Two vibrating tuning forks produce progressive waves given by , `y_(1) = 4 sin (500 pi t) and y_(2) = 2 sin (506 pi t)`. These tuning forks are held near the ear of person . The person will hear

Options

A. ` 3 beats//s` with intensity ratio between maxima and minima equal to `2`
B. ` 3 beats//s` with intensity ratio between maxima and minima equal to `9`
C. ` 6 beats//s` with intensity ratio between maxima and minima equal to `2`
D. ` 6 beats//s` with intensity ratio between maxima and minima equal to `9`

Explanation

To solve the problem, we need to analyze the two waves produced by the tuning forks and determine the beat frequency and the intensity ratio.

### Step 1: Identify the wave equations
The equations for the two waves are given as:
- \( y_1 = 4 \sin(500 \pi t) \)
- \( y_2 = 2 \sin(506 \pi t) \)

### Step 2: Determine the angular frequencies
The angular frequency \( \omega \) is related to the frequency \( f \) by the formula:
\[ \omega = 2 \pi f \]
From the equations:
- For \( y_1 \): \( \omega_1 = 500 \pi \)
- For \( y_2 \): \( \omega_2 = 506 \pi \)

### Step 3: Calculate the frequencies
Using the relationship \( \omega = 2 \pi f \):
- For \( y_1 \):
\[ 500 \pi = 2 \pi f_1 \implies f_1 = \frac{500 \pi}{2 \pi} = 250 \text{ Hz} \]
- For \( y_2 \):
\[ 506 \pi = 2 \pi f_2 \implies f_2 = \frac{506 \pi}{2 \pi} = 253 \text{ Hz} \]

### Step 4: Calculate the beat frequency
The beat frequency \( \Delta f \) is given by the absolute difference between the two frequencies:
\[ \Delta f = |f_2 - f_1| = |253 - 250| = 3 \text{ Hz} \]

### Step 5: Determine the ratio of intensities
The intensity \( I \) is proportional to the square of the amplitude. The amplitudes from the wave equations are:
- Amplitude of \( y_1 \) (A1) = 4
- Amplitude of \( y_2 \) (A2) = 2

The maximum and minimum intensities can be calculated as follows:
- Maximum intensity:
\[ I_{\text{max}} \propto (A_1 + A_2)^2 = (4 + 2)^2 = 6^2 = 36 \]
- Minimum intensity:
\[ I_{\text{min}} \propto (A_1 - A_2)^2 = (4 - 2)^2 = 2^2 = 4 \]

### Step 6: Calculate the ratio of maximum to minimum intensity
The ratio of maximum to minimum intensity is:
\[ \text{Ratio} = \frac{I_{\text{max}}}{I_{\text{min}}} = \frac{36}{4} = 9 \]

### Conclusion
The person will hear a beat frequency of 3 Hz and the ratio of maximum to minimum intensity is 9:1.

### Final Answer
- The person will hear **3 beats per second**.
- The ratio of maximum intensity to minimum intensity is **9:1**.