Physics 1B03summer-Lecture 9 Test 2 - Today 9:30 am in CNH-104 Class begins at 11am.
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Transcript of Physics 1B03summer-Lecture 9 Test 2 - Today 9:30 am in CNH-104 Class begins at 11am.
![Page 1: Physics 1B03summer-Lecture 9 Test 2 - Today 9:30 am in CNH-104 Class begins at 11am.](https://reader031.fdocuments.in/reader031/viewer/2022032517/56649cc15503460f949884aa/html5/thumbnails/1.jpg)
Physics 1B03summer-Lecture 9
Test 2 - Today
• 9:30 am in CNH-104
• Class begins at 11am
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Physics 1B03summer-Lecture 9
Wave MotionWave Motion
• Sinusoidal waves• Standing Waves
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Physics 1B03summer-Lecture 9
Sine Waves
f(x) = y(x,0) = A sin(kx)
For sinusoidal waves, the shape is a sine function, eg.,
Then, at any time: y (x,t) = f(x – vt) = A sin[k(x – vt)]
A
-A
y
x
(A and k are constants)
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Physics 1B03summer-Lecture 9
Sine wave:
A
-A
y
x
v
(“lambda”)is the wavelength (length of one complete wave); and so (kx) must increase by 2π radians (one complete cycle) when x increases by . So k = 2, or
k = 2π / λ
y (x,t) = A sin[kx – t] since kv=ω
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Physics 1B03summer-Lecture 9
Rewrite: y = A sin [kx – kvt]=A sin [kx – t]
ω = 2πf
y = A sin [ constant – t]
“angular frequency”radians/sec
frequency: cycles/sec (=hertz)
The displacement of a particle at location x is a sinusoidal function of time – i.e., simple harmonic motion:
The “angular frequency” of the particle motion is =kv; the initial phase is kx (different for different particles).
Review: SHM is described by functions of the form y(t) = A cos(t+) = A sin(/2 – –t), etc., with
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Physics 1B03summer-Lecture 9
Example A
-A
y
x
Shown is a picture of a wave, y=A sin(kxt), at time t=0 .
a
b
cd
e
i) Which particle moves according to y=A cos(t) ?
A B C D E
ii) Which particle moves according to y=A sin(t) ?
A B C D E
iii ) Sketch a graph of y(t) for particle e.
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Physics 1B03summer-Lecture 9
y(x,t) = A sin (kx ± t –)
angular wave numberk = 2π / λ (radians/metre)
angular frequencyω = 2πf (radians/second)
amplitude “phase”
The most general form of sine wave is y = Asin(kx ± ωt – )
The wave speed is v = 1 wavelength / 1 period, so
v = fλ = ω / k
phase constant
![Page 8: Physics 1B03summer-Lecture 9 Test 2 - Today 9:30 am in CNH-104 Class begins at 11am.](https://reader031.fdocuments.in/reader031/viewer/2022032517/56649cc15503460f949884aa/html5/thumbnails/8.jpg)
Physics 1B03summer-Lecture 9
Transverse waves on a string:
(proof from Newton’s second law – Pg.625)
Electromagnetic wave (light, radio, etc.):
v = c 2.998108 m/s (in vacuum)
v = c/n (in a material with refractive index n)
T
wave lengthmass/unit
tension Fv
(proof from Maxwell’s Equations for E-M fields )
The wave velocity is determined by the properties of the medium:
Wave Wave VelocityVelocity
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Physics 1B03summer-Lecture 9
A) it will decrease by 4B) it will decrease by 2C) it will decrease by √2D) it will stay the sameE) it will increase by √2
You double the diameter of a string. How will the speed of the waves on the string be affected?
QuizQuiz
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Physics 1B03summer-Lecture 9
Exercise (Wave Equation)
What are and k for a 99.7 MHz FM radio wave?
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Physics 1B03summer-Lecture 9
Particle Particle VelocitiesVelocities
Particle displacement, y (x,t)
Particle velocity, vy = dy/dt (x held constant)
(Note that vy is not the wave speed v – different directions! )
Acceleration, 2
2
dt
yd
dt
dva y
y
This is for the particles (move in y), not wave (moves in x) !
