The superposition of the two harmonic waves results in the wave equation:

$ x = a \cos(1.5t) \cos(50.5t) $

This equation can be expanded using the trigonometric identity for the product of cosines:

$ x = \frac{a}{2} \cos[(1.5 + 50.5)t] + \frac{a}{2} \cos[(50.5 - 1.5)t] $

Simplifying, we find:

$ x = \frac{a}{2} \cos(52t) + \frac{a}{2} \cos(49t) $

This represents two waves with angular frequencies $52$ and $49$, respectively.

To find the beat frequency, we calculate the differences in frequencies:

$ f_1 = \frac{52}{2\pi}, \quad f_2 = \frac{49}{2\pi} $

The beat frequency is then:

$ f_{\text{Beat}} = f_1 - f_2 = \frac{3}{2\pi} \text{ Hz} $

The period of the beats, $T_{\text{Beat}}$, is the reciprocal of the beat frequency:

$ T_{\text{Beat}} = \frac{1}{f_{\text{Beat}}} = \frac{2\pi}{3} \text{ sec} $

Approximating this gives:

$ T_{\text{Beat}} \approx 2.09 \text{ sec} \approx 2 \text{ sec} $