The Dppler EffectDue: 3:00pm on Monday, October 24, 2011

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Doppler Shift

Learning Goal: To understand the terms in the Doppler shift formula.

The Doppler shift formula gives the frequency at which a listener L hears the sound

emitted by a source S at frequency :

,

where is the speed of sound in the medium, is the velocity of the listener, and is the velocity of source.

Part A

The velocity of the source is positive if the source is ______________. Note that this equation may not use the sign convention you are accustomed to. Think about the physical situation before answering.

Hint A.1 Relating the frequency and the source velocity

Hint not displayed

ANSWER:

traveling in the +x direction

traveling toward the listener

traveling away from the listener

Correct

Part B

The velocity of the source is measured with respect to the ________.

ANSWER:

medium (such as air or water)

listener

Correct

Part C

The velocity of the listener is positive if the listener is _____________.

Hint C.1 Relating the frequency and the listener's velocity

Hint not displayed

ANSWER:

traveling in the +x direction

traveling toward the source

traveling away from the source

Correct

Part D

The velocity of the listener is measured with respect to the ________.

ANSWER:

source

medium

Correct

Here are two rules to remember when using the Doppler shift formula:

1. Velocity is measured with respect to the medium.2. The velocities are positive if they are in the direction from the listener to the source.

Part E

Imagine that the source is to the right of the listener, so that the positive reference direction

(from the listener to the source) is in the direction. If the listener is stationary, what

value does approach as the source's speed approaches the speed of sound moving to the right?

ANSWER:

0

It approaches infinity.

Correct

Part F

Now, imagine that the source is to the left of the listener, so that the positive reference

direction is in the direction. If the source is stationary, what value does approach as

the listener's speed (moving in the direction) approaches the speed of sound?

ANSWER:

0

It approaches infinity.

Correct

Basically in this case the listener doesn't hear anything since the sound waves cannot catch up with him or her.

Part G

In this last case, imagine that the listener is stationary and the source is moving toward the listener at the speed of sound. (Note that it is irrelevant whether the source is moving to the

right or to the left.) What is when the sound waves reach the listener?

ANSWER:

0

It approaches infinity.

Correct

This case involves what is called a sonic boom. The listener will hear no sound ( ) until the sonic boom reaches him or her (just as the source passes by). At that instant, the frequency will be infinite. There is no time between the passing waves--they are literally right on top of each other. That's a lot of energy to pass by the listener at once, which explains why a sonic boom is so loud.

 

Two Traveling Waves Beating Together

Learning Goal: To see how two traveling waves of nearly the same frequency can create

beats and to interpret the superposition as a "walking" wave.Consider two similar traveling transverse waves, which might be traveling along a string for example:

and .

They are similar because we assume that and are nearly equal and also that and are nearly equal.

Part A

Which one of the following statements about these waves is correct?

ANSWER:

Both waves are traveling in the direction.

Both waves are traveling in the direction.

Only wave is traveling in the direction.

Only wave is traveling in the direction.

Correct

The principle of superposition states that if two waves each separately satisfy the wave equation then the sum (or difference) also satisfies the wave equation. This follows from the fact that every term in the wave equation is linear in the amplitude of the wave.

Consider the sum of the two waves given in the introduction, that is,

.

These waves have been chosen so that their sum can be written as follows:

where is a constant, and the functions and are trigonometric functions of and

. This form is especially significant because the first function, called the envelope, is a slowly varying function of both position (

) and time ( ), whereas the second varies rapidly with both position ( ) and time ( ).

Traditionally, the overall amplitude is represented by the constant , while the functions and are trigonometric functions with unit amplitude.

Part B

Find , , and .

Hint B.1

A useful trigonometric identity

Hint not displayed

Hint B.2

Which is the envelope and which is the carrier wave?

Hint not displayed

Express your answer in terms of , , , , , , and . Separate the three terms with commas. Recall that (the second term) varies slowly whereas (the third term) varies quickly. Both and should be trigonometric functions of unit amplitude.

ANSWER:

  ,

,

 =

, ,All attempts used; correct answer displayed

Part C

Which of the following statements about is correct?

