# Doppler Effect Submarine Problem

[SOLVED] Doppler Effect Problem

## Homework Statement

A submarine (sub 1) travels through water at a speed of 8.00 m/s, emitting a sonar wave at a frequency of 1400 Hz. The speed of sound in the water is 1533 m/s. A second submarine (sub 2) is located such that both submarines are traveling directly toward each other. The second submarine is moving at 9.00 m/s. While the subs are approaching each other, some of the sound from sub 1 reflects from sub 2 and returns to sub A. If this sound were to be detected by an observer on sub A, what is its frequency?

2. Relevent equations:

Doppler Equations:
Observed frequency = Actual frequency[(velocity of sonar + velocity of observer)/(velocity of sonar- velocity of source)]

Using the above equation, I found that the frequency observed by sub 2 was 1385 Hz. I noted that this frequency was also the frequency of the wave reflected off of sub 2. I substituted this frequency for the actual frequency in a new Doppler equation taking the velocity of sub B (the source) to be 0 m/s and found that the frequency observed by sub 1 was 1392 Hz. Would you please advise whether I'm correct?

I'm only asking because the textbook and my professor says I'm wrong.

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mjsd
Homework Helper
firstly, I can't see how you get a freq at sub 2 to be smaller than 1400 Hz when the subs are approaching each other.

I mistakenly wrote that the observed frequencies of the second and first subs were "1385 Hz" and "1392 Hz" respectively. I actually found that the observed frequencies of the second and first subs were 1415.6 Hz and 1423 Hz respectively. Would you please advise whether this answer is correct as opposed to the textbook answer of 1431.4 Hz? Thanks for your help.

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mjsd
Homework Helper
i think 1431Hz is what I've got

But isn't the frequency that the second sub observes equal to the frequency that it reflects, i.e. 1415.6 Hz?

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mjsd
Homework Helper
sure, the reflected wave is 1415Hz.
You problem is getting the relative velocity incorrect (from looking at your answer of 1423 Hz). Probably confused with the reference frame that you are using.
remember, the first and foremost thing in these type of problems is not go to the formula straight away but to first select your coordinate system/ref frame and then express all variables in that frame before proceeding.

My frame of reference was a stationary point on the Earth. To have any other reference point would needlessly complicate things. I took the speed of the sound, relative to my reference point, to be a constant 1533 m/s. Likwise, I took the speeds of the first and second subs, relative to my reference point, to be 8 and 9 m/s respectively.

If you agree that the frequency reflected by the second sub is about 1415 Hz, then why do you disagree that the frequency observed by the first sub is 1423 Hz?

All I need is a definite authoritative snswer on this problem before this thread makes it to the circular file..."Doc Al", please help...

The sonar frequency reaches sub 2 when both sub 1 and sub 2 are moving:

$$f_1=f\frac {V + V_{sub1}}{V-V_{sub2}}$$

The reflected frequency reaches sub 1 when both sub 2 and sub 1 are moving:

$$f_2=f_1\frac {V + V_{sub2}}{V-V_{sub1}}$$

Your answer of 1415.6 Hz would be right if they asked what the frequency was at sub 2 without any reflection. Since it is going two ways with both ships moving, you need two equations.

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But again, isn't the frequency that is observed by sub 2 equal to the frequency that is reflected by sub 2? In that case, the I assume the velocity of sub 2 would not be relevant to the second Doppler equation.

But again, isn't the frequency that is observed by sub 2 equal to the frequency that is reflected by sub 2? In that case, the I assume the velocity of sub 2 would not be relevant to the second Doppler equation.

But both ships are moving during the initial transmission, and during the reflected transmission. You can't do relative velocities with Doppler questions; you'll get an answer close to the actual answer, but not quite.

Do we agree that the frequency observed by sub 2 (f prime) is equal to the frequency that is reflected by sub 2? This question is asked to identify where it is that we disagree.

I agree, but it won't be the same frequency as it would be if sub 2 was just sitting there.

So you're saying that if sub 1 were to put the brakes on all of a sudden it wouldn't detect the same frequency that is oberved by sub 2?

That's right, mate.

So this is where we disagree. I have been trying to argue for the past week that the frequency that is detected by any object will be the SAME frequency that the object emits/reflects. This is the case whether the object is moving or not (so long as the object is moving on an axis parallel to the velocity of the wave). Please give this some thought before responding.

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mjsd
Homework Helper
My frame of reference was a stationary point on the Earth. To have any other reference point would needlessly complicate things. I took the speed of the sound, relative to my reference point, to be a constant 1533 m/s. Likwise, I took the speeds of the first and second subs, relative to my reference point, to be 8 and 9 m/s respectively.

If you agree that the frequency reflected by the second sub is about 1415 Hz, then why do you disagree that the frequency observed by the first sub is 1423 Hz?

Sorry, i've been away.

the reason you get 1423 Hz, "I think" is because you used 9 m/s as the speed your moving source rather than 8+9=17m/s. You have to be ultra carefull when doing this in the frame of the Earth.. I used the frame of the first sub (ie. frame moving at 8 m/s towards sub 2), and so the relative velocity between the two subs was always 17m/s.

assuming that the reflected wave has the same freq as the incident wave is ok i think.

First, I'd like to thank you Snazzy and Mjhd for your assistance with this problem.

