Transcript Physics 121C Mechanics

Physics 102
Superposition
Lecture 6
Moza M. Al-Rabban
Professor of Physics
[email protected]
The Goal of this chapter is to understand
and use the idea of superposition
• Apply the principle of superposition.
• Understand how standing waves are generated.
• Calculate the allowed wavelengths and frequencies of
standing waves.
• Understand how waves cause constructive and
destructive interference.
• Calculate the beat frequency between two nearly
equal frequencies.
•
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Waves vs. Particles
3
The Principle
of Superposition
Colliding
Waves
When two or more waves are simultaneously present at a single
point in space, the displacement of the medium at that point is
the sum of the displacements due to each individual wave.
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The Principle of Superposition
When two or more waves are simultaneously present at a single
point in space, the displacement of the medium at that point is
the sum of the displacements due to each individual wave.
Dnet  D1  D2     Di
i
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Stop To Think
Two pulses on a string approach each other at speed
of 1 m/s. What is the shape of the string at t= 6 s?
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Standing Waves
A standing wave is the superposition of two waves.
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Standing Waves
We will assume that the two
waves have the same
frequency, the same
wavelength, and the same
amplitude.
They are identical waves
except that one travel to the
right and the other to the left.
What happens as these two
waves pass through each
other?
The blue dot shows that the wave in (b) is moving neither right nor left.
This is a wave, but it is not traveling wave. It is a standing wave.
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Nodes and Antinodes
The points that never move! Spaced by /2, are called nodes.
Halfway between nodes are points where the particle of medium
oscillate with maximum displacement. These points of maximum
amplitude are called antinodes, and they are also spaced /2 apart.
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Two waves 1 and 2 are said to be in phase at a point where D1 is always equal to D2.
This is called a point of constructive interference.
The antinodes of standing wave are points of constructive interference between
the two traveling waves.
In contrast, two waves are said to be out
of phase at points where D1 is always equal
to –D2.
Their superposition gives a wave with zero
amplitude== no wave at all!
This is a point of destructive interference.
The nodes of a standing wave are points of
destructive interference.
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I  CA
2
The points of maximum intensity occur
where the standing wave oscillates with
the largest amplitude (the antinodes)
and that the intensity is zero at the
nodes.
If this is a sound wave, the loudness
periodically varies from zero( no sound)
to maximum and back to zero.
The key idea is that the intensity is
maximum at points of constructive
interference and zero ( if the
waves have equal amplitude) at
points of destructive interference.
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The Mathematics of Standing
Waves
A sinusoidal wave traveling top the
right along the x-axis
DR  a sinkx  t 
An equivalent wave traveling to
the left is
sin     sin  cos   cos sin 
DL  a sinkx  t 
D x , t   DR  DL  a sinkx  t   a sinkx  t 
D x , t   a (sin kx cost  coskx sin t )  a (sin kx cost  coskx sin t )
 (2a sin kx) cost
D (x , t )  A(x ) cost
Where the amplitude function A(x) is defined as
A(x )  2a sin kx
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Notes
A(x )  2a sin kx
The amplitude reaches a maximum value
Amax = 2a at points where sin kx =1.
D (x , t )  A(x ) cost
The displacement is neither a function of (x-vt) or (x+vt) , hence it is not
a traveling wave.
The cos t , describes a medium in which each point oscillates in simple
harmonic motion with frequency f= /2.
The function A(x) =2a sin kx determines the amplitude of the oscillation
for a particle at position x.
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D (x , t )  A(x ) cost
The amplitude of oscillation,
given by A(x), varies from point
to point in the medium.
The nodes of the standing wave
are the points at which the
amplitude is zero. They are
located at positions x for which
A(x )  2a sin kx  0
That is true if
kx m 
2x m

 m
m  0,1, 2, 3, 
Thus the position xm of the mth node is
xm  m
Where m is an integer.

