A wave is travelling along x-axis . The disturbance at x=0 and t=0 is A/2 and is increasing , where A is amplitude of the wave. If `Y= sin (kx-omegat+phi)`, then the initial phase is `alpha pi` . Find `alpha`

Answer

To solve the problem, we need to analyze the wave function given by \( Y = A \sin(kx - \omega t + \phi) \) under the conditions provided.

### Step 1: Substitute the initial conditions
We start by substituting the initial conditions \( x = 0 \) and \( t = 0 \) into the wave equation.

\[ Y(0, 0) = A \sin(0 - 0 + \phi) = A \sin(\phi) \]

According to the problem, at \( x = 0 \) and \( t = 0 \), the disturbance is \( \frac{A}{2} \). Therefore, we have:

\[ A \sin(\phi) = \frac{A}{2} \]

### Step 2: Solve for \( \sin(\phi) \)
Dividing both sides by \( A \) (assuming \( A \neq 0 \)) gives:

\[ \sin(\phi) = \frac{1}{2} \]

### Step 3: Determine possible values of \( \phi \)
The values of \( \phi \) that satisfy \( \sin(\phi) = \frac{1}{2} \) are:

\[ \phi = \frac{\pi}{6} + 2n\pi \quad \text{and} \quad \phi = \frac{5\pi}{6} + 2n\pi \]

for any integer \( n \).

### Step 4: Check the increasing condition
Next, we need to check the condition that the disturbance is increasing at \( x = 0 \) and \( t = 0 \). To do this, we differentiate \( Y \) with respect to \( t \):

\[ \frac{\partial Y}{\partial t} = -A\omega \cos(kx - \omega t + \phi) \]

At \( x = 0 \) and \( t = 0 \):

\[ \frac{\partial Y}{\partial t}(0, 0) = -A\omega \cos(\phi) \]

For the disturbance to be increasing, this derivative must be greater than 0:

\[ -A\omega \cos(\phi) > 0 \]

Since \( A \) and \( \omega \) are positive, we have:

\[ \cos(\phi) < 0 \]

### Step 5: Determine the range of \( \phi \)
The cosine function is negative in the intervals:

\[ \frac{\pi}{2} < \phi < \frac{3\pi}{2} \]

### Step 6: Analyze possible values of \( \phi \)
From our earlier solutions for \( \phi \):

1. \( \phi = \frac{\pi}{6} + 2n\pi \) does not fall within the range where \( \cos(\phi) < 0 \).
2. \( \phi = \frac{5\pi}{6} + 2n\pi \) does fall within the range where \( \cos(\phi) < 0 \).

### Step 7: Conclusion
Thus, the valid solution for \( \phi \) is:

\[ \phi = \frac{5\pi}{6} \]

Now, we can express \( \phi \) in terms of \( \alpha \):

\[ \phi = \alpha \pi \implies \alpha = \frac{5}{6} \]

### Final Answer
\[ \alpha = \frac{5}{6} \]