So I used the first Doppler equation to find the wave frequency that is DETECTED by sub 2...

$$f_2=f_1\frac {V + V_{sub2}}{V-V_{sub1}}$$ which equals 1416 Hz

I noted that this 1416 Hz is also the frequency that is REFLECTED by sub 2. So now we're dealing with waves of f= 1416 Hz directed toward a moving object, i.e., sub 1. Sub 2 now has NOTHING to do with the second Doppler equation to find the wave frequency DETECTED by sub 1, which is...

$$f_1=f_1\frac {V + V_{sub1}}{V}$$ which equals 1423 Hz

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mjsd
Homework Helper
I noted that this 1416 Hz is also the frequency that is REFLECTED by sub 2. So now we're dealing with waves of f= 1416 Hz directed toward a moving object, i.e., sub 1. Sub 2 now has NOTHING to do with the second Doppler equation to find the wave frequency DETECTED by sub 1, which is...

$$f_1=f_1\frac {V + V_{sub1}}{V}$$ which equals 1423 Hz

As I have predicted, the issue here is the second part. Sub 2 has everything to do with the doppler equation, it is a moving emitter (in the frame of the Earth)! If you do things like I did and pick the frame of one of the Subs, things would be a lot easier and less confusing. If the reflection mechanism is confusing you, use the frame of Sub 2 and so it is stationary in that frame, and Sub 1 becomes a moving emitter (at 17m/s in) in the first part and a moving receiver (at 17m/s in) in the second part)... you will get the same answer no matter which frame you choose.

The fact that the reflected wave has same frequency as the (doppler shifted) incoming wave, does not mean that Sub 2 is effectively stationary (in the Earth frame), all it is telling you is that Sub 2, as an emitter, is emitting waves of a constant (doppler shifted) frequency, BUT it is still a moving emitter in the Earth frame and as far as Sub 1 is concerned.

Choosing a reference frame so that sub 2 is stationary and sub 1 is moving at 17 m/s is a bad move. You get close to the right answer though because the speeds of the subs are very small compared to the speed of the waves. Consider this same system where the speed of the wave is lets say, 100 m/s and you'll see what I mean. Using you're method, you'll get the frequency detected by sub 2 as 1638 Hz as opposed to 1811 Hz using the Doppler method. This should clear up what you're talking about in the second paragraph of you're response too.
Finally, the objective of this thread is to gain some authority with which to challenge my professor who marked my answer of 1423 Hz as incorrect on my midtem exam. ???

Doc Al
Mentor
All I need is a definite authoritative snswer on this problem before this thread makes it to the circular file..."Doc Al", please help...
Sorry, but I just saw this. If ever you want my attention, feel free to PM me.

FYI: Snazzy's response is correct.

So this is where we disagree. I have been trying to argue for the past week that the frequency that is detected by any object will be the SAME frequency that the object emits/reflects. This is the case whether the object is moving or not (so long as the object is moving on an axis parallel to the velocity of the wave). Please give this some thought before responding.
Is the frequency detected by sub 2 the same as the frequency emitted by sub 2? Yes! And that's true whether sub 2 is moving or not. But the fact that it's moving will certainly impact the frequency of the reflected sound that is detected by sub 1.

First, I'd like to thank you Snazzy and Mjhd for your assistance with this problem.

So I used the first Doppler equation to find the wave frequency that is DETECTED by sub 2...

$$f_2=f_1\frac {V + V_{sub2}}{V-V_{sub1}}$$ which equals 1416 Hz
So far, so good.

I noted that this 1416 Hz is also the frequency that is REFLECTED by sub 2. So now we're dealing with waves of f= 1416 Hz directed toward a moving object, i.e., sub 1.
Right.
Sub 2 now has NOTHING to do with the second Doppler equation to find the wave frequency DETECTED by sub 1, which is...

$$f_1=f_1\frac {V + V_{sub1}}{V}$$ which equals 1423 Hz

Using that logic, you could also argue that the speed of sub 1 has nothing to do with the first Doppler equation. (After all, sub 1 emits waves of f = 1400 Hz toward a moving object.) But for some reason you didn't fall for that!

The speed of both source and observer count in both Doppler shifts.

Hi Doc Al! Thanks for the response. Bear with me though, b/c I'm still thinking about my reply.

Doc Al, how do you do this so fast? As always you're correct except for when you agreed with me on the following...
QUOTE: "I noted that this 1416 Hz is also the frequency that is REFLECTED by sub 2. So now we're dealing with waves of f= 1416 Hz directed toward a moving object, i.e., sub 1.

Right."

Using your logic you probably meant to agree that waves of f = 1424 Hz are now directed toward a moving object, i.e. sub 1. In that case we would get the textbook answer of 1431 Hz as the detected frequency of sub 1. Thanks for you're help. By the way how do I "PM" you (I'm not very computer literate)

Doc Al
Mentor
As always you're correct except for when you agreed with me on the following...
QUOTE: "I noted that this 1416 Hz is also the frequency that is REFLECTED by sub 2. So now we're dealing with waves of f= 1416 Hz directed toward a moving object, i.e., sub 1.

Right."
No, I think you were correct in that statement, but it depends on how you interpret it. Here's what it means to me: Sub 2 detects the incoming sound as having a frequency of 1416 Hz, and it reflects that sound right back at 1416 Hz. That means that Sub 2 is now a source of 1416 Hz sound. (Exactly like Sub 1 was a source of the original 1400 Hz sound.) And I can use the Doppler equation to find out the observed frequency according to any detector of that sound, namely Sub 1. Of course, Sub 2 is a moving source, since it moves with respect to the water. (Exactly like Sub 1 was a moving source of the original sound.)

Using your logic you probably meant to agree that waves of f = 1424 Hz are now directed toward a moving object, i.e. sub 1. In that case we would get the textbook answer of 1431 Hz as the detected frequency of sub 1.
How did you get that answer of 1424 Hz? Looks like you found the frequency that would be observed by a stationary observer, and then Doppler shifted a 3rd time to get the textbook answer. That's the hard way!

By the way how do I "PM" you (I'm not very computer literate)
One way is to click on someone's name. You'll see an option: "send a private message to..."