2
m  0,1, 2, 3, 
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Example 1: Node spacing on a string
A very long string has a linear density of 5.0 g/m and is stretched with a
tension of 8.0 N. 100 Hz waves with amplitudes of 2.0 mm are
generated at the ends of string.
a. What is the node spacing along the resulting standing wave?
b. What is the maximum displacement of the string?
MODEL Two counter-propagation waves of equal frequency create a standing wave.
SOLVE
a. The speed of the waves on the string is
And thus the wavelength is
v
Ts


8.0 N
 40m / s
0.0050kg / m
v 40m / s

 0.40m  40cm
f 100Hz
Consequently, the spacing between adjacent nodes is /2 = 20 cm.

b. The maximum displacement, at the antinodes where sin kx = 1, is
Amax  2a  4.0mm
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Reflection and Transmission
When a traveling wave moving on a string reaches a point where the
mass density ( and velocity) change, the wave “splits” into a reflected
wave and an ongoing wave.
If the mass density goes down and the velocity increases, the
amplitude of the reflected wave is positive.
If the mass density goes up and the velocity decreases, the amplitude
of the reflected wave is negative.
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Reflection from a Boundary
When a traveling wave moving
on a string reaches a fixed
boundary, the wave is reflected.
Because there is no transmitted
wave, all the wave’s energy is
reflected. Hence, the amplitude
of a wave reflected from a
boundary is unchanged.
With respect to the incident
wave, the amplitude of the
reflected wave is equal in
magnitude and opposite sign.
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Standing Waves on a String
If a string is plucked at the center, traveling waves move in both
directions, are reflected at the boundaries, and a standing wave results.
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Boundary Condition
• A boundary condition is a mathematical
statements of any constraint that must be
obeyed at the boundary or edge of a medium.
Because the string is tied down at the ends, the displacement at
x = 0 and x = L must be zero at all times.
Thus the standing-wave boundary conditions are
D(x = 0,t)=0 and D(x = L,t)=0 . Or,
we require nodes at both ends of the string.
But D x , t   (2a sin kx) cost
,
It satisfies the boundary condition D(x = 0,t)=0
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The second boundary condition, at x = L, requires
D(x = L,t)=0
This condition will be met at all times if
2a sin kL = 0
( boundary condition at x = L)
This will be true if sin kL =0, i.e.
kL 
2L

 m
m  1, 2, 3, 
kL must be multiple of m, but m=0 is excluded because L can’t be zero.
For a string of fixed length L, the only quantity in this equation that can
be vary is . That is, the boundary condition can be satisfied only if the
wavelength has one of the values
2L
m 
m
m  1,2,3,
A standing wave can exist on the string only if its wavelength is
one of the values given by this equation.
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Standing Wave Normal Modes
2L
m
m 
fm 
v
m

m  1,2,3,
v
v
 m ; m  1, 2,3, 4,
2L / m
2L
The lowest allowed frequency which
correspond to wavelength 1:
f1 
f1 
v
;
2L
v
2L
Fundamental frequency
f m  mf1 ; m  1, 2,3, 4,
The allowed standing wave frequencies are all
integer multiples of fundamental frequency.
The higher frequency standing waves are called
harmonics, with m=2 wave called second harmonic,
the m=3 wave called third harmonic, and so on.
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Standing Wave Normal Modes
Here are the first four possible standing waves
on a string of fixed length L.
The possible standing waves are called the
normal modes of the string.
Each mode, numbered by the integer m, has a
unique wavelength and frequency.
Keep in mind that
These drawings simply show the envelope, or
outer edge, of oscillations. The string is
continuously oscillating at all positions between
these edges.
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Notes about the normal modes of a string
• m is the number of antinodes on the standing
wave, not the number of nodes.
• The fundamental mode, with m=1, has 1 = 2L,
not 1 =L.
• The frequencies of the normal modes form an
arithmetic series:
f 1 , 2 f 1 , 3 f 1 , 4 f 1 ,
The fundamental frequency f 1 can be found as
the difference between the frequencies of any
two adjacent modes.
f1  f  f m 1  f m
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Example 2: a standing wave on a string
A 2.50 m long string vibrates as a 100 Hz standing wave with nodes 1.00 m
and 1.50 m from one end of the string and at no points in between these two.
Which harmonic is this, and what is the string’s fundamental frequency?
Model: The nodes of a standing wave are spaced /2 apart.
SOLVE:
If there are no nodes between the two at 1.0 m
and 1.5 m. then the node spacing is /2 =0.50 m.
The number of 1.50 m wide segments that fit into
a 2.50 m length is five, so this is m=5 mode and
100 Hz is the fifth harmonic.
The harmonic frequencies are
f m  mf1
Hence, the fundamental frequency is
f5
f1 
 20 Hz
5
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Clicker Question 2
A standing wave on a string vibrates as shown.
If the tension is quadrupled while the frequency and length remain
the same, which diagram represents the new vibration?
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Standing Electromagnetic Waves
Another example of a
standing wave are the
electromagnetic waves in a
laser cavity that is bounded
by two reflecting mirrors.
Suppose that a laser cavity has length L=30 cm containing visible
light with wavelength =600 nm. Then the mode number m is:
2L
2(0.3 m)
m

