phys202 –spring 2008 - home - university of delaware dept. of physics...

2/8/2009

1

PHYS202 – SPRING 2008Lecture notes ‐Waves

Nature of Waves

• A wave is a travelling disturbance that carries energy.

• The disturbance is a displacement from the rest position.

• One type of wave is a mechanical disturbance in the media the wave is propagating through.

2/8/2009

2

Mechanical Waves

• Mechanical waves come in two types (or sometimes a combination of the two)– Transverse: Displacement is perpendicular to the direction of propagation.

– Longitudinal: Displacement is parallel to the direction of propagation.

Wave on a string

• Let us take a string as an example: an external force moves a portion of the string. Due to the molecules of the string being bonded, the portion(s) of the string are pulled from the rest position as well. Each portion of the string exerts a force on the adjacent portions (Tension).

• The disturbance reaches a maximum and then returns towards the rest position. Notice that each portion of the string has mass, therefore momentum, so there is a delay in the movement in adjacent portions of the string.

• This causes the disturbance to move along the string (propagation) with a definite velocity.

F

F

v

2/8/2009

3

Wave on a string

• The velocity of the wave is determined by the laws of mechanics.

• A strict derivation requires calculus and can be found on the web, but for our purposes we will just use the result.

v

/Tension F

linear density m L= =υ

Periodic Wave Motion

0.5 1.0 1.5 2.0 2.5 3.0

-1.0

-0.5

0.5

1.0y

t

When a wave repeats in time and space, it is called Periodic.

y

x

The Period (T ) is from any point to where the wave repeats in time.

The Wavelength (λ ) is from any point to where the wave repeats in space.

Note that the wavelength and period do not depend on where or when itis measured.

A is the Amplitude, which is the maximum displacement from the “rest” position.

λ

T

TT

λ

λ

A

A

2/8/2009

4


A

B

C

v

AB

C

•A is moving UP•B is moving UP•C is moving DOWN

A transverse wave moves down as string as shown.Which way are the points moving?


The speed of sound in a certain metal block is 3.00 × 103 m/s. The graph shows the amplitude (in meters) of a wave traveling through the block versus time (in milliseconds). What is the wavelength of this wave?

y (m)

0

1

–1

t (ms)0.5 1.5 3.52.5 4.5

f or Tf

λ λ λ= ⇒ = =υ

υ υ

By inspection the period is 2.0 x 10‐3 sec

( ) ( )3 3s m / s 6.0 mλ −= × × =2.0 10 3.00 10( )( )3 33.00 10 m / s 2.0 10 s 6mTλ −= = × × =υ

2/8/2009

5


The displacement of a vibrating string versus position along the string is shown in the figure. The periodic waves have a speed of 10.0 cm/s. A and B are two points on the string.

1.5 4.5 7.5 10.5 13.5 16.50

2

−2

4

−4

y(mm) x (cm)

A

B

4 mm6.0 cmf = v /λ =1.7 Hz

π radians

•What is the Amplitude of the Wave?•What is the Wavelength of the Wave?•What is the frequency of the Wave?•What is the difference in phase between the points A and B?

Mathematical Wave Equation

• Describes a wave is space (x, y) and time (t)• Three parameters

– Amplitude– Wavenumber– Angular frequency

Akω

2sin 2Ay tfxλπ π⎛ ⎞= ±⎜ ⎟

⎝ ⎠

( )sinA ty kx ω= ±

2 22k fPeriod

π πω πλ

= = =

2/8/2009

6


• Why the factors of 2π?

• For one complete wavelength and one complete period, there are 2π radians.

• Often seen as converted to wave number k = 2π/λ and angular frequency ω = 2πf.


⎝ ⎠

0.5 1.0 1.5 2.0 2.5 3.0

-1.0

-0.5

0.5

1.0

y

x

at t = 0

( )sinAy tkx ω= ±


• Let A = 1, λ = 1 and f = 1/3 and let us choose the minus sign.

• Now, let time progress.


⎝ ⎠

2sin 23

y x tππ⎛ ⎞= −⎜ ⎟⎝ ⎠

Wave travelsTo the right (+ x)direction

2/8/2009

7


• Again, let λ = 1 and f = 1, now we choose the plus sign.

• Letting time progress.


⎝ ⎠

2sin 23

y x tππ⎛ ⎞= +⎜ ⎟⎝ ⎠

Wave travelsTo the left (‐ x)direction


• Use the standard form of the wave equation:

• Amplitude is 0.010 m.

