Following two wave trains are approaching each other. $y_1 = a \sin 2000 \pi t, y_2 = a \sin 2008 \pi t$ The…
Following two wave trains are approaching each other.
$y_1 = a \sin 2000 \pi t, y_2 = a \sin 2008 \pi t$
The number of beats heard per second is
- 8
- 4
- 1
- zero
Solution
Beat frequency $= f_2 - f_1 = \frac{\omega_2 - \omega_1}{2\pi} = \frac{2008\pi - 2000\pi}{2\pi} = 4 \text{ Hz}$
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