# Doppler Effect questions (EK wave)

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#### Jwinsler7

##### Full Member

Q1. The source of a sound wave is stationary. The observer is moving toward the source. There's a steady wind blowing from the observer to the source. How does the wind change the observed frequency?

A. The wind magnifies the Doppler effect and increases the frequency
B. The wind minimizes the Doppler effect and increases the frequency
C. The wind magnifies the Doppler effect and decreases the frequency
D. The wind minimizes the Doppler effect and decreases the frequency

Answer is A. How does the wind increase the observed frequency? Could anyone explain this for me? EK explanation is "add the opposite direction of velocity to both source and observer.."

Q2. An interstellar gas circles the core of earth's galaxy. If the wavelength of the light reflecting off the gas coming toward the earth is 499nm, and the wavelength of light reflecting off the gas moving away from earth is 501nm, what is the speed of gas?
A. 4.2 x 10^4 m/s
B. 1.2 x 10^5 m/s
C. 6.0 x 10^5 m/s
D. 1.5 x 10^11 m/s

I have no idea how to approach this problem. Help?

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#### LoLCareerGoals

##### Full Member
5+ Year Member
Q1, I would answer D. Doppler effect in this instance increases the experienced frequency, so decrease in Doppler effect (because I guess it slows down the sound propagation?) should decrease the frequency. Are they saying the sound somehow affects the speed of the moving observer??? Wtf?

Edit Q1: So wait Doppler Effect = speed(observer)/speed(source) X freq(source) <- Only speed of source is changed(lowered) here. So Doppler effect is larger!
Observed Freq = freq(source) + Doppler Effect <- Since only Doppler effect increased and freq(source) didn't, so frequency increased!
So is it A? I don't understand how it can be B or C. Doppler effect and observed frequency are proportional to each other when moving toward the source. Are they treating doppler effect as some directional vector or smth?

Edit Q1 #2: Ok even if they say add a negative value to both velocities, then we get Doppler effect = (v(observer) - delta)/(v(source)-delta) X f. Doppler can actually change the sign here depending on the difference between v(observer) and wind speed. How do we know what their relative magnitudes are?

Q2: I would go with C. I don't remember the exact formulas, but clearly lambda changes by 0.2% due to speed, so I would pick whatever is 0.2% of the speed of light, which is C.

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#### thelullaby99

##### RIP
Q1, I would answer D. Doppler effect in this instance increases the experienced frequency, so decrease in Doppler effect (because I guess it slows down the sound propagation?) should decrease the frequency. Are they saying the sound somehow affects the speed of the moving observer??? Wtf?

Edit Q1: So wait Doppler Effect = speed(observer)/speed(source) X freq(source) <- Only speed of source is changed(lowered) here. So Doppler effect is larger!
Observed Freq = freq(source) + Doppler Effect <- Since only Doppler effect increased and freq(source) didn't, so frequency increased!
So is it A? I don't understand how it can be B or C. Doppler effect and observed frequency are proportional to each other when moving toward the source. Are they treating doppler effect as some directional vector or smth?

Edit Q1 #2: Ok even if they say add a negative value to both velocities, then we get Doppler effect = (v(observer) - delta)/(v(source)-delta) X f. Doppler can actually change the sign here depending on the difference between v(observer) and wind speed. How do we know what their relative magnitudes are?

Q2: I would go with C. I don't remember the exact formulas, but clearly lambda changes by 0.2% due to speed, so I would pick whatever is 0.2% of the speed of light, which is C.

According to my EK the answer is A? It's Q#750 I believe, based on source stationary, observer towards source, wind from observer to source. From other similar questions (749-751) where the variables are changed, none of them is a B. OP are you sure you're reading from the same book as I am?

#### Jwinsler7

##### Full Member
According to my EK the answer is A? It's Q#750 I believe, based on source stationary, observer towards source, wind from observer to source. From other similar questions (749-751) where the variables are changed, none of them is a B. OP are you sure you're reading from the same book as I am?

Ah sorry, yeah, it's A. Still don't understand why the wind TO the sound source would increase frequency

#### thelullaby99

##### RIP
Ah sorry, yeah, it's A. Still don't understand why the wind TO the sound source would increase frequency

Well, according to EK, the wind speed will be added to both source and observer speed, but in opposite direction. IF the wind speed is the same as observer, then we should subtract that from observer speed (because it becomes opposite). Thus we also add to the source speed (in this case we determine that we add to its original speed).

