Police car siren (waves)

  • #1
~christina~
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[SOLVED] police car siren (waves)

Homework Statement


A police car speeds toward a warehouse door with its siren emitting sinusoidal waves of frequency fs = 300 Hz, intending to crash through the door. The car moves at 30 m / s and the ambient temperature is 35º C.

(a) What is the wavelength of the wave if the siren is stationary?

(b) Find the wavelengths of the waves in front of and behind the source when the siren is moving at 30 m / s.

(c) What frequency does the driver of the police car hear reflected from the warehouse?

(d) Does the driver detect a beat frequency? If yes, then find the beat frequency; if no, explain why not.

Homework Equations


[tex]v= (331m/s) \sqrt{1 + Tc/ 273} [/tex]

[tex] \lambda = v/f [/tex]

[tex] \lambda' = \lambda \pm vs/f [/tex]

[tex] f '= (v/(v-vs))f [/tex]

The Attempt at a Solution



a) wavelength if the siren is stationary

temp = [tex] 35 ^oC [/tex]
f= 300Hz

[tex]v= (331m/s) \sqrt{1 + Tc/ 273} [/tex]

[tex]v= (331m/s) \sqrt{1 + 35 ^oC [/tex]/ [tex] 273 ^oC [/tex]

v= 351.578m/s

[tex] \lambda = v/f [/tex]
[tex] \lambda[/tex] = 351.578m/s / 300Hz = 1.172m

b) find the wavelengths of the waves in front of and behind the source if the siren is moving at 30m/s

Not sure if this is the right equation for what they are asking but I used it for both in front of and behind the source wavelengths => [tex] \lambda' = \lambda \pm vs/f [/tex]

ALSO NOT SURE IF LAMBDA (without ') is the wavelength I found in part a) but I used it....could be incorrect but I don't see where I'd get lambda if I didn't use that one for lambda in equation.

front of car wavelength:
[tex] \lambda'_{front} = \lambda \pm vs/f [/tex]

[tex] \lambda'_{front} = 1.172m - (30m/s)/ 300Hz = 1.072m[/tex]

back of car wavelength:

[tex] \lambda'_{back} = \lambda \pm vs/f [/tex]

[tex] \lambda'_{back} = 1.172m + (30m/s)/ 300Hz = 1.272m [/tex]

c) frequency that the driver of the police car hears reflected from the wareheouse
I looked it up and I think that the frequency of the reflected wave is the same of the wave that hits it thus...(not sure though)

[tex] f '= (v/(v-vs))f [/tex]

v= velocity of sound in air = 351.578m/s (35 deg celcius)
f= 300Hz
vs= 30m/s

[tex] f '= [(351.578m/s)/ (351.578m/s -30m/s)]*(300Hz) = 327.9869 Hz [/tex]


d) does the driver detect a beat frequency? Yes then find the beat frequency if no explain why not.

I'm not even sure what that is however I tried looking online and found that a beat frequency is made up of more than one wave and also each wave has different frequencies...thus I think that since the police car siren has one frequency and the reflected wave has the same frequency I think and so my conclusion would be there is no beat frequency

IS this right?

(I saw an example however where a police car(stationary) had a radar wave which was directed toward a moving vehicle and the reflected waves created a beat frequency but this is the opposite in my case with the police car moving and instead of a stationary vehicle there is a wall Thus I'm really not sure now)


Can someone help me out?

THANKS
 
Last edited:

Answers and Replies

  • #2
Doc Al
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a) wavelength if the siren is stationary

temp = [tex] 35 ^oC [/tex]
f= 300Hz

[tex]v= (331m/s) \sqrt{1 + Tc/ 273} [/tex]

[tex]v= (331m/s) \sqrt{1 + 35 ^oC [/tex]/ [tex] 273 ^oC [/tex]

v= 351.578m/s

[tex] \lambda = v/f [/tex]
[tex] \lambda[/tex] = 351.578m/s / 300Hz = 1.172m
Looks good.
b) find the wavelengths of the waves in front of and behind the source if the siren is moving at 30m/s

Not sure if this is the right equation for what they are asking but I used it for both in front of and behind the source wavelengths => [tex] \lambda' = \lambda \pm vs/f [/tex]

