<< Chapter < Page Chapter >> Page >

Solution for (a)

(1) Enter known values into f obs = f s v w v w v s . size 12{f rSub { size 8{"obs"} } =f rSub { size 8{s} } left ( { {v rSub { size 8{w} } } over {v rSub { size 8{w} } +- v rSub { size 8{s} } } } right )} {}

f obs = f s v w v w v s = 150 Hz 340 m/s 340 m/s – 35.0 m/s size 12{f rSub { size 8{"obs"} } =f rSub { size 8{s} } left ( { {v rSub { size 8{w} } } over {v rSub { size 8{w} } - v rSub { size 8{s} } } } right )= left ("150"" Hz" right ) left ( { {"340"" m/s"} over {"340 m/s-35" "." "0 m/s"} } right )} {}

(2) Calculate the frequency observed by a stationary person as the train approaches.

f obs = ( 150 Hz ) ( 1.11 ) = 167 Hz size 12{ {}= \( "150" ital "Hz" \) \( 1 "." "11" \) ="167" ital "Hz"} {}

(3) Use the same equation with the plus sign to find the frequency heard by a stationary person as the train recedes.

f obs = f s v w v w + v s = 150 Hz 340 m/s 340 m/s + 35.0 m/s size 12{f rSub { size 8{"obs"} } =f rSub { size 8{s} } left ( { {v rSub { size 8{w} } } over {v rSub { size 8{w} } +v rSub { size 8{s} } } } right )= left ("150"" Hz" right ) left ( { {"340"" m/s"} over {"340 m/s-35" "." "0 m/s"} } right )} {}

(4) Calculate the second frequency.

f obs = ( 150 Hz ) ( 0.907 ) = 136 Hz size 12{ {}= \( "150" ital "Hz" \) \( 0 "." "97" \) ="136" ital "Hz"} {}

Discussion on (a)

The numbers calculated are valid when the train is far enough away that the motion is nearly along the line joining train and observer. In both cases, the shift is significant and easily noticed. Note that the shift is 17.0 Hz for motion toward and 14.0 Hz for motion away. The shifts are not symmetric.

Solution for (b)

(1) Identify knowns:

  • It seems reasonable that the engineer would receive the same frequency as emitted by the horn, because the relative velocity between them is zero.
  • Relative to the medium (air), the speeds are v s = v obs = 35.0 m/s.
  • The first Doppler shift is for the moving observer; the second is for the moving source.

(2) Use the following equation:

f obs = [ f s v w ± v obs v w ] v w v w ± v s . size 12{f rSub { size 8{"obs"} } = left [f rSub { size 8{s} } left ( { {v rSub { size 8{w} } +- v rSub { size 8{"obs"} } } over {v rSub { size 8{w} } } } right ) right ] rSub { size 8{s} } left ( { {v rSub { size 8{w} } } over {v rSub { size 8{w} } +v rSub { size 8{s} } } } right )} {}

The quantity in the square brackets is the Doppler-shifted frequency due to a moving observer. The factor on the right is the effect of the moving source.

(3) Because the train engineer is moving in the direction toward the horn, we must use the plus sign for v obs ; however, because the horn is also moving in the direction away from the engineer, we also use the plus sign for v s . But the train is carrying both the engineer and the horn at the same velocity, so v s = v obs . As a result, everything but f s cancels, yielding

f obs = f s . size 12{f rSub { size 8{s} } } {}

Discussion for (b)

We may expect that there is no change in frequency when source and observer move together because it fits your experience. For example, there is no Doppler shift in the frequency of conversations between driver and passenger on a motorcycle. People talking when a wind moves the air between them also observe no Doppler shift in their conversation. The crucial point is that source and observer are not moving relative to each other.

Sonic booms to bow wakes

What happens to the sound produced by a moving source, such as a jet airplane, that approaches or even exceeds the speed of sound? The answer to this question applies not only to sound but to all other waves as well.

Suppose a jet airplane is coming nearly straight at you, emitting a sound of frequency f s . The greater the plane’s speed v s , the greater the Doppler shift and the greater the value observed for f obs . Now, as v s approaches the speed of sound, f obs approaches infinity, because the denominator in f obs = f s v w v w ± v s size 12{f rSub { size 8{"obs"} } =f rSub { size 8{s} } left ( { {v rSub { size 8{w} } } over {v rSub { size 8{w} } +- v rSub { size 8{s} } } } right )} {} approaches zero. At the speed of sound, this result means that in front of the source, each successive wave is superimposed on the previous one because the source moves forward at the speed of sound. The observer gets them all at the same instant, and so the frequency is infinite. (Before airplanes exceeded the speed of sound, some people argued it would be impossible because such constructive superposition would produce pressures great enough to destroy the airplane.) If the source exceeds the speed of sound, no sound is received by the observer until the source has passed, so that the sounds from the approaching source are mixed with those from it when receding. This mixing appears messy, but something interesting happens—a sonic boom is created. (See [link] .)

Questions & Answers

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point 1 the cross-sectional area of the pipe is 0.077 m2, and the magnitude of the fluid velocity is 3.50 m/s. (a) What is the fluid speed at points in the pipe where the cross
fagbeji Reply
what's the period of velocity 4cm/s at displacement 10cm
Andrew Reply
What is physics
LordRalph Reply
the branch of science concerned with the nature and properties of matter and energy. The subject matter of physics includes mechanics, heat, light and other radiation, sound, electricity, magnetism, and the structure of atoms.
Aluko
and the word of matter is anything that have mass and occupied space
Aluko
what is phyices
Aurang Reply
Whats the formula
Okiri Reply
1/v+1/u=1/f
Aluko
what aspect of black body spectrum forced plank to purpose quantization of energy level in its atoms and molicules
Shoaib Reply
a man has created by who?
Angel Reply
What type of experimental evidence indicates that light is a wave
Edeh Reply
double slit experiment
Eric
The S. L. Unit of sound energy is
Chukwuemeka Reply
what's the conversation like?
ENOBONG Reply
some sort of blatherring or mambo jambo you may say
muhammad
I still don't understand what this group is all about oo
ENOBONG
no
uchenna
ufff....this associated with physics ..so u can ask questions related to all topics of physics..
muhammad
what is sound?
Bella
what is upthrust
Mercy Reply
what is upthrust
Olisa
Up thrust is a force
Samuel
upthrust is a upward force that acts vertical in the ground surface.
Rodney
yes rodney's answer z correct
Paul
what is centre of gravity?
Paul
you think the human body could produce such Force
Anthony
what is wave
Bryan Reply
mirobiology
Angel
what is specific latent heat
Omosebi Reply
the total amount of heat energy required to change the physical state of a unit mass of matter without a corresponding change in temperature.
fitzgerald
is there any difference between specific heat and heat capacity.....
muhammad
what wave
Bryan
why medical physics even.we have a specoal branch of science biology for this.
Sahrrr Reply
what is physics
AbleGod Reply
Practice Key Terms 4

Get the best College physics course in your pocket!





Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics' conversation and receive update notifications?

Ask