Doppler's Effect in Sound waves.

The problem is all about finding the maximum component of the velocity towards (and away) from the observer. That is when the frequency will be maximum (and minimum). (The maximum component of velocity towards the observer will be the same as the maximum component of velocity away from the observer.) From there you can use a basic understanding of the doppler effect.

I put the origin of a coordinate system at the center of the circle that the source was traveling in (radius = 10 meters). I also put the observer at the fixed (x,y) coordinate (5,0). Call the point on the circumference where the source is at "point S" and the point where the observer is at "point B" and we will call the origin "point O" I can't really explain without a diagram, but you will find that the component of velocity ( $V$ ) in the direction of the observer is ( $V$ ) times the sine of angle OSB. So the problem comes down to finding when angle OSB is maximum.

I will call the angle SOB theta ( $\theta$ ) (which only needs to be between zero and pi) and then we can express angle OSB in terms of $\theta$

First I used the law of cosines to find the length of the unknown side in terms of $\theta$

$L=\sqrt{10^2+5^2-2(5)(10)\cos(\theta)}=5\sqrt{5-4\cos(\theta)}$

Then we can use the law of sines to find the sine of angle OSB in terms of $\theta$ which is

$\sin(OSB)=\frac{5\sin(\theta)}{5\sqrt{5-4\cos(\theta)}}=\frac{\sin(\theta)}{\sqrt{5-4\cos(\theta)}}$

Now we differentiate sin(OSB) with respect to $theta$ and set it equal to zero to find where the maximum (and minimum) frequencies will occur. Doing this yields $\theta=\frac{\pi}{3}$ and thus

$\sin(OSB)=\frac{\sin(\theta)}{\sqrt{5-4\cos(\theta)}}=0.5$

This means that the maximum component of the velocity towards and away from the observer will be half the speed of the source, which is then a quarter of the speed of sound.

I won't explain the doppler effect, but if you know about it then you will know that this means the maximum and minimum frequencies heard by the observer (respectively) are $\frac{4}{3}f$ and $\frac{4}{5}f$ which means the sum of them (and the answer to this problem) is $600(\frac{4}{3}+\frac{4}{5})=1280$

Imgur

The positions (S' & S) for max velocity of approach and max velocity of separation will occur symmetrically on the circular path relative to the line joining center (C) and observer (O) as shown.

These these velocities to be maximum, $\alpha$ must be maximum. $\frac{5}{\sin \alpha} = \frac{10}{\sin (\alpha+\theta)}$ Differentiating and equating $\frac{d\alpha}{d\theta} = 0$ to get max $\alpha$ at $\theta = 60$

Thus max velocity of both approach and separation = $\frac{C}{2} \cos 60$

Giving max frequency $a = \frac{4}{3}f_{0}=\frac{4}{3}600 = 800$ and min frequency $b = \frac{4}{5}f_{0}=\frac{4}{5}600 = 480$

And a + b = 1280 Hz