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 \( y_1 = 10 \sin(\omega t - Kx) \) and \( y_2 = 5 \sin(\omega t - Kx + \frac{\pi}{3}) \), we can follow these steps:

### Step 1: Identify the Amplitudes and Phase Difference
From the equations of the waves:
- The amplitude of the first wave \( A_1 = 10 \, \text{m} \)
- The amplitude of the second wave \( A_2 = 5 \, \text{m} \)
- The phase difference \( \phi = \frac{\pi}{3} \)

### Step 2: Use the Resultant Amplitude Formula
The formula for the resultant amplitude \( A \) when two waves with amplitudes \( A_1 \) and \( A_2 \) interfere with a phase difference \( \phi \) is given by:
\[ A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\phi)} \]

### Step 3: Substitute the Values
Substituting the known values into the formula:
\[ A = \sqrt{10^2 + 5^2 + 2 \cdot 10 \cdot 5 \cdot \cos\left(\frac{\pi}{3}\right)} \]

### Step 4: Calculate Each Term
- Calculate \( A_1^2 \):
\[ A_1^2 = 10^2 = 100 \]
- Calculate \( A_2^2 \):
\[ A_2^2 = 5^2 = 25 \]
- Calculate \( \cos\left(\frac{\pi}{3}\right) \):
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]
- Calculate \( 2 A_1 A_2 \cos\left(\frac{\pi}{3}\right) \):
\[ 2 \cdot 10 \cdot 5 \cdot \frac{1}{2} = 50 \]

### Step 5: Combine the Terms
Now combine all the terms:
\[ A = \sqrt{100 + 25 + 50} = \sqrt{175} \]

### Step 6: Simplify the Result
The final result can be expressed as:
\[ A = \sqrt{175} = 5\sqrt{7} \, \text{m} \]

Thus, the amplitude of the resultant wave is \( 5\sqrt{7} \, \text{m} \).