A particle executing SHM is described by the displacement function `x(t)=Acos(omegat+phi)`, if the initial (t=0) position of the particle is 1 cm, its initial velocity is `pi" cm "s^(-1)` and its angular frequency is `pis^(-1)`, then the amplitude of its motion is

Options

A. `picm`
B. 2 cm
C. `sqrt(2)`cm
D. 1 cm

Explanation

To solve the problem step by step, we will use the information provided about the particle executing Simple Harmonic Motion (SHM).

### Given:
- Displacement function: \( x(t) = A \cos(\omega t + \phi) \)
- Initial position at \( t = 0 \): \( x(0) = 1 \, \text{cm} \)
- Initial velocity: \( v(0) = \pi \, \text{cm/s} \)
- Angular frequency: \( \omega = \pi \, \text{s}^{-1} \)

### Step 1: Find the initial position equation
At \( t = 0 \):
\[ x(0) = A \cos(\phi) = 1 \, \text{cm} \]
This gives us our first equation:
\[ A \cos(\phi) = 1 \quad \text{(Equation 1)} \]

### Step 2: Find the velocity function
The velocity \( v(t) \) is the derivative of the displacement \( x(t) \):
\[ v(t) = \frac{dx}{dt} = -A \omega \sin(\omega t + \phi) \]
At \( t = 0 \):
\[ v(0) = -A \omega \sin(\phi) = \pi \, \text{cm/s} \]
Substituting \( \omega = \pi \):
\[ -A \pi \sin(\phi) = \pi \]
Dividing both sides by \( \pi \):
\[ -A \sin(\phi) = 1 \quad \text{(Equation 2)} \]

### Step 3: Solve for \( A \) using Equations 1 and 2
From Equation 1:
\[ A \cos(\phi) = 1 \implies \cos(\phi) = \frac{1}{A} \]
From Equation 2:
\[ -A \sin(\phi) = 1 \implies \sin(\phi) = -\frac{1}{A} \]

### Step 4: Use the Pythagorean identity
Using the identity \( \sin^2(\phi) + \cos^2(\phi) = 1 \):
\[ \left(-\frac{1}{A}\right)^2 + \left(\frac{1}{A}\right)^2 = 1 \]
This simplifies to:
\[ \frac{1}{A^2} + \frac{1}{A^2} = 1 \]
\[ \frac{2}{A^2} = 1 \]
\[ A^2 = 2 \implies A = \sqrt{2} \, \text{cm} \]

### Conclusion
The amplitude of the motion is:
\[ A = \sqrt{2} \, \text{cm} \]

### Final Answer
The amplitude of the motion is \( \sqrt{2} \, \text{cm} \).

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