# What frequency do you hear?

1. Mar 20, 2016

### ThePhysicsXV

1. The problem statement, all variables and given/known data
Temperature is 28C, a police car with a siren blazing at 988Hz is coming toward you at 23m/s so you run away at
7m/s.

2. Relevant equations

f=f(v+-o/v+-s)
3. The attempt at a solution
I'm confuse on the formula.

#### Attached Files:

• ###### image.jpg
File size:
35 KB
Views:
64
2. Mar 20, 2016

### vela

Staff Emeritus
What specifically is confusing you?

3. Mar 20, 2016

### ThePhysicsXV

I know how to solve to get the speed of sound but are they moving forward (+) or backward (-) and what's is the observer and source. Is the source the police siren?

4. Mar 20, 2016

### ThePhysicsXV

Do I use the formula I gave or the one from the picture?

5. Mar 20, 2016

### vela

Staff Emeritus
The observer is what detects the sound. The source is what produces the sound the observer hears.

Consider the two motions separately. What effect does the motion of the police car have on the frequency detected, and what effect does your movement have on the frequency detected?

6. Mar 20, 2016

### ThePhysicsXV

So is V = 988(348+7/ 348+8)
Or V=988(348-7/348-23)?

7. Mar 20, 2016

### ThePhysicsXV

348m/s is the speed of sound

8. Mar 20, 2016

### vela

Staff Emeritus

9. Mar 20, 2016

### ThePhysicsXV

10. Mar 20, 2016

### haruspex

How would you "solve to get the speed of sound"? The speed of sound is a given.
As vela says, the source is the siren, you are the observer.
Are the source and observer moving in the same direction or in opposite directions?

As vela suggests, try it in two parts. First, what frequency did you hear before starting to run? Can you quote an equation for the Doppler effect?

11. Mar 20, 2016

### ThePhysicsXV

The source and observer are moving on the same direction since the police car coming toward you, and you getting away.

V sound in air = 331.4 + 0.6T
T is temperature so temp=28
And you will get 348m/s

12. Mar 20, 2016

### haruspex

Ok, but what about the Doppler effect? So far, I see no evidence that you've even heard of it.

13. Mar 20, 2016

### ThePhysicsXV

Doppler Effect for Moving Source and Observer
ƒ'=ƒ((1±u₀/v)/(1±u(sound)/v))

14. Mar 20, 2016

### haruspex

Ok, so apply that.

15. Mar 20, 2016

### ThePhysicsXV

Actually they both moving at different speeds. Police at 23m/s and "you" at 7m/s

16. Mar 20, 2016

### haruspex

Yes, I replied a little too quickly, then edited it.
I anticipated that you would quote a formula for only one movement, so I was trying to get you to solve it in the two stages, as vela had suggested. When I realised your equation allowed for both moving I edited my post, but you were too quick for me.
So, apply the equation.

17. Mar 20, 2016

### ThePhysicsXV

f=988(348+7/348+23)
f= 945Hz

18. Mar 20, 2016

### ThePhysicsXV

#### Attached Files:

• ###### image.jpg
File size:
35.3 KB
Views:
63
19. Mar 20, 2016

### haruspex

I did have some concerns about the equation you posted previously, but I wanted to see how you applied it. The equations in the image you have attached look fine, except that you should ignore the references to "Stationary" observer. That makes no sense given the rest of the text. It's probably a cut-and-paste error from some preceding cases.
So, which of the two cases in the image do you think applies here? What do you get when you apply it?

20. Mar 20, 2016

### ThePhysicsXV

f= f(v-o/v-s)
I got 1036.64Hz