Example: observer 10 m/s towards source, source stationary, wind 5 m/s TO SOURCE. Because the wind is from the observer to source, based on EK reasoning the wind speed is now opposite of what is is. So, now wind is calculated as going from source to observer FOR BOTH CASES. I think this is the key: just reverse the wind direction and consider it as additional component of observer and source speed.

for numerator: v + (10-5)
for denominator: v - 5

From above we can conclude that the value now is going to be larger than the original.

PS. Just hoping something ridiculous like this will never show up on the real test. Even though I understand the logic, never ever in my MCAT prep have I seen wind speed on Doppler effect.

#### Jwinsler7

##### Full Member
Well, according to EK, the wind speed will be added to both source and observer speed, but in opposite direction. IF the wind speed is the same as observer, then we should subtract that from observer speed (because it becomes opposite). Thus we also add to the source speed (in this case we determine that we add to its original speed).

Example: observer 10 m/s towards source, source stationary, wind 5 m/s TO SOURCE. Because the wind is from the observer to source, based on EK reasoning the wind speed is now opposite of what is is. So, now wind is calculated as going from source to observer FOR BOTH CASES. I think this is the key: just reverse the wind direction and consider it as additional component of observer and source speed.

for numerator: v + (10-5)
for denominator: v - 5

From above we can conclude that the value now is going to be larger than the original.

PS. Just hoping something ridiculous like this will never show up on the real test. Even though I understand the logic, never ever in my MCAT prep have I seen wind speed on Doppler effect.

I am wondering how is the wind going from the observer to the source even going to affect velocity of the observer himself? That doesn't make any sense to me. Shouldn't it be the other way around? Increasing the Vobs?

#### thelullaby99

##### RIP
I am wondering how is the wind going from the observer to the source even going to affect velocity of the observer himself? That doesn't make any sense to me. Shouldn't it be the other way around? Increasing the Vobs?

Yeah I'm confused about this too, I'm going to try and search if there's anything like this for Doppler effect. To me right now it just sounds like some random stuffs that EK brought up.

#### Jwinsler7

##### Full Member
Yeah, the whole concept seems counter-intuitive.

#### Eastfolding

##### Full Member
This question is one of those tough connection questions, where you're connecting two principles that are not usually asked together. It's not only about the Doppler effect, but the frame of reference.

You know that sound propagates through air. You also know that the speed of sound is in the reference frame of the material that it travels through (Air). However, the observer is in the reference frame of the ground. The problem with this is that the doppler effect is calculated in the same reference frame! (Or more generally, you have to be in the same reference frame to do ANY calculations).

So in order to make the sound in reference to the ground, you have to factor in the wind speed. Since the wind is going in the same direction as the sound, you add the two speeds together.

It's the same concept as walking on a conveyor belt while someone on the ground watches you.

#### Eastfolding

##### Full Member
So for the second question, you should know that the doppler effect affects all waves.

The question gives you the observed frequency based on the direction of the gas. So think of the question this way instead - A speaker is moving towards you, and the frequency of the sound is 501 nm. It then starts moving away at the same speed, and the frequency changes to 499 nm.

Can you solve it now?

I believe the answer is C.

#### richkaj

##### Full Member
10+ Year Member
The second question states the gas is moving toward the observer at 499nm and away at 501 nm. This intuitively means the frequency is higher when it is approaching and lower when it is leaving the observer given that it is traveling at the same speed (there is no mention of frequency). By the equation, (d wavelenth/wavelength) = v source/ velocity of medium since both wavelength and frequency are directly related to the speed. If wavelength increases or decreases, so does the speed.

Edit: The gas velocity will be lower than the velocity of light but it is giving off light which has a velocity of 3.0x10^8m/s. It's velocity will be lower. The change in wavelength should help us to determine the gas velocity.

This will equal out to (2/501) = (vs/3x10^8m/s). I recieved 1.2x10^6m/s. But in your question you had 1.2x10^5m/s. Please check the answer to see if it is a typo. I do not have the passage/question in front of me...

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