Mon Physics 152 Janphysics.valpo.edu/courses/p152/lectures/lecture07notes.pdf · An open organ pipe...

1

Physics 152Monday,

January 29, 2007

• BBeats

• MMusical instruments

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• Hwk #1 due Friday

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.29.

Phys

152

An open organ pipe corresponds to middle C

(261.6 Hz) on the chromatic scale. The third

resonance of a pipe that’s closed at one end and

open at the other has the same frequency. Whatare the lengths of the two pipes? What is the

fundamental frequency of the pipe that is closed at

one end? Use 343 m/s for the speed of sound.

Worksheet Problem #1

Ch. 21: SuperpositionLecture

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152

So, what about two arbitrarily phased wavessuperimposed on one another? What will theresults wave look like?

We’ve now looked at a couple of special cases

In which the waves interfere to cause total

destruction and maximum construction.

Most cases do not fit neatly into one of

these two extreme cases.


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152

We can use the superposition principle to finda functional form of the resulting wave:

Let’s say we have two waves with the same

wavenumber and the same frequency traveling

in the same direction in the same medium. The

only difference will be a phase difference.

(That’s a LOT of caveats, but it makes for

A simple case that we can solve exactly.)

1= Asin(kx t)

2= Asin(kx t + )

= 1 + 2


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152

(x,t) = A[sin(kx t) + sin(kx t + )]

Recalling a principle from trigonometry that

sin + sin = 2cos2

sin+

2

(x,t) = 2Acos2

sin kx t +2

Which implies that in this case, we find that


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152

2

• Notice that the resulting wave has exactly thesame wave number and angular frequency asthe waves from which it is composed.

(x,t) = 2Acos2

sin kx t +2

• When = 0 or 2n (where n is any integer) theamplitude of the resulting wave is 2A.

• That’s the maximum constructive interference

case we examined before.


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152

• When = n (where n = …-3, -1, 1, 3, …),the amplitude of the resulting wave is 0.

• That’s the totally destructive interference

case we examined before.

• Notice that the resulting wave has exactly thesame wave number and angular frequency asthe waves from which it is composed.

(x,t) = 2Acos2

sin kx t +2


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152

• In general, the resulting wave will not be sosimple, but as we saw earlier, we can alwaysfind the resulting wave by using thesuperposition principle and adding the two(or more) waveforms together point by point.

• Let’s look at one more special case that

we can solve exactly. I’ll use sound

waves in air for this example.

• Recall that sound waves are really longitudinal

waves of variable pressure.


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152

We stand in a room at a fixed location andlisten as two tuning forks, each with aslightly different frequency, are struck.

What do we hear?

Wooowwww…..wwwooooowwww….wwwooo…

We hear a sound with varying periodically

varying amplitude. That is to say, the sound

from the two tuning fork combines to get

alternately louder and softer and louder…

Let’s listen….


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152

Since we’re fixed at a single location inthe room, we can describe the pressurevariation arriving at our ears from eachtuning fork as follows:

p

1= A

1sin(

1t)

p

2= A

2sin(

2t)

And to keep life simple, let’s assume that

the two tuning forks are struck so that the

sound emanating from each results in the

same magnitude pressure variation (i.e.,

both forks result in the same volume).


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152

By the superposition principle, theresulting pressure wave (as a function oftime) at your ear is

p = p

1+ p

2= A[sin(

1t) + sin(

2t)]

Using the same trig identity as before, we find

p = 2Acos1 2

2t sin

1+

2

2t

p

1= A

1sin(

1t)

p

2= A

2sin(

2t)


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152

3

Let’s rewrite this as follows

p = 2Acos

1

2t( )sin t( )

where

=1 2

= +1

2 1 2( )

p = 2Acos1 2

2t sin

1+

2

2t


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152

=1 2 = +

1

2 1 2( )

The pitch your

ear hears is the

average of the

two pitches.

The frequency

with which the

amplitude

varies is given

by / 2 .

p = 2Acos

1

2t( )sin t( )


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152

f = / (2 ) =

1 2/ (2 )

f = f

1f

2

Visual Version! Moire Pattern


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152

p

1= sin(2 (440Hz)t)

p

2= sin(2 (445Hz)t)

The resultant sound should have an apparent

pitch equal to the average of these two, which

is 442.5 Hz, and WOW with a frequency of 5 Hz.

Let’s look at the graphs

of pressure vs time.


