It is a beautiful summer night and you are sitting on the shore of a quiet lake. There is a light breeze coming from the lake to the shore. You see a distant light coming from a party boat, where people listen to loud music. But the boat is far away and you do not hear anything. Suddenly, as the breeze changes, you hear the music pretty well. The effect fades and comes back with the changes of the wind. Obviously, the wind “carries” the sound.

How can the wind carry the sound, when the velocity of the wind (a few m/s) is much less than the velocity of the sound ( \(c=340\)m/s)? Here is a model that can explain the effect.

Assume that close to the ground the wind velocity \(v_s\) changes as a function of height from the ground \(y\) as \(v_s = v_0 (y/h_0)^2\), and the wind blows in the direction from the boat to the observer. Let us take \(v_0=0.2\)m/s at the elevation of \(h_0=10\)m. Show that the sound waves that are originally emitted slightly upwards will bend towards the ground due to this “wind shear”. Show that at a certain distance the sound waves that would normally pass well over the observer’s head will be, in fact, focused to the point where the observer is standing. Calculate the distance between the observer and the boat where this effect occurs. Round your result to the nearest \(0.5\)km.

Hints: Consider the vertical plane connecting the ship and the observer. In this plane sound waves are emitted from the ship in all (upwards) angles relative to the horizontal. We are concerned with small angles only. For small angles approximate the path of the sound waves with parabolas. Calculate the time for the sound to travel from the ship to the shore and use Fermat’s principle to determine the distance where the sound waves emitted in the various directions will all come back to the same point. In calculating the integrals, use first-order expansion in \(v_0/c\).

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