Stationary sound waves problem

  • Thread starter Simplyjack
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  • #1

Homework Statement


A car is driven at 30m/s between two sources, each producing sound waves of 1.50x10^6 Hz. The sound heard by driver rises and fades as he travels from one sound source to the other. Find the period of regular fading.


Homework Equations



No relevant equations provided.

The Attempt at a Solution



Since the sound rise and fades along the line connecting the two sources, I believe a standing wave has been formed, where the two end points are open ends. I figure that in one second, the car moves 30m and the waves make 1.50x10^6 cycles, so the wavelength (which is the distance covered in one cycle) must equal 2x10^-5 m (or λ=v/f). Since fading takes place at every node, that is λ/2, distance between adjacent nodes = 0.00001 m. how do I translate
that into time interval between successive nodes? The answer given is 3.3 seconds.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
TSny
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Hi, Simplyjack. Are you sure the frequency that you gave is correct? I don't think even batman could hear that high of a frequency.

I suspect this is a problem where you are meant to use the concepts of the "Doppler effect" and "beats". Are those things that you've studied?
 
  • #3
TSny
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Actually, your approach to the problem will yield the correct answer. You just need to find the time for the traveler to go from one node to the next. You know the distance (λ/2). So, how much time does it take the car to move this distance?

I still think the frequency given in the problem is strange. Also, the answer of 3.3 s seems odd. For such a high frequency, I would expect the answer to be of the order of microseconds.
 
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  • #4
TSny
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Since the sound rise and fades along the line connecting the two sources, I believe a standing wave has been formed, where the two end points are open ends. I figure that in one second, the car moves 30m and the waves make 1.50x10^6 cycles, so the wavelength (which is the distance covered in one cycle) must equal 2x10^-5 m (or λ=v/f).

In the expression λ = v/f, v is not the speed of the car. What speed is it?
 
  • #5
Hi TSny,

Thanks for your comments. I got this question from a friend but as i did not copy it word for word, I did not remember correctly and misinterpreted the problem. I think it's not sound waves but rather radio waves, hence the high frequency. Therefore I can use v = 3.0x10^8 m/s and f = 1.50x10^6 Hz and obtain λ as 200m.

Since distance between successive nodes = λ/2 = 100m, and car velocity is 30m/s, I can equate 30 to 100/T, giving T to be 3.33 s. But Im not sure if driver will still be able to "hear" the rise and fall of such high frequencies.

Thanks for helping me to solve my own problem.

Simplyjack
 

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