PHYS 1443 – Section 501 Lecture #1yu/teaching/.../lectures/phys1443-fall03-120303-post.pdf ·...

PHYS 1443 – Section 003Lecture #24

Wednesday, Dec. 3, 2003Dr. Jaehoon Yu

1. Sinusoidal Waves2. Rate of Wave Energy Transfer3. Reflection and Transmission4. Superposition and Interference

Wednesday, Dec. 3, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu

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Announcements• The final exam

– On Monday, Dec. 8, 11am – 12:30pm in SH103.– Covers: Chap. 10 not covered in Term #2 – Ch15

section 8.• Lab grades are available


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Sinusoidal WavesEquation of motion of a simple harmonic oscillation is a sine function.


3

A

-A

But it does not travel. Now how does wave form look like when the wave travels?

2( ) siny x A xπλ

⎛ ⎞= ⎜ ⎟⎝ ⎠

The wave form can be described by the y-position of the particle at x, of the medium through which the sinusoidal wave is traveling can be written at t=0

Wave Length

The wave form of the wave traveling at the speed v in +x at any given time t

( )2( , ) siny x t A x vtπλ

⎛ ⎞= −⎜ ⎟⎝ ⎠

Amplitude

x

After time t the original wave form (crest in this case) has moved to the right by vt thus the wave form at x becomes the wave form which used to be at x-vt

vt

Sinusoidal Waves cont’d

Tv λ

=By definition, the speed of wave in terms of wave length and period T is

Thus the wave form can be rewritten ( , ) sin 2 x ty x t A

Tπ

λ⎡ ⎤⎛ ⎞= −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

Defining, angular wave number k and angular frequency ω, λ

π2≡k

( )( , ) siny x t A kx tω= −

Frequency, f,


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The wave form becomesT

f 1= Wave

speed, v v

( )( , ) siny x t A kx tω φ= ± +

Tπω 2

=

Tλ

=kω

=

Travels to left

Travels to right

Generalized wave form

Example for WavesA sinusoidal wave traveling in the positive x direction has an amplitude of 15.0cm, a wavelength of 40.0cm, and a frequency of 8.00Hz. The vertical displacement of the medium at t=0 and x=0 is also 15.0cm. a) Find the angular wave number k, period T, angular frequency ω, and speed v of the wave.

Using the definition, angular wave number k is

b) Determine the phase constant φ, and write a general expression of the wave function.

mradk /7.1500.540.0

22==== ππ

λπ

sradfT

/3.5022=== ππωAngular

frequency is sec125.0

00.811

===f

TPeriod is

smfT

v /2.300.8400.0 =×=== λλUsing period and wave length, the wave speed is

At x=0 and t=0, y=15.0cm, therefore the phase φ becomes

( )0.150sin 0.150y φ= =

( ) ⎟⎠⎞

⎜⎝⎛

2+−=+−=

πφω txtkxAy 3.507.15sin150.0sin

sin 1;φ = πφ =2

Thus the general wave function is


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Sinusoidal Waves on StringsLet’s consider the case where a string is attached to an arm undergoing a simple harmonic oscillation. The trains of waves generated by the motion will travel through the string, causing the particles in the string to undergo simple harmonic motion on y-axis.

⎟⎠⎞

⎜⎝⎛= xAy

λπ2sin What does this mean?

( )tkxAy ω−= sin

0=φIf the wave at t=0 is

The wave function can be written

This wave function describes the vertical motion of any point on the string at any time t. Therefore, we can use this function to obtain transverse speed, vy, and acceleration, ay.

xconstdtdy

=ty

∂∂

= ( )tkxA ωω −−= cosxconst

y

dtdv

=t

vy

∂

∂= ( )tkxA ωω −−= sin2

yv ya

The maximum speed and the acceleration of the particle in the medium at position x at time t are

max,yv How do these look for simple harmonic motion?

These are the speed and acceleration of the particle in the medium not of the wave.

Aω=

max,ya A2ω=Wednesday, Dec. 3, 2003 PHYS 1443-003, Fall 2003

Dr. Jaehoon Yu6

Example for Wave FunctionsA string is driven at a frequency of 5.00Hz. The amplitude of the motion is 12.0cm, and the wave speed is 20.0m/s. Determine the angular frequency ω and angular wave number k for this wave, and write and expression for the wave function.Using frequency, the angular frequency is

Angular wave number k is

ω 2π=

Τ2 fπ= 2 5.00 31.4 /rad sπ= ⋅ =

2 fvπ

=2πλ

=2vTπ

=31.4 1.57 /20.0

rad m= =vω

=k

Thus the general expression of the wave function is

( ) ( )txtkxAy 4.3157.1sin120.0sin −=−= ω


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Rate of Energy Transfer by Sinusoidal Waves on StringsWaves traveling through a medium carry energy.

When an external source performs work on the string, the energy enters into the string and propagates through the medium as wave.