• Wavelength is 0.040 meters.

• Period is 0.20 s, so frequency is (1 / 0.20) = 5 Hz.

• Looking at how the point x = 0 behaves, we se that it becomes positive, so the wave moves to the right (minus sign).


⎝ ⎠

Problem 16.4 – Write the mathematical expression for this wave.

( ) ( )0.010 sin 50 10y m x tπ π= −

2/8/2009

8

Sound

• Sound is a longitudinal wave

• One moment in time: Condensation (high pressure)

Rarefaction (low pressure)

Sound

• Graphing a sound wave:

2/8/2009

9

Speed of Sound in an Ideal Gas

• Based on Thermodynamics.

• Use Kelvin scale for Temperature( Kelvins = 273.15 + ˚C ).

• Based on average particle velocity:

• So when the particles have no velocity, the temperature is absolute zero.

• Due to compression and expansion laws, the sound speed is not the same as particle speed.

3kTm

υ=

Speed of Sound in an Ideal Gas

• Actual speed of sound in an ideal gas:

• Some texts will use c for speed of sound.

• T is temperature in Kelvins

• m is (average) particle mass.

• k is Boltzmann’s constant (1.38 x 10‐23 J/K)

• γ is a constant dependent on the gas (5/3 for monatomic gases, 7/5 for diatomic gases)

kTmγ

υ=

2/8/2009

10

Speed of Sound in Air

• Let’s assume air is made of 78% N2 and 22% O2.

• Checking the atomic mass of Nitrogen (14 u ) and Oxygen (16 u) and using the conversion 1 u =1.0665 x 10‐27 kg.

• There are two atoms per molecule, so, γ = 7/5.• Boltzmann’s constant is k = 1.38 x 10-23 J/K.• If we take air to be 0˚ C then

kTmγ

υ=

( )( )( )( )( )

23

27

7 / 5 1.38 10 J / K 273 0

0.78 28 0.22 32 u 1.6605 10 kg/u

m / s

−

−

× +

× + × ×υ=

υ=332

vsound in Liquids and Solids

• Similar to the string, has two components

• For sounddensity = inertial property

• From section 10.7 (as seen in PHYS201), when an increase in pressure is applied to all sides of some material (liquid or solid), the relative amount of deformation is called the Bulk Modulus.

Tension Elastic propertylinear density inertial property

=υ=

2/8/2009

11


• The Bulk Modulus is

• So,

where B is the Bulk modulus and ρ is the density of the material.

pressureBVolume

Volume

Δ=

Δ⎛ ⎞⎜ ⎟⎝ ⎠

Bρ

=υ


• For a liquid, we use the adiabatic Bulk modulus

• For a solid, we use Young’s modulus.

adBρ

=υ

Yρ

=υ

2/8/2009

12


35.04 10 m / s 344 m / srail air= × =υ υ

Problem 16.32 ‐ Have you ever listened for an approaching train by kneeling next to a railroad track and putting your ear to the rail? Young's modulus for steel is Y = 2.00 x 1011 N/m2, and the density of steel is = 7860 kg/m3. On a day when the temperature is 21.3o C, how many times greater is the speed of sound in the rail than in the air?

So, the ratio of 35.04 10 m / s 14.6

344 m / srail

air

×= =

υ

υ

rail airY kT

mγ

ρ=υ υ =

( )( )( )( )( )

2311 2

3 27

7 / 5 1.38 10 J / K 273 21.3 K2.00 10 N / m7860 kg / m 28.9 u 1.6605 10 kg/urail air

−

−

× +×= =

×υ υ

Sound Intensity

• How much Amplitude does the sound wave have?

• Sound radiates in three dimensions.

• Conservation of Energy – what goes through one surface must pass through the next surface.

2/8/2009

13

Sound Intensity

• Intensity is Power per unit area:

• For sound from a point source, with no reflection from any surface, the area is that of a sphere:

• So the Intensity at any point is

PIA

=

24A rπ=

24PIrπ

=

Sound Intensity

( ) ( )2 22 2 1 14 4I r I rπ π=

( )( )

( ) ( )( )

21

2 1 22

222 21

2 1 222

4

4

22 m3.0 10 W / m

78m

rI I

r

rI Ir

π

π

−

=

= = ×

Problem 16.51 ‐ Suppose that sound is emitted uniformly in all directions by a public address system. The intensity at a location 22 m away from the sound source is 3.0 x 10‐2 W/m2. What is the intensity at a spot that is 78 m away?

