Two waves $Y_1=0.25 \sin 316 t$ and $\mathrm{Y}_2=0.25 \sin 310 \mathrm{t}$ are propagating along the same…
Two waves $Y_1=0.25 \sin 316 t$ and $\mathrm{Y}_2=0.25 \sin 310 \mathrm{t}$ are propagating along the same direction. The number of beats produced per second are
$\frac{\pi}{3}$
$\frac{3}{\pi}$
$\frac{2}{\pi}$
$\frac{\pi}{2}$
Solution
$\begin{array}{ll}
& y=A \sin \omega t \\
& \text {Given, } y_1=0.25 \sin 316 \mathrm{t} \\
& \text { Comparing with (i), } \omega_1=316 \\
& \omega_1=2 \pi \mathrm{f}_1 \\
\therefore \quad & \frac{316}{2 \pi}=\mathrm{f}_1 \\
& \text { Given, } \mathrm{y}_2=0.25 \sin 310 \mathrm{t} \\
& \text { Comparing with (i), } \omega_2=310 \\
& \omega_2=2 \pi \mathrm{f}_2 \\
\therefore \quad & \frac{310}{2 \pi}=\mathrm{f}_2
\end{array}$ Number of beats produced per second is
$=f_1-f_2=\frac{316}{2 \pi}-\frac{310}{2 \pi}=\frac{3}{\pi}$