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

The given equation can be written as
\(\mathrm{x}=\frac{\mathrm{a}}{2} \cos [1.5+50.5] \mathrm{t}+\frac{\mathrm{a}}{2} \cos [50.5-1.5]\)
\(x=\frac{a}{2} \cos [52 t]+\frac{a}{2} \cos [49 t]\)
Here, \(2 \pi \mathrm{f}_1 \& 2 \pi \mathrm{f}_2=49\)
\(\mathrm{f}_1=\frac{52}{2 \pi}, \mathrm{f}_2=\frac{49}{2 \pi}\)
\(\therefore \mathrm{f}_{\text {Bat }}=\mathrm{f}_1-\mathrm{f}_2=\frac{3}{2 \pi} \mathrm{~Hz}\)
\(\therefore \mathrm{T}_{\text {Beat }}=\frac{1}{\mathrm{f}_{\text {Beat }}}=\frac{2 \pi}{3} \mathrm{sec}\)
\(=2.09 \mathrm{sec} \approx 2 \mathrm{sec}\)