test 2 - today

Physics 1B03summer-Lecture 9

Test 2 - Today

• 9:30 am in CNH-104

• Class begins at 11am


Wave MotionWave Motion

• Sinusoidal waves• Standing Waves


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)


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


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


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.


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


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


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


Exercise (Wave Equation)

What are and k for a 99.7 MHz FM radio wave?


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) !


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 !


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


10 min rest


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


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:


λ2

12

2

2 fLv

f

L

Second Harmonic

1

33

32

3

323

3

23

f

Lvv

f

L

L

Third Harmonic . . . .

λ3


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.).


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:


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


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


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


SolutionSolution


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


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


L

141 L

343 L

545 L

Air Columns: column with one closed end, one open end


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

test 2 - today

Documents

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