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152

Component Waves

-3

-2

-1

0

1

2

3

0 0.01 0.01 0.02 0.02 0.03 0.04 0.04 0.05 0.05 0.06 0.07 0.07 0.08 0.08 0.09 0.1

Time (s)

Pre

ssu

re



445 Hz Wave

440 Hz Wave


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Beat Frequency

-3

-2

-1

0

1

2

3

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35

Time (s)

Pre

ssu

re

0.2 s

Yup! That’s 5 Hz!


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4

Worksheet Problem #2


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152

The difference in sound quality has to do

with the permitted harmonics and the

degree to which those harmonics are

supported on the instrument.

How can you tell different instruments

apart even though they may be playing

the same note?


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152

How can you tell different instruments

apart even though they may be playing

the same note?

The violin, on the other hand, produces a

much richer set of supported harmonics

and a more complicated sound pattern.

Our ears easily detect the difference in

these sounds, and hence, we differentiate a

violin’s A note from a flute’s A note.


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152

Here’s a graph of the

sounds associated with

a variety of instruments

all playing the same note.

The image is taken from a paper by

Jean-Marc Bonard, The Physicist’s

Guide to the Orchestra (submitted

to the European Journal of Physics

in 2000).

The “A” note at 440 Hz.

It’s all the extra wiggles

that distinguish the

instruments.

v

r

f

c

s


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152

This chapter is a study of the wave properties

of light. As we saw with other waves, light

waves can interfere with one another, leading

to constructive and destructive interference.

We’ll examine a couple of ways in which such

interference can be produced and demonstrated.

Wave Optics

Ch. 22: Wave OpticsLecture

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152

AND


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5

Recall our example of interference with sound.

Depending on where you are on the field,the voice can sound louder or softer.


07Phys

152Ch. 22: Wave Optics

Lecture

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152

I’ve frequently walked around in a room with 2light bulbs, and I’m gonna tell you, I never evernotice places in the room that are lighter anddarker (unless there are shadows in the room).


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Well, you’re right...

You’ll never detect constructive or destructiveinterference of light waves in a room with twolight bulbs.

Technically, one reason has to do with therelationship between the phase of the 2 bulbs.

For the sound waves from the two speakers, anamplifier is generally attached to both speakers.The amplifier keeps the phase differencebetween the two speakers constant. Sourceswith constant phase differences are known as


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152

For the case of the two light bulbs, however,no single device controls the emission ofphotons from the filament in the two bulbs.

In fact, random fluctuations and unevennessin the heating of the filaments pretty muchguarantee that no matter how hard we mighttry, the light coming from the two bulbs willNEVER have a constant phase difference!The phase of the light from these two sourceswill be changing constantly!

Sources like the light bulbs are called


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Non-coherent sources cannot producea recognizable interference pattern.

If we’re clever, however, we just might beable to think of a way to produce coherentsignals from a light source...

Put up a wall with twoslits in it between youand the light source


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6

In this configuration, it looks like to melike there are two light sources on thewall, one at the location of each slit.

Furthermore, the light from the two sourcesoriginates from the same light bulb! So the“two sources” must be coherent, right?


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152

So, I’ve punched a couple of holes inthis black posterboard. And I’ll setup the light source behind theposter board.Let

’s T

ry!

Will I now get an interference pattern? 1) YES 2) NO

Mak

e a

Predic

tion! Worksheet Problem #3


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152

A black posterwith two dots ofwhite lightshining through.

Which is to say, as I wander around the roomon the far side of the curtain (provided I stayat the same distance from the slits), I don’t findparts of the room brighter and other parts ofthe room darker. In general, the room is darkerthe further away from the slits I am (which iswhat I expected anyway)!


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We might also note that we really never findplaces in the field at which we hear no sound,as suggested by our addition of the waves.

d(le

ft) d(right)

+There really isno place on thefield where wehear nothing…unless…


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Our example assumed that the speakerswere emitting a SINGLE frequency.

A similar thing is happening with thelight from the incandescent bulb...

In reality, a human voice or piece of musiccontains many frequencies that are transmittedsimultaneously.

Because each of these frequencies has an asso-ciated wavelength, all of which will be slightlydifferent from one another, the addition of thesewaves cannot produce 0 for all frequencies atthe same time in the same place!


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White light from the bulb actually containslight of a variety of wavelengths (justrecall what happens when white lightpasses through a prism).

To best observe the interference, therefore,we really need both sources of light tocoherently emit the same, single wavelength.

A light source that emits a single wavelengthof light is known as a monochromatic source.

One good example of a nearly monochromaticlight source is a laser (e.g. my laser pointer).


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152

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