Find the speed of the source (Doppler Effect)

[mastermind1]

A kid is riding a bike and is ringing bell of 3000 Hz. You are standing still on the sidewalk.When the bicycle moves toward you, you hear a shift in the frequency of the bell. In addition, when the bicycle moves away from you, you hear a different shift in the frequency of the bell. If there is a 30 Hz change in the frequencies that you hear for the bell as the bicycle approaches you and then moves away from you, what is the speed (in m/s) of the bicycle? (Write your answer to the nearest 0.01 m/s.)

1. Homework Statement

Frequency(source)(fs) = 3000 Hz
speed(observer)(vo) = 0 m/s
Frequency(towards)(ft)= 3030 Hz (NOT SURE)
Frequency(away)(fa) = 2970 Hz (NOT SURE)
speed of sound(v) = 343 m/s

Homework Equations

fo = ((v+vo)/(v-vs)) * fs Bike is moving TOWARDS me
fo = ((v-vo)/(v+vs)) * fs Bike is moving AWAY from me

The Attempt at a Solution

Towards
------------
3030Hz/3000Hz = (343 m/s) / (343 m/s - vs)
(1.01)(343 m/s - vs) = 343 m/s
345.20 m/s - 1.01 vs = 343 m/s
vs = 2.18 m/s

I do the same thing using the AWAY equation, and i get the same answer.
However, that answer is incorrect. I do not know what I am doing wrong.

 

[mfb]

The 30 Hz difference are between "towards you" and "away from you". They are not relative to the 3000 Hz.

 

[mastermind1]

The 30 Hz difference are between "towards you" and "away from you". They are not relative to the 3000 Hz.

Could you please elaborate on that. And, give me a hint.

 

[mfb]

If the bike approaches you, the frequency is x, a bit higher than 3000 Hz.
If the bike moves away from you, the frequency is y, a bit lower than 3000 Hz.
x - y = 30 Hz.
Find formulas for x and y and solve the equations, there is no trick involved.

 

[mastermind1]

If the bike approaches you, the frequency is x, a bit higher than 3000 Hz.
If the bike moves away from you, the frequency is y, a bit lower than 3000 Hz.
x - y = 30 Hz.
Find formulas for x and y and solve the equations, there is no trick involved.

That's what I am doing. I added and subtracted 30Hz.

 

[mfb]

You used x - 3000 Hz = 30 Hz and 3000 Hz - y = 30 Hz, but that's not right. It would imply x-y = 60 Hz.

 

[mastermind1]

You used x - 3000 Hz = 30 Hz and 3000 Hz - y = 30 Hz, but that's not right. It would imply x-y = 60 Hz.

So you are saying that 30Hz is the total change in frequency I hear for towards and away. That means that I hear 15 Hz for towards and 15 Hz for away.
If so, then it is 3000-15=2985(away) and 3000+15=3015(towards). Then I just use the equation I used above to get the answer.

 

[mfb]

So you are saying that 30Hz is the total change in frequency I hear for towards and away.

That's what the problem statement says.

That means that I hear 15 Hz for towards and 15 Hz for away.

It is not, the doppler shift is not symmetric.

 

[mastermind1]

That's what the problem statement says.It is not, the doppler shift is not symmetric.

I don't think I get it yet. You will need to help me with equations.

 

[mfb]

All necessary equations are in the thread already, you just have to combine them.

 

[mastermind1]

All necessary equations are in the thread already, you just have to combine them.

You know what I am trying to tell you. If I had understood what you were trying to tell me, I would have done it by now.
I don't get it.

 

[mastermind1]

You know what I am trying to tell you. If I had understood what you were trying to tell me, I would have done it by now.
I don't get it. It's just one question, and I am dedicated enough to waste hours on it. It's certainly not going to make any difference to my overall grade in class if I get it wrong. I am not good at physics, and that is why I am having trouble with this.

 

[mfb]

You know what I am trying to tell you.

No I don't know. The problem got reduced from a problem about the Doppler effect to a topic your classes covered several years ago: plugging in equations and numbers into each other.