Relate Power to Intensity: ( )22 4

4PI P I rr

ππ

= ⇒ =

The Power of the source is constant, thus

3 22 2.4 10 W / mI −= ×

2/8/2009

14

Decibels

• Compares two sound intensities.

• 0 dB is set at the “threshold of human hearing” (10‐12

W/m2).

• Human hearing responds logarithmically to sound intensity.

• Generally speaking, if you increase the sound intensity by a factor of 10, it seems to be twice as loud.

( )0

10dB log II

β⎛ ⎞

= ⎜ ⎟⎝ ⎠

Decibels

( ) ( )2 12 1

0 0

10dB log 10dB logI II I

β β⎛ ⎞ ⎛ ⎞

− = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

How much more intense in dB is 5.0 x 10‐8 W/m2 than 7.0 x 10‐9 W/m2?

( ) ( ) ( )2 1 2 1 2 0 1 010dB log / log /I I I Iβ β β β ⎡ ⎤− = − = −⎣ ⎦

( ) 2 02 1

1 0

/10dB log/

I II I

β β⎛ ⎞

− = ⎜ ⎟⎝ ⎠

( ) 22 1

1

10dB log II

β β⎛ ⎞

− = ⎜ ⎟⎝ ⎠

( )8 2

2 1 9 2

5 10 W / m10dB log 8.5dB7 10 W / m

β β−

−

⎛ ⎞×− = =⎜ ⎟×⎝ ⎠

2/8/2009

15

Decibels

( )0

10dB log II

β⎛ ⎞

= ⎜ ⎟⎝ ⎠

A stereo speaker produces 100 watts of power. What is the sound level in dB at a distance of 14 m (assuming no reflected sound)?

24PIrπ

=

( )2 20

10dB log4

Pr I

βπ

⎛ ⎞= ⎜ ⎟

⎝ ⎠

( )( ) ( )2 12 2

100 W10dB log4 14 m 10 W / m

βπ −

⎛ ⎞⎜ ⎟=⎜ ⎟⎝ ⎠

106dBβ =

Human HearingThe response of the human ear changes with frequency. The phon scale is purely psychological and measures perceived loudness. The dB and phon scales are normalized at 1 kHz.

2/8/2009

16

The Doppler Effect

• Occurs when an source and/or observer is in motion relative to the medium transmitting the sound.

• This causes a shift in frequency (and thus wavelength) observed.

• For mechanical waves, the shift is different for observer in motion versus source in motion.

The Doppler Effect – moving source

λ0 λ0 λ0

Source at rest Source in motion

rest wavelengthThe wavelengths in the direction of motion are shorter than the rest wavelength.The wavelengths in the opposite direction of motion are shorter thanthe rest wavelength.

2/8/2009

17

The Doppler Effect –moving source, stationary observer

• In the medium, each successive wave is closer (farther) than the previous by the velocity of the wave times its period.

• So the new wavelength is , where vs is the velocity of the source.

• It is customary to work with frequency, so

• T = 1/f and λ = v/f,

• Multiply by

• The negative sign is for an approaching source the positive for a receding source.

sTλ λ′ = ∓ υ

s

fTλ λ

′ = =′ ∓

υ υυ

sf

f f

′ =∓

υυυ

/1/ s

f fff

′= ⇒ =∓

υυυ 1υ

The Doppler Effect –stationary source, moving observer

• In the medium, the wavelength remains unchanged, but the observer encounters each wave faster (slower), so the frequency is higher (lower).

• So the new frequency is , where vo is the velocity of the observer.

• Rewrite the equation:

• In the medium, we have, fλ = v, the velocity of sound.

• The positive sign is for an observer approaching the source and the negative for a moving away from the source.

of fλ

′ = ±υ

1 of ff λ

⎛ ⎞′ = ±⎜ ⎟⎝ ⎠

υ

1 of f ⎛ ⎞′ = ±⎜ ⎟⎝ ⎠

υ

υ

2/8/2009

18

The Doppler Effect –moving source, moving observer

• If the source and observer are both moving with respect to the medium, then apply the equations sequentially:

• Source moving: becomes the frequency f in the observer moving

equation:

• Rewriting:

• Note that the substitution of moving observer frequency into moving source frequency works the same!