 1, 000, 000
-7
 (6.0 10 m)
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Example:
Standing Light Wave in a Laser
A helium-neon laser emits red light of  = 632.9924 nm. The distance
between the laser mirrors is 310.372 mm.
(a) In what standing-wave mode does the laser operate?
(b) What is the next longest wavelength that could make
standing waves in the laser cavity?
  2L / m
2L
2(310.372 mm)
m


(632.9924 nm)
 980, 650
m '  m  1  980,649
2L
m

'


m 1
m  1 1  1/ m
  (1  1/ m)    ( / m)
 632.9924 nm  0.00064 nm
 632.9930 nm
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Example:
Cold Spots in a Microwave Oven
“Cold spots”, i.e. locations where objects are not adequately
heated in a microwave oven are found to be 1.25 cm apart.
What is the frequency of the microwaves?
dnode  1.25 cm   / 2 so   2.50 cm
(3.00 108 m/s)
f  
 1.20 1010 Hz  12.0 GHz

(0.0250 m)
c
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Standing Sound Waves and Musical
Acoustics
A long narrow column of air such as
the air in a tube or pipe can support a
longitudinal standing sound wave. Such
a tube may be open or closed at the
ends.
The closed end of a column of air
must be a node.
An open end of an air column is
required to be an antinode.
In the example shown here, both ends
are closed and the standing wave mode
is m=2. There are nodes at each end
and one in the center, and there are two
antinodes at the quarter wave locations.
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Example 5: Singing in the Shower
A shower stall is 2.45 m tall. For what frequencies less than 500 Hz
can there be vertical standing sound waves in the shower stall?
Model: The shower stall, at least to a first approximation, is a column of
air 2.45 m long. It is closed at the ends by the ceiling and floor. Assume
a room temperature speed of sound.
f1 
v
(343 m/s)

 70 Hz
2 L 2(2.45 m)
The possible standing wave frequencies are integer multiples
of the fundamental frequency.
fm  70 Hz, 140 Hz, 210 Hz, 280 Hz, 350 Hz, 420 Hz, and 490 Hz,
m  1, 2, 3, 4, 5, 6, and 7
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Pipes and Modes
Open-Open or Closed-Closed
2L

 1 / m 

m
 m  1, 2,3, 4,
v
fm  m
 mf1 

2L
m 
Open-Closed
4L

 1 / m 

m
 m  1,3,5, 7,
v
fm  m
 mf1 

4L
m 
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Clicker Question 2
An open-open tube of air supports standing waves of frequencies
of 300 Hz and 400 Hz, with no frequencies between these two.
The second harmonic (m=2) of this tube has frequency:
(a) 100 Hz; (b) 200 Hz;
(c) 400 Hz;
(d) 600 Hz;
(e) 800 Hz.
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Musical Instruments
Many woodwind instruments are effectively an open-closed
pipe. This means they have only odd harmonics. Their
fundamental frequency will be:
vsound
f1 
4L
The vibrating string of a stringed instrument is the
equivalent of a closed-closed pipe. This means it will have both
odd and even harmonics. Its fundamental frequency is:
vstring
1 Ts
f1 

2L 2L 
Note that for wind instruments, L is the only adjustable
parameter, while for stringed instruments, L, Ts and  can, in
principle, be varied. However, wind instruments can be played at
relatively pure harmonic frequencies, while strings cannot.
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Example:
The Notes of a Clarinet
A clarinet (an open-closed instrument) is 66 cm long. The speed of
sound in warm air is 350 m/s.
What are the frequencies of the lowest note on a clarinet and of
the next highest harmonic?
f1 
v
(350 m/s)

 133 Hz
4 L 4(0.66 m)
f3  3 f1  399 Hz
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End of Lecture 6