Double doppler shift??

[Ari1584]

I have a question in berkeley review's physics part 2 question #61 (passage 9). I know we don't need to come up with exact answers, and i got the question right...but i looked at the answer in the back as to how someone would solve a mathematical double doppler shifted problem and i need some help. I am confused on what is considered the source and what is considered the listener and when.

For example, the question said: "Submarina A is moving at 10 m/s toward submarine B (stationary). If Sub A sends out signal to sub B and then listens to the echo, how would the echo frequency detected by sub A compare to the source frequency? [source freq = 500 hz, v of sound = 1500 m/s]

Where do we put the speeds of each submarine and which sign do i use (the plus or minus sign) in the numerator and denominator? It's kind of confusing. Thanks!

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[Bernoull]

I have a question in berkeley review's physics part 2 question #61 (passage 9). I know we don't need to come up with exact answers, and i got the question right...but i looked at the answer in the back as to how someone would solve a mathematical double doppler shifted problem and i need some help. I am confused on what is considered the source and what is considered the listener and when.

For example, the question said: "Submarina A is moving at 10 m/s toward submarine B (stationary). If Sub A sends out signal to sub B and then listens to the echo, how would the echo frequency detected by sub A compare to the source frequency? [source freq = 500 hz, v of sound = 1500 m/s]

Where do we put the speeds of each submarine and which sign do i use (the plus or minus sign) in the numerator and denominator? It's kind of confusing. Thanks!

I'll take a bite at this. But i'm interested in others' opinions and answers.
I'm looking at this in 2 ways and I think approach 2 is correct, da 1st approach is too simplistic and so it's suspect. I don't have the book, so i'm relying entirely on the info u gave.

1. First and easiest solution is that since submarine A is both the source and receiver/observer of the signal, it's velocity relative to itself is 0m/s therefore no doppler effect happens. fo=fs

2. Second approach is to think of this problem like one thinks of optics. In optics, generally there's a light source (different from the object) and its photons bounce of the object and later on in space they intersect to form an image. To simplify, we assume that the photons emanate from the object and not the light source. Here, since the sub A produces the signal which is then reflected by sub B back to sub A (scenario unnecessarily complicates things), to simplify I'll assume sub B produces the signal. Now we have our typical Doppler effect problem:

fo=fs((v+-vo)/(v+-vs)) = 500hz(1510m/s/1500) = 503.3Hz


About ur question of signs:

My system for signs is intuitive. fs is multiplied by a fraction to get fo. There's 3 possibilities
a) fraction=1; therefore numerator = denominator
b) fraction>1; therefore numerator > denominator
c) fraction<1; therefore numerator < denominator

a) If source and observer are stationary relative to one another, fraction=1

b) If source and observer are getting closer to each other, fo must increase therefore fraction > 1

c) If source and observer are getting farther apart, fo must decrease therefore fraction < 1

I hope this helps.

 

[Bernoull]

I have a question in berkeley review's physics part 2 question #61 (passage 9). I know we don't need to come up with exact answers, and i got the question right...but i looked at the answer in the back as to how someone would solve a mathematical double doppler shifted problem and i need some help. I am confused on what is considered the source and what is considered the listener and when.

For example, the question said: "Submarina A is moving at 10 m/s toward submarine B (stationary). If Sub A sends out signal to sub B and then listens to the echo, how would the echo frequency detected by sub A compare to the source frequency? [source freq = 500 hz, v of sound = 1500 m/s]

Where do we put the speeds of each submarine and which sign do i use (the plus or minus sign) in the numerator and denominator? It's kind of confusing. Thanks!

Wat was da book's answer anyway??

 

[Bernoull]

I'm curious about what others think of this problem... how would u handle it?? anyone???

 

[Sach]

The answer in the book is A. f(echo) > f(source)

f(echo) = 507 Hz
f (source) = 503 Hz

 

[rocketbooster]

Here's my advice to you...

do NOT use BR's method for doppler effect using the +/- formula. you're right. the +/- gets confusing. use EK's method for everything doppler and you will get every doppler question correct, won't have to use that stupid equation, and solve the problem in half of the time.