1 1o o

s

ff f f⎛ ⎞ ⎛ ⎞′′ ′ ′′= ± ⇒ = ±⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∓

υ υυυ υ1υ

s

ff ′ =∓ υ

1υ

1 o

sf f

⎛ ⎞±⎜ ⎟′′ = ⎜ ⎟

⎜ ⎟⎝ ⎠∓

υυυ

1υ

The Doppler Effect

16.72 ‐ You are flying in an ultra‐light aircraft at a speed of 39 m/s. An eagle, whose speed is 18 m/s, is flying directly toward you. Each of the given speeds is relative to the ground. The eagle emits a shrill cry whose frequency is 3400 Hz. The speed of sound is 330 m/s. What frequency do you hear?

Use the Doppler shift equation:

( ) 3

39 m / s11 m / s3400 Hz 4.0 10 Hz18m / sm / s

o

sf f

⎛ ⎞⎛ ⎞ +± ⎜ ⎟⎜ ⎟′ ⎜ ⎟= = = ×⎜ ⎟

⎜ ⎟⎜ ⎟ −⎜ ⎟⎝ ⎠ ⎝ ⎠∓

υ330υ

υ11

330υ

Note the signs – the observer is approaching the source and the source isapproacing the observer!

Your calculator may have said the answer is 4021 Hz, but remember we only havetwo significant digits!

2/8/2009

19

Sonic Boom

• A sonic boom occurs when a source is movingfaster than the speed of sound in themedium.

• This causes wave fronts to “build up”creating a Mach cone.

• The wavelengths behind are longerthe same way as for sub‐sonic.

• The wavelengths in front are nownegative!

λ0 λ0

Sonic Boom

• The blue line indicates where the wave fronts from each successive period are the same!

• The angle θ, can be calculated. θ

2/8/2009

20

Sonic Boom

• The blue line indicates where the wave fronts from each successive period are the same!

• The angle of the Mach cone, θ, canbe calculated.

• The Mach cone and direction ofsound propagation is perpendicular, so using a little trigonometry:

• Note that if the velocity of the source equals the velocity of sound, the angle is a right angle and if the velocity of the source is less than sound, there is no Mach cone (i.e. it is undefined).

θvs

v

1sin sins s

θ θ − ⎛ ⎞= ⇒ = ⎜ ⎟

⎝ ⎠

υ υυ υ

Sonic Boom

A jet moving slightly faster than the speed of sound shows several Mach cones (two visible from this angle).

The white material is water vapor condensed out of the air.

2/8/2009

21

When Waves Meet

• When two or more waves coincide, add their Amplitudes to get the result.

• The purple and blue waves coincide at some point in space. Each wave has the same wavelength and amplitude.

• The green wave is the result.

• This is known as the principle of Linear Superposition.

When Waves Meet – examples

• Purple and blue are the waves with identical wavelength and amplitude that meet – green is the result.

Completely in phase and Completely out of phase

2/8/2009

22

When Waves Meet – examples

• Purple and blue are the waves with the same amplitude but different wavelengths that meet – green is the result.

• The result can be very complex waves.

• In problems, we will deal with the interference at a position in space from two different sources with identical wavelengths only (and usually assume comparable or identical amplitudes).

Determining points of Constructive and Destructive Interference.

• Let us have two sources of sound.

• Waves radiate in all directions from each source.

• If a point is located such that the difference in the distances to each source is equal to an integer number multiple of wavelength, there is completely constructive interference.

• If, however, a point is located such that the difference in the distances to each source is equal to (an integer number + ½) multiple of the wavelength, there is completely constructive interference.

• If neither of these conditions apply it is considered partially constructive (or destructive) interference.

Source 1 Source 2

d1 d2( )

( )

1 2

1 2

0, 1, 2, 3,1 0, 1, 2, 3,2

Completly Constructive Interference d d n n

Completly Destructive Interference d d n n

λ

λ

− = =

⎛ ⎞− = + =⎜ ⎟⎝ ⎠

…

…

2/8/2009

23

Diffraction

• Occurs when waves (e.g. sound) “bends” around objects.

Diffraction

• When a wave passes through an hole in a barrier, it experiences diffraction.

• This is due to every point in the gap acting as a point source of the wave.

• Shown is a plane wave encountering a rectangular opening.

• It is easy to see that near the center of the opening, the waves will interfere such that it will still be “plane‐like”.

• Away from the slit (above and below here), there is less interference and the waves travel up and down.

2/8/2009

24

Diffraction

• Normally the cancel, but only portion of wave comes through slit, yielding constructive and destructive interference.

• Where is first angle of destructive interference?

θ

Diffraction

• This calculation assumes that the distance we are from the slit is much larger than the width of the slit .

