Two harmonic waves moving in the same direction superimpose to form a wave $x=\mathrm{a} \cos (1.5 \mathrm{t}) \cos (50.5 \mathrm{t})$ where t is in seconds. Find the period with which they beat. (close to nearest integer)

<p>The superposition of the two harmonic waves results in the wave equation:</p> <p>$ x = a \cos(1.5t) \cos(50.5t) $</p> <p>This equation can be expanded using the trigonometric identity for the product of cosines:</p> <p>$ x = \frac{a}{2} \cos[(1.5 + 50.5)t] + \frac{a}{2} \cos[(50.5 - 1.5)t] $</p> <p>Simplifying, we find:</p> <p>$ x = \frac{a}{2} \cos(52t) + \frac{a}{2} \cos(49t) $</p> <p>This represents two waves with angular frequencies $52$ and $49$, respectively.</p> <p>To find the beat frequency, we calculate the differences in frequencies:</p> <p>$ f_1 = \frac{52}{2\pi}, \quad f_2 = \frac{49}{2\pi} $</p> <p>The beat frequency is then:</p> <p>$ f_{\text{Beat}} = f_1 - f_2 = \frac{3}{2\pi} \text{ Hz} $</p> <p>The period of the beats, $T_{\text{Beat}}$, is the reciprocal of the beat frequency:</p> <p>$ T_{\text{Beat}} = \frac{1}{f_{\text{Beat}}} = \frac{2\pi}{3} \text{ sec} $</p> <p>Approximating this gives:</p> <p>$ T_{\text{Beat}} \approx 2.09 \text{ sec} \approx 2 \text{ sec} $</p>