Two waves are simultaneously passing through a string and their equations are :
`y_1 =A_1 sin k(x - vt), y_2=A_2 sink(x- vt+x_0)`. Given amplitudes `A_1 =12 mm` and `A_2 =5 mm, x_0 = 3.5` cm and wave number `k = 6.28 cm^( - 1)`. The amplitude of resulting wave will be ___________mm.

Answer

To find the amplitude of the resulting wave when two waves are superimposed, we can use the principle of superposition. The equations of the two waves are given as:

1. \( y_1 = A_1 \sin(k(x - vt)) \)
2. \( y_2 = A_2 \sin(k(x - vt + x_0)) \)

### Step 1: Identify the given values
- Amplitude of the first wave, \( A_1 = 12 \, \text{mm} \)
- Amplitude of the second wave, \( A_2 = 5 \, \text{mm} \)
- Phase shift \( x_0 = 3.5 \, \text{cm} = 35 \, \text{mm} \) (converted to mm)
- Wave number \( k = 6.28 \, \text{cm}^{-1} = 62.8 \, \text{mm}^{-1} \) (converted to mm)

### Step 2: Calculate the phase difference
The phase difference \( \phi \) can be calculated using the formula:
\[ \phi = k x_0 \]
Substituting the values:
\[ \phi = 62.8 \, \text{mm}^{-1} \times 35 \, \text{mm} = 2198 \, \text{radians} \]

### Step 3: Simplify the phase difference
Since the sine function is periodic, we can simplify the phase difference by taking it modulo \( 2\pi \):
\[ \phi \approx \frac{2198}{2\pi} \approx 349.5 \, \text{cycles} \]
This means we can reduce it to a phase difference of \( \phi \approx 0 \) or \( \pi \) based on the sine function's periodicity.

### Step 4: Determine the amplitude of the resultant wave
The formula for the amplitude \( A \) of the resultant wave when two waves with amplitudes \( A_1 \) and \( A_2 \) interfere with a phase difference \( \phi \) is given by:
\[ A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\phi)} \]
For \( \phi = \pi \):
\[ \cos(\pi) = -1 \]
Substituting the values:
\[ A = \sqrt{12^2 + 5^2 + 2 \cdot 12 \cdot 5 \cdot (-1)} \]
Calculating each term:
\[ A = \sqrt{144 + 25 - 120} \]
\[ A = \sqrt{49} = 7 \, \text{mm} \]

### Final Result
The amplitude of the resulting wave is \( 7 \, \text{mm} \).

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