.
vs
Observer
Source
s
vT
m
vm
Source stationary wave fronts
spaced by s
vs
Source
a
Source now moves with speed vm.
Successive wave fronts spaced by
a. The time taken for successive
wave fronts to be emitted is T,
the period of the source. In this
time the source moves towards
Observer
the observer a distance vmT.
Hence the distance between the
new wave font and old wave front
is reduced.
2.11(II) The Doppler effect- Source moving and
observer stationary
The apparent wavelength is given by a = s - vmT.
But a = vs/fa and s = vs/fs and T = 1/fs
So
Hence

vs vs vm
 
fa fs fs

 v
s

fa  

fs
v s  v m 
If source was moving away from the
source
the the apparent frequency

shift is

 v
s

fa  

fs
v s  v m 
2.11(II) The Doppler effect- Source moving observer
moving
If both the source and the observer are moving then the
apparent change in frequency is given by
v  v 
s
o

fa  

fs
v s v m 
