Two harmonic waves moving in the same direction superimpose to form a wave $\mathrm{x}=\mathrm{a} \cos (1.5…
- 6 s
- 4 s
- 1 s
- 2 s
Solution
$\mathrm{x}=\frac{\mathrm{a}}{2} \cos [1.5+50.5] \mathrm{t}+\frac{\mathrm{a}}{2} \cos [50.5-1.5]$
$x=\frac{a}{2} \cos [52 t]+\frac{a}{2} \cos [49 t]$
Here, $2 \pi \mathrm{f}_1 \& 2 \pi \mathrm{f}_2=49$
$\mathrm{f}_1=\frac{52}{2 \pi}, \mathrm{f}_2=\frac{49}{2 \pi}$
$\therefore \mathrm{f}_{\text {Bat }}=\mathrm{f}_1-\mathrm{f}_2=\frac{3}{2 \pi} \mathrm{~Hz}$
$\therefore \mathrm{T}_{\text {Beat }}=\frac{1}{\mathrm{f}_{\text {Beat }}}=\frac{2 \pi}{3} \mathrm{sec}$
$=2.09 \mathrm{sec} \approx 2 \mathrm{sec}$
Asked in: JEE Main 2025 (07 Apr Shift 1)