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Physics 1B03summer-Lecture 9
y
tkxAdt
dva
tkxAdtdy
v
tkxAy
yy
y
2
2 )sin(
)cos(
)sin(
“Standard” sine wave:
maximum displacement, ymax = A maximum velocity, vmax = A maximum acceleration, amax = 2 A
Same as before for SHM !
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Physics 1B03summer-Lecture 9
ExampExamplele
string: 1 gram/m; 2.5 N tension
Oscillator:50 Hz, amplitude 5 mm
y
x
Find: y (x, t)vy (x, t) and maximum speed
ay (x, t) and maximum acceleration
vwave
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Physics 1B03summer-Lecture 9
10 min rest
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Physics 1B03summer-Lecture 9
1) Identical waves in opposite directions: “standing waves”
2) 2 waves at slightly different frequencies:“beats”
3) 2 identical waves, but not in phase: “interference”
Superposition of Waves
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Physics 1B03summer-Lecture 9
Practical Setup: Fix the ends, use reflections.
TF
vf
nodenode
L
Lv
f
LL
2
2
1
1121
(“fundamental mode”)
We can think of travelling waves reflecting back and forth from the boundaries, and creating a standing wave. The resulting standing wave must have a node at each fixed end. Only certain wavelengths can meet this condition, so only certain particular frequencies of standing wave will be possible.
example:
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Physics 1B03summer-Lecture 9
λ2
12
2
2 fLv
f
L
Second Harmonic
1
33
32
3
323
3
23
f
Lvv
f
L
L
Third Harmonic . . . .
λ3
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Physics 1B03summer-Lecture 9
In this case (a one-dimensional wave, on a string with both ends fixed) the possible standing-wave frequencies are multiples of the fundamental: f1, 2f1, 3f1, etc. This pattern of frequencies depends on the shape of the medium, and the nature of the boundary (fixed end or free end, etc.).
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Physics 1B03summer-Lecture 9
Sine Waves In Opposite Directions:Sine Waves In Opposite Directions:
y1 = Aosin(kx – ωt)
Total displacement, y(x,t) = y1 + y2
y2 = Aosin(kx + ωt)
Trigonometry :
2cos
2sin2sinsin
bababa
""
""
btkx
atkx
cos sin2),( 0 tkxAtxy Then:
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Physics 1B03summer-Lecture 9
Example8
mm
y
1.2 m
f = 150 Hz
x
a) Write out y(x,t) for the standing wave.
b) Write out y1(x,t) and y2(x,t) for two travelling waves which would produce this standing wave.
wave at t=0
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Physics 1B03summer-Lecture 9
ExampleExample
When the mass m is doubled, what happens to
a) the wavelength, andb) the frequency
of the fundamental standing-wave mode?
What if a thicker (thus heavier) string were used?
m
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Physics 1B03summer-Lecture 9
ExamplExamplee
a) m = 150g, f1 = 30 Hz. Find μ (mass per unit length)
b) Find m needed to give f2 = 30 Hz
c) m = 150g. Find f1 for a string twice as thick, made of the same material.
m
120 cm
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Physics 1B03summer-Lecture 9
SolutionSolution
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Physics 1B03summer-Lecture 9
Standing sound wavesSound in fluids is a wave composed of longitudinal vibrations of molecules. The speed of sound in a gas depends on the temperature. For air at room temperature, the speed of sound is about 340 m/s. At a solid boundary, the vibration amplitude must be zero (a standing wave node).
nodenode nodeantinodeantinode
Physical picture of particle motions (sound wave in a closed tube)
graphical picture
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Physics 1B03summer-Lecture 9
Standing sound waves in tubes – Boundary Conditions-there is a node at a closed end
-less obviously, there is an antinode at an open end (this is only approximately true)
node antinodeantinode
graphical picture
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Physics 1B03summer-Lecture 9
L
141 L
343 L
545 L
Air Columns: column with one closed end, one open end
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Physics 1B03summer-Lecture 9
Exercise: Sketch the first three standing-wave patterns for a pipe of length L, and find the wavelengths and frequencies if:
a) both ends are closedb) both ends are open