A particle is executing SHM given by `x = A sin (pit + phi)`. The initial displacement of particle is `1 cm` and its initial velocity is `pi cm//sec`. Find the amplitude of motion and initial phase of the particle.

Explanation

To solve the problem step by step, we will use the given information about the simple harmonic motion (SHM) of the particle.

### Step 1: Write the equation of SHM
The equation of SHM is given as:
\[ x = A \sin(\pi t + \phi) \]

### Step 2: Use initial conditions for displacement
At \( t = 0 \), the initial displacement \( x \) is given as \( 1 \, \text{cm} \). Substituting \( t = 0 \) into the equation:
\[ x = A \sin(\phi) \]
Thus, we have:
\[ 1 = A \sin(\phi) \tag{1} \]

### Step 3: Use initial conditions for velocity
The initial velocity \( v \) is given as \( \pi \, \text{cm/s} \). The velocity in SHM is given by the derivative of displacement with respect to time:
\[ v = \frac{dx}{dt} = A \pi \cos(\pi t + \phi) \]
At \( t = 0 \):
\[ v = A \pi \cos(\phi) \]
Substituting the value of \( v \):
\[ \pi = A \pi \cos(\phi) \]
Dividing both sides by \( \pi \) (assuming \( \pi \neq 0 \)):
\[ 1 = A \cos(\phi) \tag{2} \]

### Step 4: Solve the equations simultaneously
Now we have two equations:
1. \( A \sin(\phi) = 1 \)
2. \( A \cos(\phi) = 1 \)

### Step 5: Square and add the equations
Squaring both equations and adding them gives:
\[ (A \sin(\phi))^2 + (A \cos(\phi))^2 = 1^2 + 1^2 \]
This simplifies to:
\[ A^2 (\sin^2(\phi) + \cos^2(\phi)) = 2 \]
Using the Pythagorean identity \( \sin^2(\phi) + \cos^2(\phi) = 1 \):
\[ A^2 \cdot 1 = 2 \]
Thus:
\[ A^2 = 2 \implies A = \sqrt{2} \]

### Step 6: Find the initial phase \( \phi \)
From equation (1):
\[ A \sin(\phi) = 1 \]
Substituting \( A = \sqrt{2} \):
\[ \sqrt{2} \sin(\phi) = 1 \implies \sin(\phi) = \frac{1}{\sqrt{2}} \implies \sin(\phi) = \frac{\sqrt{2}}{2} \]
The angle \( \phi \) that satisfies this is:
\[ \phi = \frac{\pi}{4} \text{ or } \phi = \frac{3\pi}{4} \]
However, we also need to satisfy equation (2):
\[ A \cos(\phi) = 1 \implies \sqrt{2} \cos(\phi) = 1 \implies \cos(\phi) = \frac{1}{\sqrt{2}} \implies \cos(\phi) = \frac{\sqrt{2}}{2} \]
The angle \( \phi \) that satisfies this is:
\[ \phi = \frac{\pi}{4} \]

### Final Result
Thus, the amplitude \( A \) is \( \sqrt{2} \, \text{cm} \) and the initial phase \( \phi \) is \( \frac{\pi}{4} \).