ANSWER:

It is a rapidly oscillating wave traveling in the direction.

It is a rapidly oscillating wave traveling in the direction.

It is slowly oscillating in time but is standing still.

It is traveling rapidly but oscillating slowly.

Correct

Part D

Which of the following statements about is correct if you assume that

and are both positive?

ANSWER:

It is a slowly oscillating wave traveling in the direction.

It is a slowly oscillating wave traveling in the direction.

It is slowly oscillating in time but is standing still.

It is traveling rapidly but oscillating slowly.

Correct

Part E

The envelope function can be written simply in terms of and . If you do so, what is , the velocity of propagation of the envelope?

Hint E.1 Substituting and

Hint not displayed

Hint E.2

Traveling wave velocity

Hint not displayed

Express your result in terms of and .

ANSWER:    =

Correct

is called the group velocity because this ratio is the velocity at which the group of waves under a maximum of the envelope function appears to travel. It is technically defined as a derivative (although you found it for a finite difference):

.

This contrasts with the velocity of propagation of the waves themselves:

.

Information (e.g., applied as amplitude modulation of the wave, that is, as a variation in the envelope) travels at the group velocity. No information can be sent at the phase velocity of a wave, which therefore can exceed the speed of light. (Einstein's Special Theory of Relativity implies that neither a physical body nor information can travel faster than light.)

 

Beat Frequency Ranking Task

An all female guitar septet is getting ready to go on stage. The lead guitarist, Kira,who is always in tune, plucks her low E string and the other six members, sequentially, do the same.

Each member records the initial beat frequency between her low E string and Kira's low E string.

Part A

Rank each member on the basis of the frequency of her low E string.

Hint A.1

Beat frequency

Hint not displayed

Hint A.2

Find the frequency of Aiko's E string

Hint not displayed

Rank from largest to smallest. To rank items as equivalent, overlap them.

ANSWER:

View  Correct

Because the beat frequency between Kira's guitar and Diane's guitar is 0 , these guitars play the exact same note and are in tune.

To tune an instrument using beats, more information than just the beat frequency is needed.

In addition to recording the initial beat frequency , each member, except Diane, also

records the change in the frequency (increase or decrease) when they increase the tension in their low E string.

Part B

Rank each member on the basis of the initial frequency of their low E string.

Hint B.1

Determine the relationship between tension and beat frequency

Hint not displayed

Hint B.2

Determine the initial frequency of Aiko's E string

Hint not displayed

Rank from largest to smallest. To rank items as equivalent, overlap them.

ANSWER:

View  Correct

 

The Doppler Effect on a Train

A train is traveling at 30.0 relative to the ground in still air. The frequency of the note

emitted by the train whistle is 262 .

The speed of sound in air should be taken as 344 .

Part A

What frequency is heard by a passenger on a train moving at a speed of 18.0 relative to the ground in a direction opposite to the first train and approaching it?

Hint A.1

How to approach the problem

Hint not displayed

Hint A.2

Doppler shift equations for moving source or observer

Hint not displayed

Hint A.3

Doppler equations when both the source and the listener are in motion

Hint not displayed

Hint A.4

Determine the appropriate signs

Hint not displayed

Express your answer in hertz.

ANSWER:    =

302Correct  

Part B

What frequency is heard by a passenger on a train moving at a speed of 18.0 relative to the ground in a direction opposite to the first train and receding from it?

Hint B.1

How to approach the problem

Hint not displayed

Hint B.2

Doppler shift equations for moving source or observer

Hint not displayed

Hint B.3

Doppler equations when both the source and the listener are in motion

Hint not displayed

Hint B.4

Determine the appropriate signs

Hint not displayed

Express your answer in hertz.

ANSWER:

   =

228Correct  

 

± The Hearing of a Bat

Bats are mainly active at night. They have several senses that they use to find their way about, locate prey, avoid obstacles, and "see" in the dark. Besides the usual sense of vision, bats are able to emit high-frequency sound waves and hear the echo that bounces back when these sound waves hit an object. This sonar-like system is called echolocation. Typical frequencies emitted by bats are between 20 and 200 kHz. Note that the human ear is sensitive only to frequencies as high as 20 kHz.

