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Two waves $y_1=0.35 \sin (316 t)$ and $y_2=0.35 \sin (310 t)$ are propagating along the same direction. The number of beats produced per second are
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Verified Answer
The correct answer is:
$\frac{3}{\pi}$
According to question:
$f_1=\frac{316}{2 \pi}$
and,
$f_2=\frac{310}{2 \pi}$
Now,
The number of beats heard per second $=f_1-f_2=\frac{316}{2 \pi}-\frac{310}{2 \pi}=\frac{3}{\pi}$
$f_1=\frac{316}{2 \pi}$
and,
$f_2=\frac{310}{2 \pi}$
Now,
The number of beats heard per second $=f_1-f_2=\frac{316}{2 \pi}-\frac{310}{2 \pi}=\frac{3}{\pi}$
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