What is the frequency of a police siren when you run away from it?

  • Thread starter ThePhysicsXV
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In summary, the conversation discusses the application of the Doppler effect to a scenario involving a police car with a siren and a person running away. The formula for the Doppler effect is given and there is confusion about which values to use for the observer and source, as well as the direction of motion. The conversation ends with the correct application of the formula to determine the frequency heard by the observer, which is approximately 1036.64Hz.
  • #1
ThePhysicsXV
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Homework Statement


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.

Homework Equations



f=f(v+-o/v+-s)

The Attempt at a Solution


I'm confuse on the formula.
 

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  • #2
What specifically is confusing you?
 
  • #3
vela said:
What specifically is confusing you?
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
vela said:
What specifically is confusing you?
Do I use the formula I gave or the one from the picture?
 
  • #5
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
vela said:
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?

So is V = 988(348+7/ 348+8)
Or V=988(348-7/348-23)?
 
  • #7
ThePhysicsXV said:
So is V = 988(348+7/ 348+8)
Or V=988(348-7/348-23)?
348m/s is the speed of sound
 
  • #8
Instead of plugging numbers in randomly, please answer the questions.
 
  • #9
vela said:
Instead of plugging numbers in randomly, please answer the questions.
I have no idea how to answer your question.
 
  • #10
ThePhysicsXV said:
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?
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
haruspex said:
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?
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
ThePhysicsXV said:
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
Ok, but what about the Doppler effect? So far, I see no evidence that you've even heard of it.
 
  • #13
haruspex said:
Ok, but what about the Doppler effect? So far, I see no evidence that you've even heard of it.

Doppler Effect for Moving Source and Observer
ƒ'=ƒ((1±u₀/v)/(1±u(sound)/v))
 
  • #14
ThePhysicsXV said:
Doppler Effect for Moving Source and Observer
ƒ'=ƒ((1±u₀/v)/(1±u(sound)/v))
Ok, so apply that.
 
  • #15
haruspex said:
Ok, so apply that to the case where the car is approaching you but you are standing still.
Actually they both moving at different speeds. Police at 23m/s and "you" at 7m/s
 
  • #16
ThePhysicsXV said:
Actually they both moving at different speeds. Police at 23m/s and "you" at 7m/s
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 realized your equation allowed for both moving I edited my post, but you were too quick for me.
So, apply the equation.
 
  • #17
haruspex said:
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 realized your equation allowed for both moving I edited my post, but you were too quick for me.
So, apply the equation.

f=988(348+7/348+23)
f= 945Hz
 
  • #18
haruspex said:
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 realized your equation allowed for both moving I edited my post, but you were too quick for me.
So, apply the equation.
 

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  • #19
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
haruspex said:
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?
f= f(v-o/v-s)
I got 1036.64Hz
 
  • #21
ThePhysicsXV said:
f= f(v-o/v-s)
I got 1036.64Hz
I get the same. Round that to the appropriate number of digits.
 
  • #22
haruspex said:
I get the same. Round that to the appropriate number of digits.
Okay, I guess that's the correct answer. But Thank you sir for your help
 

FAQ: What is the frequency of a police siren when you run away from it?

1. What is frequency and how is it measured?

Frequency is the number of times a sound wave repeats itself in a second, measured in Hertz (Hz). It is typically measured using specialized equipment such as an oscilloscope or frequency counter.

2. How does frequency affect what we hear?

The frequency of a sound wave determines the pitch of the sound. Higher frequencies correspond to higher pitches, while lower frequencies correspond to lower pitches. The human ear can typically hear sounds ranging from 20 Hz to 20,000 Hz.

3. Can everyone hear the same frequencies?

No, a person's ability to hear different frequencies can vary due to age, genetics, and exposure to loud noises. As we age, our ability to hear higher frequencies decreases. Additionally, some individuals may have a condition called high frequency hearing loss, which affects their ability to hear sounds above a certain frequency.

4. What is the frequency range of human speech?

The frequency range of human speech typically falls between 250 Hz to 8,000 Hz. This range allows us to hear the different tones and nuances in someone's voice.

5. How do animals hear different frequencies?

Animals have different hearing abilities and ranges compared to humans. For example, dogs can hear frequencies up to 45,000 Hz, while bats can hear frequencies up to 100,000 Hz. Some animals, like whales, can also communicate using very low frequencies that humans cannot hear.

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