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, we need to find the amplitude \( A \) and the initial phase \( \phi \) of a particle executing simple harmonic motion (SHM) described by the equation:

\[ x = A \sin(\pi t + \phi) \]

Given:
- Initial displacement \( x(0) = 1 \, \text{cm} \)
- Initial velocity \( v(0) = \pi \, \text{cm/s} \)

### Step 1: Write the equations for initial displacement and velocity

At \( t = 0 \):
1. The displacement equation becomes:
\[ x(0) = A \sin(\phi) = 1 \quad \text{(Equation 1)} \]

2. The velocity \( v(t) \) is the derivative of displacement with respect to time:
\[ v(t) = \frac{dx}{dt} = A \pi \cos(\pi t + \phi) \]
At \( t = 0 \):
\[ v(0) = A \pi \cos(\phi) = \pi \quad \text{(Equation 2)} \]

### Step 2: Simplify the equations

From Equation 1:
\[ A \sin(\phi) = 1 \quad \Rightarrow \quad A = \frac{1}{\sin(\phi)} \quad \text{(Equation 3)} \]

From Equation 2:
\[ A \pi \cos(\phi) = \pi \quad \Rightarrow \quad A \cos(\phi) = 1 \quad \Rightarrow \quad A = \frac{1}{\cos(\phi)} \quad \text{(Equation 4)} \]

### Step 3: Equate the two expressions for \( A \)

From Equations 3 and 4, we have:
\[ \frac{1}{\sin(\phi)} = \frac{1}{\cos(\phi)} \]

Cross-multiplying gives:
\[ \cos(\phi) = \sin(\phi) \]

### Step 4: Solve for \( \phi \)

The equation \( \cos(\phi) = \sin(\phi) \) implies:
\[ \tan(\phi) = 1 \quad \Rightarrow \quad \phi = \frac{\pi}{4} \quad \text{(since } 0 \leq \phi < \frac{\pi}{2}\text{)} \]

### Step 5: Substitute \( \phi \) back to find \( A \)

Now substitute \( \phi = \frac{\pi}{4} \) back into either Equation 3 or 4 to find \( A \):
Using Equation 3:
\[ A = \frac{1}{\sin\left(\frac{\pi}{4}\right)} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2} \, \text{cm} \]

### Final Answers

- Amplitude \( A = \sqrt{2} \, \text{cm} \)
- Initial phase \( \phi = \frac{\pi}{4} \)