A moth of length 1.0 is flying about 1.0 from a bat when the bat emits a sound wave at

80.0 . The temperature of air is about 10.0 . To sense the presence of the moth using echolocation, the bat must emit a sound with a wavelength equal to or less than the length of the insect.

The speed of sound that propagates in an ideal gas is given by

,

where is the ratio of heat capacities ( for air), is the absolute temperature in

kelvins (which is equal to the Celsius temperature plus 273.15 ), is the molar mass of

the gas (for air, the average molar mass is ), and is the universal gas

constant ( ).

Part A

Find the wavelength of the 80.0- wave emitted by the bat.

Hint A.1

Relating wavelength, frequency, and speed of a wave

Hint not displayed

Hint A.2

Find the speed of sound in air

Hint not displayed

Express your answer in millimeters.

ANSWER:

   =

4.23Correct

 

Part B

Will the bat be able to locate the moth despite the darkness of the night?

ANSWER:

yes

no

Correct

Part C

How long after the bat emits the wave will it hear the echo from the moth?

Hint C.1

How to approach the problem

Hint not displayed

Hint C.2

Find the time needed for the sound wave to reach the moth

Hint not displayed

Express your answer in milliseconds to two significant figures.

ANSWER:

5.9Correct

 

 

The Beat Heard by a Bat

A bat flies toward a wall, emitting a steady sound with a frequency of 20.5  . This bat hears its own sound plus the sound reflected by the wall.

Part A

How fast should the bat fly, , to hear a beat frequency of 195  ?

Take the speed of sound to be 344 .

Hint A.1

How to approach the problem

Hint not displayed

Hint A.2

Find the frequency of the wave bouncing off the wall

Hint not displayed

Hint A.3

Find the frequency of the echo that the bat hears

Hint not displayed

Hint A.4

Find the expression for the beat frequency

Hint not displayed

Hint A.5

Working the math

Hint not displayed

Express your answer numerically in meters per second to three significant figures.

ANSWER:    =

1.71Correct  

 

Problem 16.77

Horseshoe bats (genus Rhinolophus) emit sounds from their nostrils, then listen to the frequency of the sound reflected from their prey to determine the prey's speed. (The "horseshoe" that gives the bat its name is a depression around the nostrils that acts like a focusing mirror, so that the bat emits sound in a narrow beam like a flashlight.) A

Rhinolophus flying at speed emits sound of frequency ; the sound it hears reflected

from an insect flying toward it has a higher frequency .

Part A

If the bat emits a sound at a frequency of 80.7  and hears it reflected at a frequency of

83.8  while traveling at a speed of 4.5  , calculate the speed of the insect.

Use 344  for the speed of sound in air. Express your answer using two significant figures.

ANSWER:    =

2.0Correct

 

 

Exercise 16.40

Two identical taut strings under the same tension produce a note of the same fundamental

frequency . The tension in one of them is now increased by a very small amount .

Part A

If they are played together in their fundamental, show that the frequency of the beat

produced is .Essay answers are limited to about 500 words (3800 characters maximum, including spaces).

ANSWER:

My Answer:I made it halfway through the derivation before I hit the wall. Could you please go over this in class if you feel it is important

Part B

Two identical violin strings, when in tune and stretched with the same tension, have a

fundamental frequency of 448.0  . One of the strings is retuned by increasing its tension. When this is done, 1.1 beats per second are heard when both strings are plucked simultaneously at their centers. By what percentage was the string tension changed?Express your answer using two significant figures.

ANSWER:

   =0.49Correct  

 

Problem 16.74

A 1.70- sound wave travels through a pregnant woman’s abdomen and is reflected from the fetal heart wall of her unborn baby. The heart wall is moving toward the sound receiver as the heart beats. The reflected sound is then mixed with the transmitted sound, and 80 beats

per second are detected. The speed of sound in body tissue is 1490  .

Part A

Calculate the speed of the fetal heart wall at the instant this measurement is made.

ANSWER:    =

3.51×10−2

Correct  

Score Summary:Your score on this assignment is 100.5%.You received 85.43 out of a possible total of 85 points.

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