The waves, whose equations are
`y_(1)=4xx10^(-3)sin(308pix-(x)/(50)) and y_(2)=1xx10^(-3)sin(302pit-(x)/(50))`
ar superposed in a medium. Now,

Options

A. Beats are produced with a frequency 3Hz
B. Beats are produced with a frequency 6 Hz
C. The ratio of maximum to minimum intensity is 25:9
D. The ratio of maximum to minimum amplitude is 2:1

Answer

To solve the problem, we need to analyze the given wave equations and derive the required quantities step by step.

### Step 1: Identify the wave equations
The two waves are given by:
1. \( y_1 = 4 \times 10^{-3} \sin(308\pi t - \frac{x}{50}) \)
2. \( y_2 = 1 \times 10^{-3} \sin(302\pi t - \frac{x}{50}) \)

### Step 2: Determine the angular frequencies (\(\omega\))
From the wave equations, we can identify the angular frequencies:
- For \( y_1 \): \( \omega_1 = 308\pi \)
- For \( y_2 \): \( \omega_2 = 302\pi \)

### Step 3: Calculate the frequencies (\(f\))
The frequency \(f\) can be calculated using the relation \(f = \frac{\omega}{2\pi}\):
- For \( y_1 \):
\[ f_1 = \frac{308\pi}{2\pi} = 154 \text{ Hz} \]
- For \( y_2 \):
\[ f_2 = \frac{302\pi}{2\pi} = 151 \text{ Hz} \]

### Step 4: Calculate the beat frequency
The beat frequency is given by the absolute difference between the two frequencies:
\[ \text{Beat frequency} = |f_2 - f_1| = |151 - 154| = 3 \text{ Hz} \]

### Step 5: Determine the amplitudes
From the wave equations, we can see the amplitudes:
- Amplitude of \( y_1 \) (denoted as \(A_1\)): \( A_1 = 4 \times 10^{-3} \)
- Amplitude of \( y_2 \) (denoted as \(A_2\)): \( A_2 = 1 \times 10^{-3} \)

### Step 6: Calculate the maximum and minimum amplitudes
The maximum amplitude (\(A_{\text{max}}\)) occurs when the waves are in phase:
\[ A_{\text{max}} = A_1 + A_2 = 4 \times 10^{-3} + 1 \times 10^{-3} = 5 \times 10^{-3} \]
The minimum amplitude (\(A_{\text{min}}\)) occurs when the waves are out of phase:
\[ A_{\text{min}} = |A_1 - A_2| = |4 \times 10^{-3} - 1 \times 10^{-3}| = 3 \times 10^{-3} \]

### Step 7: Calculate the ratio of maximum to minimum intensity
Intensity \(I\) is proportional to the square of the amplitude:
\[ \frac{I_{\text{max}}}{I_{\text{min}}} = \left(\frac{A_{\text{max}}}{A_{\text{min}}}\right)^2 \]
Calculating the ratio:
\[ \frac{I_{\text{max}}}{I_{\text{min}}} = \left(\frac{5 \times 10^{-3}}{3 \times 10^{-3}}\right)^2 = \left(\frac{5}{3}\right)^2 = \frac{25}{9} \]

### Step 8: Calculate the ratio of maximum to minimum amplitude
The ratio of maximum to minimum amplitude is:
\[ \frac{A_{\text{max}}}{A_{\text{min}}} = \frac{5 \times 10^{-3}}{3 \times 10^{-3}} = \frac{5}{3} \]

### Summary of Results
1. Beat frequency: **3 Hz**
2. Ratio of maximum to minimum intensity: **\(\frac{25}{9}\)**
3. Ratio of maximum to minimum amplitude: **\(\frac{5}{3}\)**