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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
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$\mathrm{Y}_{2}=0 \cdot 25 \sin 310 \mathrm{t}$
are propagating along the same direction. The number of beats produced per second are
Solution:
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Verified Answer
The correct answer is:
$\frac{3}{\pi}$
No. of beats $=\mathrm{n}_{1}-\mathrm{n}_{2}$
$\mathrm{Y}_{1}=0.25 \sin 316 \mathrm{t}$
$\therefore \frac{2 \pi}{\mathrm{T}}=316 \quad \therefore \mathrm{n}_{1}=\frac{316}{2 \pi}$
$\mathrm{Y}_{2}=0.25 \sin 310 \mathrm{t}$
$\therefore \mathrm{n}_{2}=\frac{310}{2 \pi}$
$\quad \mathrm{n}_{1}-\mathrm{n}_{2}=\frac{316-310}{2 \pi}=\frac{6}{2 \pi}=\frac{3}{\pi}$
$\mathrm{Y}_{1}=0.25 \sin 316 \mathrm{t}$
$\therefore \frac{2 \pi}{\mathrm{T}}=316 \quad \therefore \mathrm{n}_{1}=\frac{316}{2 \pi}$
$\mathrm{Y}_{2}=0.25 \sin 310 \mathrm{t}$
$\therefore \mathrm{n}_{2}=\frac{310}{2 \pi}$
$\quad \mathrm{n}_{1}-\mathrm{n}_{2}=\frac{316-310}{2 \pi}=\frac{6}{2 \pi}=\frac{3}{\pi}$
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