I don't really want to type out the entire EK method. if you can, borrow the EK physics from someone. it's only 2 pages of reading and you're set for everything doppler!

 

[Bernoull]

The answer in the book is A. f(echo) > f(source)

f(echo) = 507 Hz
f (source) = 503 Hz

Thanks 4 the answer. So the frequency shifted twice. I think I benefited the most from this question, i have never seen a question where the freq changed more than once.

I hope I dont see some crazy problem on the mcat like light reflected many times in an optic cable n then we should find the final freq or something😱😱😱😱

 

[BloodySurgeon]

The frequency when it hits Submarine B is f' = 500 (1500 + 0)/ (1500 -10) = 503 Hz. Then when it hits Submarine A is f'' = 503 (1500 + 10) / (1500 + 0) = 507 Hz.

f' = f (v +/- vo) / (v +/- vs)

v = velocity of the sound in a medium
vo = observed velocity
vs = source velocity

v + vo when the observer is going towards the source making f' larger
v - vo when the observer is going away from the source making f' smaller

v + vs when the source is moving away from the observer making f' smaller
v - vs when the source is moving towards the observer making f' bigger

 

[amikhchi]

Attack it in 2 steps:

1. sub A is sending the 500Hz to sub B

Fd = Fs (v +- Vd)/(v +- Vs)

you know Fs = 500, and you know v = 1500, you also know Vd (sub B) = 0 because it's stationary

as for Vs, you know sub A is moving toward sub B, therefore the frequency detected would be *higher* and therefore the denominator should be smaller to make the frequency after multiplication higher: so throw in a (-) for denominator

doing the math you get roughly 503.4

Now step 2:

2. Now your F's = 503, your source is not moving (sub B) and your detector is (sub A) at 10m/s

so F'd = F's (v +-Vd)/(v +- Vs)

since your detector (sub A) is moving towards your source, again freq would be higher therefore to make frequency higher in the numerator use a (+)

doing the math here you'll get ~ 507Hz echo

*just saw that someone explained this pretty well right before me*

 

[JohnnytheNomad]

Attack it in 2 steps:

1. sub A is sending the 500Hz to sub B

Fd = Fs (v +- Vd)/(v +- Vs)

you know Fs = 500, and you know v = 1500, you also know Vd (sub B) = 0 because it's stationary

as for Vs, you know sub A is moving toward sub B, therefore the frequency detected would be *higher* and therefore the denominator should be smaller to make the frequency after multiplication higher: so throw in a (-) for denominator

doing the math you get roughly 503.4

Now step 2:

2. Now your F's = 503, your source is not moving (sub B) and your detector is (sub A) at 10m/s

so F'd = F's (v +-Vd)/(v +- Vs)

since your detector (sub A) is moving towards your source, again freq would be higher therefore to make frequency higher in the numerator use a (+)

doing the math here you'll get ~ 507Hz echo

*just saw that someone explained this pretty well right before me*

Just for logic reasons only, because two things have a net movement toward each other, you know the frequency is going be faster. And because you know that the source is the observer, all you have to do is double the v relative. so (delta f)/f = (2 times v the relative velocity)/c (speed of the sound wave in the media.) Let me know if I am wrong in the logic. Just to let you all know. the (delta f/f) = v relative/c is the easiest equation to memorize for this type of problem. Just use logic. the new f is going to be greater if the objects are moving toward each other and smaller if away. the wavelength is going to be exactly opposite because lambda = c/f.

 

[sankara]

Basic Question
So Double Doppler Shifts can only occur with EM waves?

 

[Phantastic]

Basic Question
So Double Doppler Shifts can only occur with EM waves?

I believe a sound echo (longitudinal wave) is also a double doppler shift...

 

[Longshanks]

Basic Question
So Double Doppler Shifts can only occur with EM waves?

No. Any wave. The most often used examples are sound, which is a mechanical wave. (i.e. a bat chasing an insect)

The other often used example is light, which of course is an EM wave. (i.e. blue shift/red shift with stars)

 

[Enoko]

A very, very common example is "speed gun" used by police on the highway. The basics of how that works is the gun calculating the double shift that happens when the police and the target car are moving with different speeds.