When is the f/f = -v/c Doppler equation okay to use?

[Trayshawn]

EK says that we should always use Δf/f = -v/c for Doppler questions over the often harder-to-use f '=f(v+/-vo)/(v+/-vs).

I like the first equation and I want to be able to use it really badly like EK says I can.
But, I'm wary of it because I havent seen it in any textbook or even any other MCAT prep books.

It was working pretty well until today when I hit two TPRH questions. It failed miserably. I'm hoping that I just applied it wrong and there is a way to reconcile everything.

Here's one of the questions:

TPRH SW #367 - A sound wave of frequency f is emitted from a stationary source and detected by a receiver moving toward the source at speed w; let f' be the frequency detected. In a separate trial, the source emits a wave of the same frequency f while moving with a speed w toward the stationary receiver; let f'' be the frequency so detected. If w is half the speed of sound, then:

A. f' < f''
B. f' = f''
C. f' > f''
D. the relationship between f' and f'' can be any of the above depending on the specific value of f.

Using the EK formula: f is the same for both trials. c is the same for both trials. and v, the relative velocity, is the same for both trials (relative velocity is the velocity one object moves if you hold the other constant. this equals w in both). thus, delta f should be the same, and f' should equal f''.

This led me to choice B, but the answer is choice A. Use of the normal Doppler formula (with quite a bit of nasty math) leads to this.

Help!

---

Members don't see this ad.

 

[brood910]

You should be aware of the fact that this equation does not always work..
For example, if there's wind blowing, you cant use that..

Just use the original Doppler's equation.

 

[polyatomic]

Yeah....I'm always wary of shortcut formulas. I find it easier to just remember the original equation and learn to apply it swiftly over time. But that's for you to decide.

For this question, the math is not too bad. Just remember for comparisons like this, you don't have to complete the math, you just need to plug in enough to get a sense for what the answer is.

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

Quick tutorial: When I use this equation, I usually go by intuition in order to figure out if you add or subtract the velocity. For instance, no matter what, if the listener moves toward the sound or if the sound moves toward the listener, the frequency ALWAYS increases. If one moves away from the other, it ALWAYS decreases.

So....if the listener (vo) moves towards the source, then you add vo on top to increase f
if the source moves towards the listener, then you subtract vs on the bottom, which also increases f. You wouldn't add vs on the bottom, otherwise you would decrease the frequency.


Now for this problem:

f ' = f(v+/-vo)/(v+/-vs) --- turns into ----> f ' = f*(v+vo) / (v)

f " = f(v+/-vo)/(v+/-vs) --- turns into ----> f " = f*(v) / (v-vs)

All you have to do is plug in fake numbers into each to see what the outcome would be

Let's plug in 1 for vo and vs; then let's plug in 2 for v.

f ' = f(v+/-vo)/(v+/-vs) --- turns into ----> f ' = f*(2+1) / (2) = f * 1.5

f " = f(v+/-vo)/(v+/-vs) --- turns into ----> f " = f*(2) / (2-1) = f * 2

Therefore f " > f ' because f " has a greater multiplier


I know this looks tedious...but if you practice enough problems, you can do this within a matter of seconds. It's all about familiarity. But that's the basics to how I approach these problems.

 

[Trayshawn]

Yeah....I'm always wary of shortcut formulas. I find it easier to just remember the original equation and learn to apply it swiftly over time. But that's for you to decide.

For this question, the math is not too bad. Just remember for comparisons like this, you don't have to complete the math, you just need to plug in enough to get a sense for what the answer is.

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

Quick tutorial: When I use this equation, I usually go by intuition in order to figure out if you add or subtract the velocity. For instance, no matter what, if the listener moves toward the sound or if the sound moves toward the listener, the frequency ALWAYS increases. If one moves away from the other, it ALWAYS decreases.

So....if the listener (vo) moves towards the source, then you add vo on top to increase f
if the source moves towards the listener, then you subtract vs on the bottom, which also increases f. You wouldn't add vs on the bottom, otherwise you would decrease the frequency.


