two waves `y_1 = 10sin(omegat - Kx)` m and `y_2 = 5sin(omegat - Kx + π/3)` m are superimposed. the amplitude of resultant wave is

Options

A. `sqrt7` m
B. `2sqrt7 m `
C. `3sqrt7` m
D. `5sqrt7` m

Answer

To find the amplitude of the resultant wave formed by the superposition of two waves given by \( y_1 = 10 \sin(\omega t - Kx) \) and \( y_2 = 5 \sin(\omega t - Kx + \frac{\pi}{3}) \), we will follow these steps:

### Step 1: Identify the amplitudes and phase differences
From the equations of the waves:
- Amplitude of the first wave, \( A_1 = 10 \)
- Amplitude of the second wave, \( A_2 = 5 \)
- Phase of the first wave, \( \phi_1 = 0 \)
- Phase of the second wave, \( \phi_2 = \frac{\pi}{3} \)

### Step 2: Calculate the phase difference
The phase difference \( \Delta \phi \) is given by:
\[ \Delta \phi = \phi_2 - \phi_1 = \frac{\pi}{3} - 0 = \frac{\pi}{3} \]

### Step 3: Use the formula for resultant amplitude
The resultant amplitude \( A \) of two waves can be calculated using the formula:
\[ A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\Delta \phi)} \]
Substituting the values:
- \( A_1 = 10 \)
- \( A_2 = 5 \)
- \( \Delta \phi = \frac{\pi}{3} \)

### Step 4: Calculate \( \cos(\Delta \phi) \)
We know that:
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

### Step 5: Substitute values into the formula
Now substituting the values into the formula:
\[ A = \sqrt{10^2 + 5^2 + 2 \cdot 10 \cdot 5 \cdot \frac{1}{2}} \]
Calculating each term:
- \( 10^2 = 100 \)
- \( 5^2 = 25 \)
- \( 2 \cdot 10 \cdot 5 \cdot \frac{1}{2} = 50 \)

### Step 6: Combine the terms
Now, combine the terms:
\[ A = \sqrt{100 + 25 + 50} = \sqrt{175} \]

### Step 7: Simplify the square root
We can simplify \( \sqrt{175} \):
\[ \sqrt{175} = \sqrt{25 \cdot 7} = 5\sqrt{7} \]

### Final Result
Thus, the amplitude of the resultant wave is:
\[ A = 5\sqrt{7} \text{ m} \]