Two waves $Y_1=0.25 \sin 316 \mathrm{t}$ and $\mathrm{Y}_2=0.25 \sin 310 \mathrm{t}$ are propagating in same…

Two waves $Y_1=0.25 \sin 316 \mathrm{t}$ and $\mathrm{Y}_2=0.25 \sin 310 \mathrm{t}$ are propagating in same direction. The number of beats produced per second are
  1. $\frac{3}{\pi}$
  2. $\frac{\pi}{3}$
  3. $\frac{\pi}{2}$
  4. $\frac{2}{\pi}$

Solution

$\begin{aligned} & \mathrm{y}_1=0.25 \sin 316 \mathrm{t}, \mathrm{y}_2=0.25 \sin 310 \mathrm{t} \\ & \therefore 2 \pi \mathrm{f}_1=316 \text { and } 2 \pi \mathrm{f}_2=310 \\ & \therefore \mathrm{f}_1=\frac{316}{2 \pi}=\frac{158}{\pi} \\ & \mathrm{f}_2=\frac{310}{2 \pi}=\frac{155}{\pi} \\ & \therefore \text { Beat frequency }=\mathrm{f}_1-\mathrm{f}_2 \\ & =\frac{158}{\pi}-\frac{155}{\pi}=\frac{3}{\pi}\end{aligned}$

Asked in: MHT CET 2021 (23 Sep Shift 2)

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