Two superimposig waves are represented by equation ` y_1=2 sin 2 pi ( 10t-0.4x) and y_2=4 sin2 pi ( 20 t- 0.8x)` . The ratio of `l_("max") " to " l_("min")` is

Options

A. `36:4`
B. `25:9`
C. `1:4`
D. `4:1`

Answer

To find the ratio of maximum intensity \( I_{\text{max}} \) to minimum intensity \( I_{\text{min}} \) for the given superimposing waves, we can follow these steps:

### Step 1: Identify the Amplitudes
The given wave equations are:
- \( y_1 = 2 \sin(2\pi(10t - 0.4x)) \)
- \( y_2 = 4 \sin(2\pi(20t - 0.8x)) \)

From these equations, we can identify the amplitudes:
- Amplitude of wave 1, \( A_1 = 2 \)
- Amplitude of wave 2, \( A_2 = 4 \)

### Step 2: Calculate Maximum Intensity
The maximum intensity \( I_{\text{max}} \) occurs when the two waves are in phase, which is given by the formula:
\[ I_{\text{max}} \propto (A_1 + A_2)^2 \]
Substituting the amplitudes:
\[ I_{\text{max}} \propto (2 + 4)^2 = 6^2 = 36 \]

### Step 3: Calculate Minimum Intensity
The minimum intensity \( I_{\text{min}} \) occurs when the two waves are out of phase, which is given by the formula:
\[ I_{\text{min}} \propto (A_1 - A_2)^2 \]
Substituting the amplitudes:
\[ I_{\text{min}} \propto (2 - 4)^2 = (-2)^2 = 4 \]

### Step 4: Find the Ratio of Intensities
Now, we can find the ratio of maximum intensity to minimum intensity:
\[ \frac{I_{\text{max}}}{I_{\text{min}}} = \frac{36}{4} = 9 \]

### Step 5: Final Ratio
Thus, the ratio of \( I_{\text{max}} \) to \( I_{\text{min}} \) is:
\[ \frac{I_{\text{max}}}{I_{\text{min}}} = 9 \]