 

[mastermind1]

No I don't know. The problem got reduced from a problem about the Doppler effect to a topic your classes covered several years ago: plugging in equations and numbers into each other.

You have been telling me to do the same thing over and over, and I don't understand it. You are an expert. And, I appreciate your help.
But, if I am your student in this case, and I am not understanding it, then you should help me with a different approach. Please, help me in a different way.

 

[mfb]

There is only one reasonable approach, and I don't know how to explain it differently, sorry.

 

[Darious Warren]

You used x - 3000 Hz = 30 Hz and 3000 Hz - y = 30 Hz, but that's not right. It would imply x-y = 60 Hz.

So would it be 3000-60?

 

[SammyS]

So would it be 3000-60?

Definitely not.

 

[Darious Warren]

Definitely not.

Wait! I have a question. When mfb said x-y=30, does he mean that the x and y values are the frequencies the observer hears?

 

[SammyS]

Wait! I have a question. When mfb said x-y=30, does he mean that the x and y values are the frequencies the observer hears?

Yes, he is using x and y for frequencies to make a statement about his understanding of the problems statement.

Well this is what @mfb wrote in Post #4.

If the bike approaches you, the frequency is x, a bit higher than 3000 Hz.
If the bike moves away from you, the frequency is y, a bit lower than 3000 Hz.​

.

 

[Darious Warren]

Okay. I got the idea of what the answer is! Thank you @SammyS

 

[SammyS]

A kid is riding a bike and is ringing bell of 3000 Hz. You are standing still on the sidewalk.When the bicycle moves toward you, you hear a shift in the frequency of the bell. In addition, when the bicycle moves away from you, you hear a different shift in the frequency of the bell. If there is a 30 Hz change in the frequencies that you hear for the bell as the bicycle approaches you and then moves away from you, what is the speed (in m/s) of the bicycle? (Write your answer to the nearest 0.01 m/s.)

1. Homework Statement
Frequency(source)(fs) = 3000 Hz
speed(observer)(vo) = 0 m/s
Frequency(towards)(ft)= 3030 Hz (NOT SURE)
Frequency(away)(fa) = 2970 Hz (NOT SURE)
speed of sound(v) = 343 m/s

Homework Equations

fo = ((v+vo)/(v-vs)) * fs Bike is moving TOWARDS me
fo = ((v-vo)/(v+vs)) * fs Bike is moving AWAY from me

The Attempt at a Solution

: Towards
3030Hz/3000Hz = (343 m/s) / (343 m/s - vs)
(1.01)(343 m/s - vs) = 343 m/s
345.20 m/s - 1.01 vs = 343 m/s
vs = 2.18 m/s

I do the same thing using the AWAY equation, and i get the same answer.
However, that answer is incorrect. I do not know what I am doing wrong.

You made some error in multiplying (1.01)(343) which is 346.43. This is a huge error b/c if hold off on that multiplication and do some manipulation, you get (1.01)(343) − 343 = (1.01)v_s ..
Of course, that left hand side is just (0.01)(343) → 3.43

If you do the AWAY carefully and use 3000 − 30 → 2970 (Hz) , you get a slightly greater bicycle speed .

That confirms the interpretation of the problem given by @mfb .

Take the two frequency equations: towards and away and introduce a way to distinguish the observed TOWARD frequency from the observed AWAY frequency.

$\displaystyle f_{o\,T} = \frac{v+v_o}{v-v_s} \cdot f_s \ \$ Bike is moving TOWARDS me

$\displaystyle f_{oA} = \frac{v-v_o}{v+v_s} \cdot f_s \ \$ Bike is moving AWAY from me

Just subtract the second equation from the first, That should give the difference in the two frequencies as pointed out by @mfb . A little trickier Algebra but do-able.

 

[mfb]

The thread is from 2015, it is unlikely that OP is still active here.

 

[SammyS]

The thread is from 2015, it is unlikely that OP is still active here.

LOL !

I suppose I saw it in the current list because of the post by @Darious Warren . I didn't pay attention to the date other than Sept.

Oh well, it kept me busy for a while.