Question 334: Waves

Question

Two waves $\mathrm{Y}_1=0.25 \sin 316 \mathrm{t} \quad$ and $\mathrm{Y}_2=0.25 \sin 310 \mathrm{t}$ are propagating along the same direction. The number of beats produced per second are

Accepted Answer

To determine the number of beats produced per second by the two waves given by $ \mathrm{Y}_1 = 0.25 \sin 316\, \mathrm{t} $ and $ \mathrm{Y}_2 = 0.25 \sin 310\, \mathrm{t} $, we'll start by identifying the angular frequencies of each wave, denoted as $ \omega_1 $ and $ \omega_2 $.
Identifying Angular Frequencies:
The general equation for a wave is $ \mathrm{y} = \mathrm{A} \sin(\omega \mathrm{t}) $.
For the wave $ \mathrm{Y}_1 = 0.25 \sin 316\, \mathrm{t} $, comparing with the general equation, we find:
$      \omega_1 = 316      $
For the wave $ \mathrm{Y}_2 = 0.25 \sin 310\, \mathrm{t} $, similarly:
$      \omega_2 = 310      $
Converting Angular Frequencies to Frequencies:
Angular frequency $ \omega $ is related to the frequency $ f $ by the equation $ \omega = 2\pi f $.
Hence, the frequency for the first wave $ f_1 $ is:
$      f_1 = \frac{316}{2\pi}      $
The frequency for the second wave $ f_2 $ is:
$      f_2 = \frac{310}{2\pi}      $
Calculating the Number of Beats per Second:
The number of beats produced per second is the difference between the two frequencies $ f_1 $ and $ f_2 $.
Therefore, the number of beats per second is:
$      f_1 - f_2 = \frac{316}{2\pi} - \frac{310}{2\pi} = \frac{6}{2\pi} = \frac{3}{\pi}      $
Thus, the number of beats produced per second is $ \frac{3}{\pi} $.

Answer

$\frac{\pi}{3}$

Answer

$\frac{3}{\pi}$

Answer

$\frac{2}{\pi}$

Answer

$\frac{\pi}{2}$

Waves

Two waves \mathrm{Y}_1=0.25 \sin 316 \mathrm{t} \quad and \mathrm{Y}_2=0.25 \sin 310 \mathrm{t} are propagating along the same direction. The number of beats produced per second are

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