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A Honking Car Hears an Echo

  1. Aug 28, 2015 #1

    B3NR4Y

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    1. The problem statement, all variables and given/known data
    A car is traveling with a constant speed v on a straight road which forms an angle α with nearby straight wall. When the car is at a distance l from the wall the driver presses the horn briefly. What distance would the car have traveled (from the spot of the honk) when the driver hears the echo? The speed of sound is c.

    2. Relevant equations
    We're learning about the Euler Equation, but I'm not sure this fits that.

    3. The attempt at a solution
    The time for the sound to go the distance, x, to the wall is x/c, in this time the car travels [itex] \frac{x}{c} v [/itex], the sound travels the new distance in [itex] \frac{xv}{c^{2}} + \frac{x}{c} [/itex], but I'm not sure how to find the point at which they both meet. Since c is travelling faster than v, it will meet.
     
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  3. Aug 28, 2015 #2

    haruspex

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    To get that new distance, you seem to be assuming the car is moving directly away from the wall, not angle alpha.
    You might find it helpful to consider the image of the car and road as though reflected in the wall as a.mirror.
     
  4. Aug 29, 2015 #3

    B3NR4Y

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    If it reflected like a mirror, wouldn't the sound never reach the car? I'm assuming the sound is a single ray, travelling behind it and reflecting off the wall at an angle α, or from normal 90-α, and by the law of reflection it would bounce off at an angle α again, going away from the car.
     
  5. Aug 30, 2015 #4

    B3NR4Y

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    So I've done some thinking, and if the car honks its horn, the sound shouldn't only go behind it, but in front of it too in all directions. This includes an angle that bounces the sound back up to the car, in a minimum amount of time. Maybe this is where the Euler equation comes in?
     
  6. Aug 30, 2015 #5

    haruspex

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    Try my mirror approach. Draw the wall and the path of the car. Project the path of the car backwards to where it meets the wall. Draw the mirror image of the car's path, taking the wall to be a mirror. Draw the path of a sound wave that leaves the car at its initial position, bounces off the wall and reaches the car at its final position. What happens if you project that path of the sound wave through the wall instead of bouncing off?
     
  7. Aug 30, 2015 #6

    B3NR4Y

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    I've drawn something up that resembles what I think you mean. I haven't drawn it perfectly, as it's in paint, but I assume there are similar triangles or something I should be noticing. But first I want to check if this is how you intended?

    IjgKoIT.png
     
  8. Aug 30, 2015 #7

    haruspex

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    That's about right, but you have drawn the rebound angle very inaccurately. Compare it with the path through the wall. Shouldn't they have something in common?
    If we label the car's path AB, its path in the reflection A'B', the point where the back-projected path meets the wall O, and set the time elapsed to be t, you should be able to deduce a lot about the triangle OAB'.
     
  9. Aug 31, 2015 #8

    B3NR4Y

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    Okay, so I've drawn on paper that setup, and have gotten some pretty good insights I think. Where the car is initially I've labelled ##O##, where the mirror image of the car is I've labelled ##P(t)##. I want to find the point at which the line segment ##OP(t) = ct##, but I've run into trouble actually describing ##OP(t)## in a usable format.

    I've gotten other advice from people that the angle at which the sound bounces off and reaches the car is 45°, but I'm not convinced, because it says that the distance the sound travels to the wall is the same as the distance it travels back to the car. Which doesn't look right, though that could be due to the way I've drawn my triangles. If the angle the sound bounces off is 45°, it is rather nice because that means the distance the car has to travel is the distance to the wall at the start, l.

    I noticed that in attempting to describe ##OP(t)##, a lot of the angles I assumed were constant actually vary with time. It also seems massively complicated to find the length of this line segment. I just want to see if my reasoning so far has been correct.
     
  10. Aug 31, 2015 #9

    haruspex

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    No, the angle will not in general be 45 degrees. It's the same light, which follows what rule?
     
  11. Aug 31, 2015 #10

    B3NR4Y

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    Light follows the law of reflection (coincidentally we proved the law of reflection for the first problem of this set), which says that the angle with respect to normal is the same going in as it is going out. So if the angle was alpha going in, it would be alpha going out. I have redrawn the triangle to say what I am thinking. I need help deducing what the distance the car travels is and how it relates to the distance the light travels (other than by t), If only these were right triangles, I could easily use the Pythagorean Theorem.


    3Mi8Tlk.jpg
     
  12. Aug 31, 2015 #11

    haruspex

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    Yes, that's the diagram we need.
    You know lengths AO and AO'. OP = vt. Don't worry for now about the fact that we do not know t. To complete the triangle, how far is O'P?
     
  13. Aug 31, 2015 #12

    B3NR4Y

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    I messed up my primes, I meant the primes to be on the left. Oopsies, but I think you understood anyway.

    O'P should be ct away. That answer seems to be too obvious though, does it have to be in terms of l, α, and v?
     
  14. Sep 1, 2015 #13

    haruspex

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    ct is right. So now you have expressions for the three sides of that triangle in terms of the parameter t. You also have an angle. What equation can you write?
     
  15. Sep 1, 2015 #14

    B3NR4Y

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    The length of P'P, by the law of cosines, is

    $$
    |P'P| = 2l+2vtsin(\alpha)
    $$

    This means the length of OP' is
    $$
    |OP'|^{2} = (ct)^2 = (2l+2vtsin(\alpha))^{2} + (vt)^{2} - 4(l + vtsin(\alpha)) (vt) sin(\alpha)
    $$

    I'm going to work on simplifying this, just want to check if this is the track I should go down.

    I've simplified it down to ## (ct)^2 = 4l^2 + 4lvtsin(\alpha)+v^2 t^2 ##
     
  16. Sep 1, 2015 #15

    haruspex

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    Yes, that looks right. How will you proceed from there?
    .
     
  17. Sep 1, 2015 #16

    B3NR4Y

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    I momentarily forgot algebra exists, I solved that for t, and multiplied by v to get the distance.

    So my final answer is
    $$
    d=v\sqrt{\frac{4l^2+4lvtsin(\alpha)}{c^2 - v^2} } = 2v \sqrt{\frac{l^2+lvtsin(\alpha)}{c^2 - v^2}}
    $$
     
  18. Sep 1, 2015 #17

    haruspex

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    Looks good. It certainly satisfies a couple of sanity checks: c=v has no solution; l=-vt sin(alpha) has a d=0 solution.
     
  19. Sep 1, 2015 #18

    B3NR4Y

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    Awesome! Thank you so much for your help, and your patience.
     
  20. Sep 1, 2015 #19

    haruspex

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    A pleasure, and well done.
     
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