ALSO NOT SURE IF LAMBDA (without ') is the wavelength I found in part a) but I used it....could be incorrect but I don't see where I'd get lambda if I didn't use that one for lambda in equation.

front of car wavelength:
[tex] \lambda'_{front} = \lambda \pm vs/f [/tex]

[tex] \lambda'_{front} = 1.172m - (30m/s)/ 300Hz = 1.072m[/tex]

back of car wavelength:

[tex] \lambda'_{back} = \lambda \pm vs/f [/tex]

[tex] \lambda'_{back} = 1.172m + (30m/s)/ 300Hz = 1.272m [/tex]
Looks good.
c) frequency that the driver of the police car hears reflected from the wareheouse
I looked it up and I think that the frequency of the reflected wave is the same of the wave that hits it thus...(not sure though)
I would agree with that statement. But note that the driver is moving towards that reflected wave.

[tex] f '= (v/(v-vs))f [/tex]

v= velocity of sound in air = 351.578m/s (35 deg celcius)
f= 300Hz
vs= 30m/s

[tex] f '= [(351.578m/s)/ (351.578m/s -30m/s)]*(300Hz) = 327.9869 Hz [/tex]
This is the frequency heard by someone standing by the warehouse. It gets reflected and then heard by the approaching car.


d) does the driver detect a beat frequency? Yes then find the beat frequency if no explain why not.

I'm not even sure what that is however I tried looking online and found that a beat frequency is made up of more than one wave and also each wave has different frequencies...thus I think that since the police car siren has one frequency and the reflected wave has the same frequency I think and so my conclusion would be there is no beat frequency

IS this right?
The question is: Does the driver hear two different frequencies? He can certainly hear his own siren. How does the frequency of the reflected wave sound to the driver? (That's the answer to the previous part.)
 
  • #3
~christina~
Gold Member
715
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Looks good.

Looks good.

I would agree with that statement. But note that the driver is moving towards that reflected wave.

This is the frequency heard by someone standing by the warehouse. It gets reflected and then heard by the approaching car.

so that creates 2 waves...I never thought of it like that.
But wouldnt the frequency heard by the driver be and average of the 2 waves thus:

[tex]f_{av}= (300Hz + 327.9869)/2= 313.99 Hz [/tex]

The question is: Does the driver hear two different frequencies? He can certainly hear his own siren. How does the frequency of the reflected wave sound to the driver? (That's the answer to the previous part.)

I say yes... but one is stationary and thus it's as if he wasn't moving at all right? thus I'd find the beat frequency fb= |f1-f2| and

f1= 300Hz (originally emitted by siren)
f2= 327.9869 Hz (reflected wave)

fb= |300Hz- 327.9869Hz|= 27.9869Hz

Thank you Doc :smile:
 
  • #4
Doc Al
Mentor
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so that creates 2 waves...I never thought of it like that.
But wouldnt the frequency heard by the driver be and average of the 2 waves thus:

[tex]f_{av}= (300Hz + 327.9869)/2= 313.99 Hz [/tex]
No. You calculated the frequency of the sound that hits the warehouse (327.98), which reflects with the same frequency. Now the warehouse is the source and the police car is the observer. Another Doppler shift!
 
  • #5
~christina~
Gold Member
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No. You calculated the frequency of the sound that hits the warehouse (327.98), which reflects with the same frequency. Now the warehouse is the source and the police car is the observer. Another Doppler shift!

wait..so the last part is incorrect?
--------
Oh...wow..

so observer is moving toward the source...thus I think it'd be..

f'= ((v+ v_o)/v)f

v= 351.578m/s => speed of sound at 35 deg celcius
[tex]v_o= 30m/s [/tex]
f= 327.9869 Hz

f'=((351.578m/s + 30m/s)/ 351.578 m/s)(327.9869Hz)= 355.973Hz

Is this right now? I think that it should but the last part where they say if there is a beat frequency...do I use this f I found and subtract that from the 300Hz that the police car has initially?

thus

fb= |f1-f2|

f1= 300Hz
f2= 355.973Hz

fb= |300Hz- 355.973Hz|= 55.973Hz

Thanks Doc Al
 
  • #6
Doc Al
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Looks good to me!
 
  • #7
~christina~
Gold Member
715
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Yay!
Thanks for your help :smile:
 

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