Two waves Y_1 = 0.25 sin 316 t and Y_2 = 0.25 | vedclass

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(A) The general equation for a wave is $Y = A \sin(\omega t)$,where $\omega = 2 \pi f$.Given $Y_1 = 0.25 \sin 316 t$,we have $\omega_1 = 316 	ext{ ra

Vedclass · Accepted Answer

$\frac{3}{\pi}$

Vedclass · Answer

(A) The general equation for a wave is $Y = A \sin(\omega t)$,where $\omega = 2 \pi f$.Given $Y_1 = 0.25 \sin 316 t$,we have $\omega_1 = 316 	ext{ rad/s}$.Thus,$2 \pi f_1 = 316 \implies f_1 = \frac{316}{2 \pi} = \frac{158}{\pi} 	ext{ Hz}$.Given $Y_2 = 0.25 \sin 310 t$,we have $\omega_2 = 310 	ext{ rad/s}$.Thus,$2 \pi f_2 = 310 \implies f_2 = \frac{310}{2 \pi} = \frac{155}{\pi} 	ext{ Hz}$.The beat frequency is the difference between the two frequencies:$	ext{Beat frequency} = f_1 - f_2 = \frac{158}{\pi} - \frac{155}{\pi} = \frac{3}{\pi} 	ext{ Hz}$.

Two waves $Y_1 = 0.25 \sin 316 t$ and $Y_2 = 0.25 \sin 310 t$ are propagating in the same direction. The number of beats produced per second is

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