What is the potential energy stored in one wave length of a traveling wave? ∆x, ∆m Elastic potential energy of a particle in a simple harmonic motion 2

21 kyU =

Since ω2=k/m 2

21 ymU 2= ω The energy ∆U of the segment ∆m is U∆


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As ∆x 0, the energy ∆U becomes dxydU 2

21 2= µω

Using the wave function,the energy is dU

2

21 ym 2∆= ω 2

21 yx 2∆= ωµ

( )dxtkxA ωµω −= 2 22 sin21

For the wave at t=0, the potential energy stored in one wave length, λ, is

∫=

=

2=λ

µωx

xkxdxA

0

22 sin21

∫=

=

2 −=

λµω

x

xdxkxA

0

2

22cos1

21

λU

λ

λπµω

=

=

2⎥⎦⎤

⎢⎣⎡ −=

x

x

xk

xA0

2 4sin41

21

21 λµω 2

41 A2=Recall k=2π/λ

Rate of Energy Transfer by Sinusoidal Waves cont’dHow does the kinetic energy of each segment of the string in the wave look?

dK

λK

Since the vertical speed of the particle is


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As ∆x 0, the energy ∆K becomes

For the wave at t=0, the kinetic energy in one wave length, λ, is

Recall k=2π/λ

yv

The kinetic energy, ∆K, of the segment ∆m is K∆

Just like harmonic oscillation, the total mechanical energy in one wave length, λ, is λE

As the wave moves along the string, the amount of energy passes by a given point changes during one period. So the power, the rate of energy transfer becomes

P P of any sinusoidal wave is proportion to the square of angular frequency, the square of the amplitude, the density of medium, and wave speed.

( )tkxA ωω −−= cos

2

21

ymv∆= ( )tkxAx ωωµ −∆= 222 cos21

( )dxtkxA ωµω −= 2 22 cos21

∫=

=

2=λ

µωx

xkxdxA

0

22 cos21

∫=

=

2 +=

λµω

x

xdxkxA

0

2

22cos1

21

λ

λπµω

=

=

2⎥⎦⎤

⎢⎣⎡ +=

x

x

xk

xA0

2 4sin41

21

21 λµω 2

41 A2=

λλ KU += λµω 2

21 A2=

tE∆

= λ

TA λµω 2

21 2=

vA2

21 2= µω

Example for Wave Energy TransferA taut string for which µ=5.00x10-2 kg/m is under a tension of 80.0N. How much power must be supplied to the string to generate sinusoidal waves at a frequency of 60.0Hz and an amplitude of 6.00cm?

µT

= sm /0.401000.50.80

2 =×

= −The speed of the wave is v

Using the frequency, angular frequency ω is

ωΤ

=π2 fπ2= srad /3770.602 =⋅= π

Since the rate of energy transfer is

Pt

E∆

= λ vA2

21 2= µω

( ) ( ) ( ) W5120.4006.03771000.521 222 =×××××= −


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Reflection and TransmissionA pulse or a wave undergoes various changes when the medium it travels changes.Depending on how rigid the support is, two radically different reflection patterns can be observed.


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1. The support is rigidly fixed: The reflected pulse will be inverted to the original due to the force exerted on to the string by the support in reaction to the force on the support due to the pulse on the string.

2. The support is freely moving: The reflected pulse will maintain the original shape but moving in the reverse direction.

Transmission Through Different MediaIf the boundary is intermediate between the previous two extremes, part of the pulse reflects, and the other undergoes transmission, passing through the boundary and propagating in the new medium.

When a wave pulse travels from medium A to B:• vA> vB (or µA<µB), the pulse is inverted upon reflection.• vA< vB(or µA>µB), the pulse is not inverted upon reflection.


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Superposition Principle of WavesSuperposition Principle

If two or more traveling waves are moving through a medium, the resultant wave function at any point is the algebraic sum of the wave functions of the individual waves.The waves that follow this principle are called linear waves which in general have small amplitudes. The ones that don’t are nonlinear waves with larger amplitudes.

Thus, one can write the resultant wave function as

y nyyy +⋅⋅⋅++= 21 ∑=

=n

iiy

1


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Wave InterferencesTwo traveling linear waves can pass through each other without being destroyed or altered.

What do you think will happen to the water waves when you throw two stones in the pond?

They will pass right through each other.

The shape of wave will change InterferenceWhat happens to the waves at the point where they meet?

Constructive interference: The amplitude increases when the waves meet


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Destructive interference: The amplitude decreases when the waves meet

In phase constructive Out of phase by π/2 destructive Out of phase not by π/2 Partially destructive


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Congratulations!!!!You all have done very well!!!

I certainly had a lot of fun with ya’ll!

Good luck with your exams!!!Happy Holidays!!

PHYS 1443 – Section 501 Lecture #1yu/teaching/.../lectures/phys1443-fall03-120303-post.pdf ·...

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