Two sources of sound placed closed to each other, are emitting...

The given equations of the progressive waves are:
$y_1 = 4\sin(600\pi t)$
$y_2 = 5\sin(608\pi t)$

Comparing these with the standard equation $y = A\sin(\omega t)$, we get the angular frequencies:
$\omega_1 = 600\pi \Rightarrow 2\pi f_1 = 600\pi \Rightarrow f_1 = 300$ Hz
$\omega_2 = 608\pi \Rightarrow 2\pi f_2 = 608\pi \Rightarrow f_2 = 304$ Hz

The beat frequency is the difference between the frequencies of the two waves:
$n = |f_1 - f_2| = |300 - 304| = 4$ beats/s

The amplitudes of the two waves are $A_1 = 4$ and $A_2 = 5$.
The maximum amplitude during waxing is A_\max = A_1 + A_2 = 4 + 5 = 9.
The minimum amplitude during waning is A_\min = |A_1 - A_2| = |4 - 5| = 1.

Since intensity is proportional to the square of the amplitude (I \propto A^2), the intensity ratio between waxing and waning is:<br>\dfrac{I_\max}{I_\min} = \left(\dfrac{A_\max}{A_\min}\right)^2 = \left(\dfrac{9}{1}\right)^2 = \dfrac{81}{1}

Thus, the observer will hear $4$ beats/s with an intensity ratio of $81 : 1$.

Answer: $4$ beats/s with intensity ratio $81 : 1$ between waxing and waning.