AAMC 4 Question 13: Doppler Effect Question

[aspiringdoc09]

---

Members don't see this ad.

Why are the percentages of the change in frequency and wavelength much greater when sound waves are used instead of radio waves in these experiments?

A) Sound waves travel more slowly.
B) Sound waves have a much higher frequency.
C) Sound waves have a much shorter wavelength.
D) Interference in the atmosphere affects sound wave much more.

I guess B and C can be eliminated because for one to be true the other must be true too (shorter wavelength = higher frequency). How do you choose between A and D.

What I don't understand, which they based their explanation (below) on, is how they transitioned from radio waves to sound? Was it implied? Is the Doppler effect only applied to sound waves and light, not other waves (i.e. radio)? Did you only use POE to arrive at the answer or did you recognize it was the doppler effect and used the approximation equation? I did recognize it as the doppler effect but forgot the approximation equation and didn't see relationship with the orignal equation.

Explanation: The Doppler equation for frequency isdf/f=-v/c for a given relative velocity v b/t source and detector. Thus, thefrequency shift change of frequency (df) depends inversely on the speed of thewave in the medium in which it propagates, c. The velocity of sound is muchsmaller than that of electromagnetic radiation, so for the same relativevelocity the frequency and wavelength shifts are much greater for sound thanfor radio waves.

 

[pfaction]

This has nothing to do with the question, but I think radio waves are EM waves. They fit on that EM spectrum of radio->visible->x-ray.

 

[MedPR]

Why are the percentages of the change in frequency and wavelength much greater when sound waves are used instead of radio waves in these experiments?

A) Sound waves travel more slowly.
B) Sound waves have a much higher frequency.
C) Sound waves have a much shorter wavelength.
D) Interference in the atmosphere affects sound wave much more.

I guess B and C can be eliminated because for one to be true the other must be true too (shorter wavelength = higher frequency). How do you choose between A and D.

What I don't understand, which they based their explanation (below) on, is how they transitioned from radio waves to sound? Was it implied? Is the Doppler effect only applied to sound waves and light, not other waves (i.e. radio)? Did you only use POE to arrive at the answer or did you recognize it was the doppler effect and used the approximation equation? I did recognize it as the doppler effect but forgot the approximation equation and didn't see relationship with the orignal equation.

Explanation: The Doppler equation for frequency isdf/f=-v/c for a given relative velocity v b/t source and detector. Thus, thefrequency shift change of frequency (df) depends inversely on the speed of thewave in the medium in which it propagates, c. The velocity of sound is muchsmaller than that of electromagnetic radiation, so for the same relativevelocity the frequency and wavelength shifts are much greater for sound thanfor radio waves.

EM waves and sound waves can both experience the doppler effect. The only info you needed to answer this question was the equation (doppler equation) and the knowledge than the speed of light is significantly faster than the speed of sound.

 

[Buttafuoco]

f=f'[(s + or - sd)/(s = or - ss)] Sorry for the crappy symbols, but I mean speed of sound or light for s, speed of detector, and speed of source. If the speed of your wave is infinity, how much does it change the frequency if the source and detector are moving 2 m/s relative to it? Not at all. If it's moving 10 m/s and the source and detector are moving say 2 m/s, it's going to be significant though.

 

[aspiringdoc09]

So basically, if there is ever a question dealing with the doppler effect that requires math, I should use the Doppler approximation df/f =-v/c. I just didn't clue that I need to do any math or intuitively use the equation for that particular question. That's where I got stuck. I will be sure to study that more and complete more questions.

 

[MedPR]

There isn't really any math involved. The only answer that is true is A.

 

[stayingpositive2014]

Why are the percentages of the change in frequency and wavelength much greater when sound waves are used instead of radio waves in these experiments?.

A) Sound waves travel more slowly..
B) Sound waves have a much higher frequency..
C) Sound waves have a much shorter wavelength..
D) Interference in the atmosphere affects sound wave much more..

I guess B and C can be eliminated because for one to be true the other must be true too (shorter wavelength = higher frequency). How do you choose between A and D. .

What I don't understand, which they based their explanation (below) on, is how they transitioned from radio waves to sound? Was it implied? Is the Doppler effect only applied to sound waves and light, not other waves (i.e. radio)? Did you only use POE to arrive at the answer or did you recognize it was the doppler effect and used the approximation equation? I did recognize it as the doppler effect but forgot the approximation equation and didn't see relationship with the orignal equation..

Explanation: The Doppler equation for frequency isdf/f=-v/c for a given relative velocity v b/t source and detector. Thus, thefrequency shift change of frequency (df) depends inversely on the speed of thewave in the medium in which it propagates, c. The velocity of sound is muchsmaller than that of electromagnetic radiation, so for the same relativevelocity the frequency and wavelength shifts are much greater for sound thanfor radio waves...

Simply know that all electromagnetic waves (UV, IR, radio waves, microwaves, gamma, etc) travel at the speed of light and nothing travels faster than the speed of light! So the sound waves have to travel more slowly...

 

[RSK25]

I think more importantly, when it comes to doppler in light, the only thing matter is relative velocity. so, that would hint at speed of light (radio is EM). Sound is slower than EM in air. so, answer has to be A.