Two waves $Y_{1}=0 \cdot 25 \sin 316 \mathrm{t}$ and $\mathrm{Y}_{2}=0 \cdot 25 \sin 310 \mathrm{t}$ are…

Two waves $Y_{1}=0 \cdot 25 \sin 316 \mathrm{t}$ and $\mathrm{Y}_{2}=0 \cdot 25 \sin 310 \mathrm{t}$ are propagating along the same direction. The number of beats produced per second are
  1. $\frac{\pi}{2}$
  2. $\frac{2}{\pi}$
  3. $\frac{\pi}{3}$
  4. $\frac{3}{\pi}$

Solution

No. of beats $=\mathrm{n}_{1}-\mathrm{n}_{2}$ $\mathrm{Y}_{1}=0.25 \sin 316 \mathrm{t}$ $\therefore \frac{2 \pi}{\mathrm{T}}=316 \quad \therefore \mathrm{n}_{1}=\frac{316}{2 \pi}$ $\mathrm{Y}_{2}=0.25 \sin 310 \mathrm{t}$ $\therefore \mathrm{n}_{2}=\frac{310}{2 \pi}$ $\quad \mathrm{n}_{1}-\mathrm{n}_{2}=\frac{316-310}{2 \pi}=\frac{6}{2 \pi}=\frac{3}{\pi}$ .

Asked in: MHT CET 2020 (16 Oct Shift 1)

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