Now for this problem:

f ' = f(v+/-vo)/(v+/-vs) --- turns into ----> f ' = f*(v+vo) / (v)

f " = f(v+/-vo)/(v+/-vs) --- turns into ----> f " = f*(v) / (v-vs)

All you have to do is plug in fake numbers into each to see what the outcome would be

Let's plug in 1 for vo and vs; then let's plug in 2 for v.

f ' = f(v+/-vo)/(v+/-vs) --- turns into ----> f ' = f*(2+1) / (2) = f * 1.5

f " = f(v+/-vo)/(v+/-vs) --- turns into ----> f " = f*(2) / (2-1) = f * 2

Therefore f " > f ' because f " has a greater multiplier


I know this looks tedious...but if you practice enough problems, you can do this within a matter of seconds. It's all about familiarity. But that's the basics to how I approach these problems.

thanks for that. I didn't have much of a problem figuring out how to do it with the normal equation. it was just annoying, ya know?

but i guess there was no flaw in my logic and the doppler approximation simply didn't work. oh well. guess i should just remember it lol.

 

[MedGrl@2022]

EK says that we should always use &#916;f/f = -v/c for Doppler questions over the often harder-to-use f '=f(v+/-vo)/(v+/-vs).

I like the first equation and I want to be able to use it really badly like EK says I can.
But, I'm wary of it because I havent seen it in any textbook or even any other MCAT prep books.

It was working pretty well until today when I hit two TPRH questions. It failed miserably. I'm hoping that I just applied it wrong and there is a way to reconcile everything.

Here's one of the questions:

TPRH SW #367 - A sound wave of frequency f is emitted from a stationary source and detected by a receiver moving toward the source at speed w; let f' be the frequency detected. In a separate trial, the source emits a wave of the same frequency f while moving with a speed w toward the stationary receiver; let f'' be the frequency so detected. If w is half the speed of sound, then:

A. f' < f''
B. f' = f''
C. f' > f''
D. the relationship between f' and f'' can be any of the above depending on the specific value of f.

Using the EK formula: f is the same for both trials. c is the same for both trials. and v, the relative velocity, is the same for both trials (relative velocity is the velocity one object moves if you hold the other constant. this equals w in both). thus, delta f should be the same, and f' should equal f''.

This led me to choice B, but the answer is choice A. Use of the normal Doppler formula (with quite a bit of nasty math) leads to this.

Help!

Hey I think that with the EK shortcut the v isn't negative is that helpful at all... let me know if it changes the answer...

 

[dudewheresmymd]

BTW does anyone know if transecho location ever shows up on any of the AAMCS? When I read it in TBR I was like "this isn't going to show up on the MCAT." Thoughts? The AAMC content outline mentions "reflection of sound from a moving object" so wasn't sure... I guess that's why TBR put it in their books, but the math would take far too long for a single MC question IMO...so perhaps just conceptually tested?

 

[dudewheresmymd]

bump.

 

[milski]

What is transecho? Doppler effect applied for a bounced signal, like radar gun? If yes, that's a fairly standard application, I can imagine it being a fair game for the exam, probably as part of a passage. You're unlikely to get a yes/no answer for such a specific question about the exam though.

 

[dudewheresmymd]

What is transecho? Doppler effect applied for a bounced signal, like radar gun? If yes, that's a fairly standard application, I can imagine it being a fair game for the exam, probably as part of a passage. You're unlikely to get a yes/no answer for such a specific question about the exam though.

Transecholation meaning like sonar or echo problems where the doplar shift is shifted twice (TBR explained it as "the sender of the outbound wave becomes the receiver of the inbound wave) Sounds good. I mean I did the TBR passages, but wasn't fully confident after doing the some of them.

 

[milski]

Transecholation meaning like sonar or echo problems where the doplar shift is shifted twice (TBR explained it as "the sender of the outbound wave becomes the receiver of the inbound wave) Sounds good. I mean I did the TBR passages, but wasn't fully confident after doing the some of them.

Yes, same thing - wales, bats, state patrol related problems, all look very possible to me.