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Bouncing ball and Doppler effect

  1. Jun 7, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider two parallel walls perfectly reflective placed at the distance ##d = 0.8 m ##. A ball, provided with a device through which are emitted continuously frequency sound waves equal to ##f_0=430 Hz##, is launched from one wall to another. It moves with constant velocity ##v##. After some time, an observer placed at the position where the ball was thrown a beat frequency equal to ##5 Hz##.
    Compute:

    a) the speed of the ball;

    b) assuming the formation of standing waves when the two walls are made the same traveling conditions (nodes) of the gas particles, determine the speed at which must travel the ball so that between the two walls standing waves will be generated with frequency equal to twice the fundamental frequency. Neglect, in this calculation, the higher frequency sound wave presence.

    Assume the speed of sound ##c = 340 m / s##.

    [Results: a) ## v = 1.98 m / s ##; b) ## v = 4 m / s ##]

    2. Relevant equations
    Doppler effect : $$f^*=f_0(\frac{c}{c\pm v_{ball}})$$

    3. The attempt at a solution

    a. I get confused because the result is correct if I use
    $$f_{beats}=|f_0(\frac{c}{c- v_{ball}})-f_0(\frac{c}{c+v_{ball}})|$$
    But this does not seem correct to me. Immediately after the first bounce the ball i moving towards the observer, so it should be
    $$f_{beats}=|f_0(\frac{c}{c- v_{ball}})-f_0(\frac{c}{c-v_{ball}})|=0$$

    Which of course is not.

    b. This is the point I cannot do at all. What is the strategy to use here?

    Any help is really appreciated.
     
  2. jcsd
  3. Jun 9, 2016 #2

    ehild

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    The ball continuously emits sound. The sound also reflects from the wall, so the observer hears the sounds emitted by both the receding and approaching balls.
     
  4. Jun 12, 2016 #3
    Thanks for the reply!

    I thought about the problem for a while, I'm totally ok with point a. now, but point b. is still a mistery for me.

    How does the speed of the ball influences the formation of standing waves?

    The frequency twice of the fundamental is ##f=2 \frac{c}{2d}## but the ball emits at ##f_0## regardless its velocity.

    So can I deal with point b?
     
  5. Jun 12, 2016 #4

    ehild

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    The
    The speed of the ball influences the frequency of the sound waves which propagate in air forward and backward between the walls. These two waves generate the standing wave. The problem says to neglect the higher frequency wave. So you have sound wave of frequency f*=f0 c/(c+v), find v.
     
  6. Jun 23, 2016 #5
    Thanks for the reply! I thought about it again and I understood your solution, except one thing.

    The ball is continuosly emitting, so the reflected waves interfere not only with the other reflected waves, but also with the waves just emitted by the moving ball, don't they? So shouldn't this be taken in to account? I mean the ball should have a velocity such that it does not emit at all when it is in the position of a node and emits when it is on an antinode?

    Maybe that's a wrong complication to the problem that I'm thinking about, but is it really only necessary to use doppler effect and neglect the fact that the ball is continously emitting?
     
  7. Jun 23, 2016 #6

    ehild

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    No, the ball emits continuously at frequency fo=430 Hz. The emitted sound makes sound waves travelling in air at higher frequency in the direction of the velocity of the ball, and at lower frequency in the opposite direction. The sound emitted by the ball is connected to the vibration of the ball or some part of it, but the sound wave travelling in the air is connected to vibration of air molecules.
    It is the sound wave travelling in air that reflects from the walls and interfere with the sound wave travelling in opposite direction.
    Yes, it would be enough to consider only the sound wave emitted at an instant, and ignoring the emitting ball. But that wave would extinct in a short time. The continuous emission of sound ensures constant intensity, making the formation of standing waves possible.
    There are sound waves with two different frequencies, but the problem says to ignore the one with higher frequency. And the conditions correspond to standing wave formation for one frequency only.
     
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