Chapter 17

Chapter 17

The Principle of Linear Superposition

and Interference Phenomena

17.1 The Principle of Linear Superposition

When the pulses merge, the Slinkyassumes a shape that is the sum ofthe shapes of the individual pulses.


When the pulses merge, the Slinkyassumes a shape that is the sum ofthe shapes of the individual pulses.


THE PRINCIPLE OF LINEAR SUPERPOSITION

When two or more waves are present simultaneously at the same place,the resultant disturbance is the sum of the disturbances from the individualwaves.

Superposition

http://img.webassign.net/ebooks/cj8/simulations/full/Concept_Simulations/sim25/sim25.html

http://img.webassign.net/ebooks/cj8/simulations/full/Concept_Simulations/sim25/sim25.html

SuperpositionThe drawing shows a graph of 2 pulses traveling toward each other at t = 0s. Each pulse has a constant speed of v = 1 cm/s. When t = 2.0, what is the height of the resultant pulse at a) x = 3 cm, and b) x = 4 cm

SuperpositionThe drawing shows a graph of 2 pulses traveling toward each other at t = 0s. Each pulse has a constant speed of v = 1 cm/s. When t = 2.0, what is the height of the resultant pulse at a) x = 3 cm, and b) x = 4 cm.

Ans: a) -3cm, b)-2 cm

17.2 Constructive and Destructive Interference of Sound Waves

When two waves always meet condensation-to-condensation and rarefaction-to-rarefaction, they are said to be exactly in phase and to exhibit constructive interference.


When two waves always meet condensation-to-rarefaction, they are said to be exactly out of phase and to exhibit destructive interference.


Pilots use noise-canceling headsets like the ones shown below


If the wave patters do not shift relative to one another as time passes,the sources are said to be coherent.

For two wave sources vibrating in phase, a difference in path lengths thatis zero or an integer number (1, 2, 3, . . ) of wavelengths leads to constructive interference; a difference in path lengths that is a half-integer number(½ , 1 ½, 2 ½, . .) of wavelengths leads to destructive interference.


Example 1 What Does a Listener Hear?

Two in-phase loudspeakers, A and B, are separated by 3.20 m. A listener is stationedat C, which is 2.40 m in front of speaker B.

Both speakers are playing identical 214-Hz tones, and the speed of sound is 343 m/s.

Does the listener hear a loud sound, or no sound?


Calculate the path length difference.

m 1.60m 40.2m 40.2m 20.3 22

Calculate the wavelength.

m 60.1Hz 214

sm343

fv

Because the path length difference is equal to an integer (1) number of wavelengths, there is constructive interference, whichmeans there is a loud sound.


Conceptual Example 2 Out-Of-Phase Speakers

To make a speaker operate, two wires must be connected between the speaker and the amplifier. To ensure that the diaphragms of the two speakers vibrate in phase, it is necessary to make these connectionsin exactly the same way. If the wires for onespeaker are not connected just as they are for the other, the diaphragms will vibrateout of phase. Suppose in the figures (next slide), the connections are made so that the speakerdiaphragms vibrate out of phase, everythingelse remaining the same. In each case, what kind of interference would result in the overlap point?


17.4 Beats

Two overlapping waves with slightly different frequencies gives rise to the phenomena of beats.

17.4 Beats

The beat frequency is the difference between the two soundfrequencies:

fbeat = | f1 - f2 |

17.5 Transverse Standing Waves

,4,3,2,1 2

nLvnfnString fixed at both ends

A standing wave will develop

or L = n (λn /2)

Why is that?


• Since T = 1/f, and v = fλ, this leads to:

L = (λn /2)and a standing wave of one segment. How about standing waves of more than one segment?

•In reflecting from the wall, aforward-traveling half-cycle (top 2 waves) becomes a backward-traveling half-cycle (bottom two waves) that is inverted.• If the timing is just right the backward-traveling reflected wave will constructively interfere with the forward-traveling wave.•When does this occur? When the wave period (T) equals the time it takes the wave to travel from the oscillation source to the reflecting point and return:

T = 2L/v


Transverse standing wave patterns.

http://www.walter-fendt.de/ph14e/stwaverefl.htm

http://www.walter-fendt.de/ph14e/stwaverefl.htm

For a string of fixed length L, the boundary conditions can be satisfied only if the wavelength has one of the values

A standing wave can exist on the string only if its wavelength is one of the values shown above. Because λf = v, the frequency corresponding to wavelength λm is

Standing Waves on a String

n (or m) is the harmonic number. You can tell a string’s harmonic number by counting the number of loops or segments on the string.

1. The fundamental mode, with n = 1, has λ1 = 2L, notλ1 = L. Only half of a wavelength is contained between the boundaries, a direct consequence of the fact that the spacing between nodes is λ/2.

2. The frequencies of the normal modes form a series: f1, 2f1, 3f1, …The fundamental frequency f1 can be found as the difference between the frequencies of any two adjacent modes. That is, f1 = Δf = fm+1 – fm.

Standing Waves on a String

Node spacing on a string

EXAMPLE 21.1 Node spacing on a string

Standing Sound Waves• A long, narrow column of air, such as the air in a tube or pipe, can support a longitudinal standing sound wave.

• It is often useful to think of sound as a pressure wave rather than a displacement wave. The pressure oscillates around its equilibrium value.

• Musical instruments use longitudinal standing waves in tubes that are either open at both ends (e.g. flute, organ) or open at one end (e.g. clarinet, trumpet).

•An open end acts as an antinode (maximum vibration) while a closed end acts as a node, just like for a wave on a string.

,4,3,2,1 2

nLvnfn

Tube open at both ends

This is the same as for a transverse standing wave on a string, except that the antinodes are at the ends.

The length of an organ pipeQUESTION:

vsound = 343 m/sAn organ pipe is open at both ends

The length of an organ pipe

,5,3,1 4

nLvnfn

Tube open at one end• The open end is an antinode and the closed end is a node.

•The distance between node and antinode is λ/4.

• Therefore the tube length (L) must be equal to an odd number of quarter wavelengths:

The notes on a clarinet in a tropical country

The notes on a clarinet

The notes on a clarinet

Check your understanding

A tube is found to have standing waves at 390, 520 and 650 Hz and no frequencies in between. The behavior of the tube at frequencies higher/lower than that is not known.

a.Is this an open-open tube or an open-closed tube?

b. How long is the tube?

Check your understanding

A tube is found to have standing waves at 390, 520 and 650 Hz and no frequencies in between. The behavior of the tube at frequencies higher/lower than that is not known.

a.Is this an open-open tube or an open-closed tube? Open-open

b. How long is the tube? – 1.32 m

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