Find beat frequency if the motion of two particles is given by
`y_1=0.25sin(310t)`
`y_2=0.25sin(316t)`

Options

A. 3
B. `3/pi`
C. `6/pi`
D. 6

Answer

To find the beat frequency of the two particles given by the equations \( y_1 = 0.25 \sin(310t) \) and \( y_2 = 0.25 \sin(316t) \), we can follow these steps:

### Step 1: Identify the angular frequencies
The angular frequency \( \omega \) is given in the equations:
- For \( y_1 \): \( \omega_1 = 310 \)
- For \( y_2 \): \( \omega_2 = 316 \)

### Step 2: Convert angular frequencies to frequencies
The relationship between angular frequency \( \omega \) and frequency \( f \) is given by:
\[ \omega = 2\pi f \]
Thus, we can express the frequencies \( f_1 \) and \( f_2 \) as follows:
- For \( y_1 \):
\[ f_1 = \frac{\omega_1}{2\pi} = \frac{310}{2\pi} \]
- For \( y_2 \):
\[ f_2 = \frac{\omega_2}{2\pi} = \frac{316}{2\pi} \]

### Step 3: Calculate the frequencies
Now, we can calculate the values of \( f_1 \) and \( f_2 \):
- \( f_1 = \frac{310}{2\pi} \)
- \( f_2 = \frac{316}{2\pi} \)

### Step 4: Find the beat frequency
The beat frequency \( f_b \) is given by the absolute difference between the two frequencies:
\[ f_b = |f_2 - f_1| \]
Substituting the values we found:
\[ f_b = \left| \frac{316}{2\pi} - \frac{310}{2\pi} \right| = \left| \frac{316 - 310}{2\pi} \right| = \left| \frac{6}{2\pi} \right| = \frac{3}{\pi} \]

### Final Answer
The beat frequency is:
\[ f_b = \frac{3}{\pi} \text{ Hz} \]

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