• If we want to find the angle of completely destructive interference, then the outgoing wave, every point in the slit must exactly cancel out another.

• Thus, the length to a wave front from one side of the slit must be an integer multiple longer than the other side!

θ

θ

2/8/2009

25

Diffraction

θ

D

θ

Let the width of the slit be D.

Using a little trigonometry wefind the difference in distanceto either side of the slit isD sin θ .

Which must be one wavelength.

So,

sin

sin

D

D

θ λλθ

=

=

Diffraction

• We see that the angle of the first destructive node is given by

• But higher order destructive nodes can exist. Again, any non‐zero integer multiple of the wavelength that gives a real result yields a node.

• But, of course , for a real answer.

• It can be the case that there are no destructive nodes if .

sinDλθ =

sin 1, 2, 3,n nDλθ = = …

n Dλ ≤

Dλ >

2/8/2009

26

When Waves Meet in Time

• Let us take two waves of identical amplitude but different wavelengths (i.e. frequencies).

• This has the effect of making a single wave change in amplitude.

• This variation in amplitude is known as Beats.

• The rate at which the amplitude changes is known as the beat frequency.

• Here, the beat frequency would be 0.05 Hz.

Beat Frequency

• The beat frequency can be calculated from the frequencies of the combining waves.

• Here, one wave has a frequency of 1 Hz, while the other is 0.95 Hz.

• The beat frequency is the absolute value of the difference of the wave frequencies:

1 2beatf f f= −

2/8/2009

27

Reflection of Waves

• When a wave encounters a hard surface (i.e. one that will not move). The wave is reflected back with inverse amplitude (i.e. upside down).

• See the animation on‐line for waves on a string approaching a fixed point and reflected back.

• If the end is free to move, the reflected wave is not inverted (remains right side up).

Standing Waves on a String

• We now take the case where a string has two fixed ends.

• The wave will travel to one end, become inverted and travel back to the other end and repeat the process.

• So, what are the wavelengths (frequencies) which cause standing waves to appear on the string?

• We look for waves with wavelengths that has “rest” points the length of the string.

• In the above case, the wavelength is exactly one‐half the length of the string.

• Note that every half of a wavelength can fit the criteria.

2/8/2009

28


• So a the wavelength of a standing wave on a string must be , where

l is the length of the string and n is a positive integer (n = 1, 2, 3, …).

• Nodes occur where the string does not deviate from rest, Antinodes occur where the string has maximum deviation from rest.

• Note the number of Antinodes equals the value n. The number of nodes is n – 1.

2n

λ =

l

Nodes

Antinodes

2λ =

λ =

23

λ =

1n =

2n =

3n =


• Often we are looking for the frequency of the wave which isagain where n is a positive integer.

• Here, the frequency is , where the wave speed would be determined by

the characteristics of the string.

2nf

λ= =

υ υ

l 23

λ =3n =32

f = υ

2/8/2009

29

Longitudinal Standing Waves

• Longitudinal waves can also be standing.

• This could be a case of a spring with fixed ends, or sound waves in a solid, or sound waves in a hollow a pipe with closed ends, open ends or one open end the other closed.

• We will look carefully at the case of the pipe with either open or closed ends (or one of each).

• As with the case of the string with fixed ends, a closed end of a pipe will not move, thus, a node will occur on that end.

• As you may expect, the wavelength and frequency of a standing wave(s) will follow the same formula: 2

nλ =

2nf

λ= =

υ υ

l


• For an open end of a pipe, an antinode must be at that point (i.e. maximum amplitude).

• As you may expect the wavelength and frequency of a standing wave(s) will follow the same formula:

2n

λ =2nf

λ= =

υ υ

l

2/8/2009

30


• With one end close and the other open, we need a node at one end and an antinode at the other.

• Here we observe that the wavelength must be one‐quarter shorter than if we had two closed (or two open) ends:

( )2 4

2 1 44 2 1n

n nλ λλ λ −

⇒ =−

= =−

( )2 1where is a positive integer .

4n n

f nλ

−= =

υυ

l


• Cutnell and Johnson don’t explain how to get to the answer, but if you take the

series (2n – 1) and count by integers (i.e. 1, 2, 3, 4,…), you simply get a series of

odd integers (i.e. 1, 3, 5, 7,…).

• Thus the book formula: where 1, 3, 5,4nf n= = …υ

l

phys202 –spring 2008 - home - university of delaware dept. of physics...

Documents