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(D) The given equation is $x = a \cos(1.5 t) \cos(50.5 t)$.Using the trigonometric identity $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,we can rewrite t

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(D) The given equation is $x = a \cos(1.5 t) \cos(50.5 t)$.Using the trigonometric identity $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,we can rewrite the equation as:$x = \frac{a}{2} [\cos(50.5 t + 1.5 t) + \cos(50.5 t - 1.5 t)]$$x = \frac{a}{2} \cos(52 t) + \frac{a}{2} \cos(49 t)$.Here,the angular frequencies are $\omega_1 = 52 \ rad/s$ and $\omega_2 = 49 \ rad/s$.The beat frequency is given by $f_{	ext{beat}} = |f_1 - f_2| = \frac{|\omega_1 - \omega_2|}{2 \pi}$.$f_{	ext{beat}} = \frac{52 - 49}{2 \pi} = \frac{3}{2 \pi} \ Hz$.The beat period is $T_{	ext{beat}} = \frac{1}{f_{	ext{beat}}} = \frac{2 \pi}{3} \ s$.Substituting $\pi \approx 3.14$,we get $T_{	ext{beat}} = \frac{2 	imes 3.14}{3} = \frac{6.28}{3} \approx 2.09 \ s$.Rounding to the nearest integer,the beat period is $2 \ s$.

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

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