16.9 The Doppler Effect

MOVING SOURCE

The effect of moving source is SIMILAR, but not identical to the case of the moving observer. They give difference relationships between the the observed frequency fO and the source frequency fS. MOVING OBSERVER

λ ′ = λ − v sT fo =

v v v = = λ ′ λ − v sT v f s − v s f s

 1   f o = f s   1 − vs v 

fo =

v + vo

λ

=

 f s v + f s vo v  = f s 1 + o  f sλ f sλ  

 vo  f o = f s 1 +  v 

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16.9 The Doppler Effect source moving toward a stationary observer

source moving away from a stationary observer

 1   f o = f s   1 − vs v   1   f o = f s   1 + vs v 

GENERAL CASE

 vo 1± v fo = f s   1  vs  v 

     

Observer moving towards stationary source

Observer moving away from stationary source

 vo  f o = f s 1 +  v   vo  f o = f s 1 −  v 

Numerator: plus sign applies when observer moves towards the source v vS vO

speed of the sound in medium speed of source rel. to medium speed of observer rel. to medium

Denominator: minus sign applies when source moves towards the observer

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16.9 The Doppler Effect

Example:

The Sound of a Passing Train

A high-speed train is traveling at a speed of 44.7 m/s when the engineer sounds the 415-Hz warning horn. The speed of sound is 343 m/s. What are the frequency and wavelength of the sound, as perceived by a person standing at the crossing, when the train is (a) approaching and (b) leaving the crossing?

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16.9 The Doppler Effect

Example:

The Sound of a Passing Train

A high-speed train is traveling at a speed of 44.7 m/s when the engineer sounds the 415-Hz warning horn. The speed of sound is 343 m/s. What are the frequency and wavelength of the sound, as perceived by a person standing at the crossing, when the train is (a) approaching and (b) leaving the crossing? This is the case of moving source ( ~100.6 mph or Mach 0.13), stationary observer . We always assume, by default in such problems, that there is no wind. (a) Train approaching

(b) Train leaving

 1   f o = f s   1 − vs v 

 1   f o = f s   1 + vs v 

  1   f o = (415 Hz )  1 − 44.7 m s  343m s   = 477 Hz

  1   f o = (415 Hz )  1 + 44.7 m s  343m s   = 367 Hz

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17.1 The Principle of Linear Superposition

When the pulses merge, the Slinky assumes a shape that is the sum of the shapes of the individual pulses.

Chapter 17: Linear Superposition and Interference

http://www.ablongman.com/mullin/ AnimaImages/ConsInterf.gif

In this case, the pulses (both upward) reinforce one another and we have “constructive interference” 5

17.1 The Principle of Linear Superposition

When the pulses merge, the Slinky assumes a shape that is the sum of the shapes of the individual pulses.

http://www.ablongman.com/mullin/ AnimaImages/DesIntef.gif

In this case, the pulses (one up, one down) work against one another, and cancel each other at one instant. Here we have “destructive interference”

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Constructive and Destructive interference Video

http://www.youtube.com/watch?v=P_rK66GFeI4

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17.2 Constructive and Destructive Interference of Sound Waves

When two waves always meet condensation-to-condensation (at a particular location) and rarefaction-to-rarefaction, they are said to be exactly in phase and to exhibit constructive interference. The locations of these “antinodes” depends on: (a) the wavelength (assumed to be the same for both sources), (b) the placement of the sources, and (c) the RELATIVE PHASE between the sources.

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17.2 Constructive and Destructive Interference of Sound Waves

When two waves always meet condensation-to-rarefaction (again, at a particular location), they are said to be exactly out of phase and to exhibit destructive interference. The locations of these “nodes” depends on: (a) the wavelength (assumed to be the same for both sources), (b) the placement of the sources, and (c) the RELATIVE PHASE between the sources.

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The nodes are relatively easy to spot in this “ripple tank” demonstration of two “in-phase” sources.

http://www.youtube.com/watch?v=5PmnaPvAvQY

If the wave patterns (e.g. the nodes) do not shift relative to one another as time passes, the sources are said to be coherent (as they are here!)

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17.2 Constructive and Destructive Interference of Sound Waves

The math of interference For two wave sources vibrating in phase (producing

condensations/crests at the same time), what matters is ∆s, the difference in the path length from the two sources to a given location, compared to the wavelength λ. (a) a difference in path lengths that is zero or an integer number (0, ±1, ± 2, ±3, . . ) of wavelengths leads to constructive interference (b) a difference in path lengths that is a half-integer number(±1/2, ±3/2, ±5/2, . .) of wavelengths leads to destructive interference. For two wave sources vibrating in opposite phase (one producing a condensation/crest while the other produces a rarefaction/trough): (a) ∆s/λ = 0, ±1, ± 2, ±3 gives destructive interference (b) ∆s/λ = ±1/2, ± 3/2, ±5/2 gives constructive interference 11

17.2 Constructive and Destructive Interference of Sound Waves Simple Example : What Does a Listener Hear? Two in-phase loudspeakers, A and B, are separated by 3.20 m. A listener is stationed at 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? Solution: very loud = constructive interference No sound = destructive interference First calculate the path length difference. ∆s

∆s = a − b =

(3.20 m )2 + (2.40 m )2 − 2.40 m = 1.60 m Next: calculate the wavelength.

v 343 m s = = 1.60 m f 214 Hz → ∆s / λ = 1.00

λ=

Because the path length difference is equal to an integer (1) number of wavelengths, there is constructive interference, which means there is a 12 loud sound.

17.2 Constructive and Destructive Interference of Sound Waves

Out-Of-Phase Speakers? To make a speaker operate, two wires (a red and a black) must be connected between each speaker and the amplifier. To ensure that the diaphragms of the two speakers vibrate in phase, it is necessary to make these connections in the same way (red-to-red, black-to-black) on both speakers. If the wires for one (and only one) of the two speaker are connected In reverse (red-toblack), the diaphragms will vibrate exactly out of phase, also said to be in opposite phase.

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