Two waves Y_{1} = 0.25 sin(316 t) and Y_{2} = | vedclass

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(D) The general equation for a wave is $Y = A \sin(\omega t)$,where $\omega = 2 \pi n$ and $n$ is the frequency in Hz.For the first wave,$Y_{1} = 0.25

Vedclass · Accepted Answer

$\frac{3}{\pi}$

Vedclass · Answer

(D) The general equation for a wave is $Y = A \sin(\omega t)$,where $\omega = 2 \pi n$ and $n$ is the frequency in Hz.For the first wave,$Y_{1} = 0.25 \sin(316 t)$,so $\omega_{1} = 316 	ext{ rad/s}$.The frequency $n_{1} = \frac{\omega_{1}}{2 \pi} = \frac{316}{2 \pi} 	ext{ Hz}$.For the second wave,$Y_{2} = 0.25 \sin(310 t)$,so $\omega_{2} = 310 	ext{ rad/s}$.The frequency $n_{2} = \frac{\omega_{2}}{2 \pi} = \frac{310}{2 \pi} 	ext{ Hz}$.The number of beats produced per second is the difference in frequencies: $n_{beat} = |n_{1} - n_{2}|$.$n_{beat} = \frac{316}{2 \pi} - \frac{310}{2 \pi} = \frac{316 - 310}{2 \pi} = \frac{6}{2 \pi} = \frac{3}{\pi} 	ext{ beats/s}$.

Two waves $Y_{1} = 0.25 \sin(316 t)$ and $Y_{2} = 0.25 \sin(310 t)$ are propagating along the same direction. The number of beats produced per second is:

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