Question

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 are

• A. \( \frac{\pi}{2} \)
• B. \( \frac{2}{\pi} \)
• C. \( \frac{\pi}{3} \)
• D. \( \frac{3}{\pi} \)

Hint:

To calculate the number of beats, always find the difference between the frequencies of the two waves.

Answer 1

The Correct Option is D

Step 1: Formula for beats.
The number of beats per second is given by the difference in the frequencies of the two waves: \[ f_{\text{beats}} = |f_1 - f_2| \] where \( f_1 \) and \( f_2 \) are the frequencies of the waves.
Step 2: Calculation of frequencies.
For the wave \( y_1 = 0.25 \sin(316 t) \), the frequency is: \[ f_1 = \frac{316}{2\pi} = 50 \, \text{Hz} \] For the wave \( y_2 = 0.25 \sin(310 t) \), the frequency is: \[ f_2 = \frac{310}{2\pi} = 49.42 \, \text{Hz} \] Thus, the number of beats per second is: \[ f_{\text{beats}} = |50 - 49.42| = 0.58 \, \text{Hz} \]
Step 3: Conclusion.
Thus, the number of beats produced per second is (D) \( \frac{3}{\pi} \).