Two waves represented by `y_1 =a sin omega t and y_2 =a sin (omega t+phi) "with " phi =(pi)/2`
are superposed at any point at a particular instant. The resultant amplitude is

Options

A. a
B. 4a
C. `sqrt(2a)`
D. zero

Answer

To find the resultant amplitude when two waves are superposed, we can follow these steps:

### Step 1: Write down the equations of the waves.
The two waves are given as:
- \( y_1 = a \sin(\omega t) \)
- \( y_2 = a \sin(\omega t + \phi) \)

Given that \( \phi = \frac{\pi}{2} \), we can rewrite the second wave:
- \( y_2 = a \sin(\omega t + \frac{\pi}{2}) \)

### Step 2: Use the sine addition formula.
The sine addition formula states that:
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]
Applying this to \( y_2 \):
\[ y_2 = a \left( \sin(\omega t) \cos\left(\frac{\pi}{2}\right) + \cos(\omega t) \sin\left(\frac{\pi}{2}\right) \right) \]
Since \( \cos\left(\frac{\pi}{2}\right) = 0 \) and \( \sin\left(\frac{\pi}{2}\right) = 1 \), we have:
\[ y_2 = a \cdot 0 + a \cos(\omega t) = a \cos(\omega t) \]

### Step 3: Write the resultant wave equation.
The resultant wave \( y \) when \( y_1 \) and \( y_2 \) are superposed is:
\[ y = y_1 + y_2 = a \sin(\omega t) + a \cos(\omega t) \]

### Step 4: Factor out the amplitude.
We can factor out \( a \):
\[ y = a \left( \sin(\omega t) + \cos(\omega t) \right) \]

### Step 5: Use the formula for the sum of sine and cosine.
The sum \( \sin A + \cos A \) can be expressed as:
\[ \sin A + \cos A = \sqrt{2} \sin\left(A + \frac{\pi}{4}\right) \]
Thus, we can rewrite:
\[ y = a \sqrt{2} \sin\left(\omega t + \frac{\pi}{4}\right) \]

### Step 6: Identify the resultant amplitude.
From the expression \( y = a \sqrt{2} \sin\left(\omega t + \frac{\pi}{4}\right) \), we see that the resultant amplitude \( A_r \) is:
\[ A_r = a \sqrt{2} \]

### Final Answer:
The resultant amplitude is \( A_r = a \sqrt{2} \).

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