A particle executes simple harmonic motion represented by displacement function as
`x(t)=A sin(omegat+phi)`
If the position and velocity of the particle at t = 0 s are 2 cm and `2omega" cm s"^(-1)` respectively, then its amplitude is `xsqrt(2)` cm where the value of x is _________.

Answer

To solve the problem step by step, we will analyze the given information and derive the required amplitude.

### Step 1: Write down the displacement function
The displacement of the particle executing simple harmonic motion (SHM) is given by:
\[ x(t) = A \sin(\omega t + \phi) \]

### Step 2: Find the position at \( t = 0 \)
At \( t = 0 \):
\[ x(0) = A \sin(\phi) \]
We are given that \( x(0) = 2 \, \text{cm} \). Therefore, we can write:
\[ A \sin(\phi) = 2 \quad \text{(Equation 1)} \]

### Step 3: Find the velocity function
The velocity \( v(t) \) is the derivative of the displacement function:
\[ v(t) = \frac{dx}{dt} = A \omega \cos(\omega t + \phi) \]

### Step 4: Find the velocity at \( t = 0 \)
At \( t = 0 \):
\[ v(0) = A \omega \cos(\phi) \]
We are given that \( v(0) = 2\omega \, \text{cm/s} \). Therefore, we can write:
\[ A \omega \cos(\phi) = 2\omega \quad \text{(Equation 2)} \]

### Step 5: Simplify Equation 2
Dividing both sides of Equation 2 by \( \omega \) (assuming \( \omega \neq 0 \)):
\[ A \cos(\phi) = 2 \quad \text{(Equation 3)} \]

### Step 6: Divide Equation 1 by Equation 3
Now, we will divide Equation 1 by Equation 3:
\[ \frac{A \sin(\phi)}{A \cos(\phi)} = \frac{2}{2} \]
This simplifies to:
\[ \tan(\phi) = 1 \]
From this, we find:
\[ \phi = 45^\circ \quad \text{or} \quad \phi = \frac{\pi}{4} \, \text{radians} \]

### Step 7: Substitute \( \phi \) back into Equation 1
Now we substitute \( \phi = 45^\circ \) back into Equation 1:
\[ 2 = A \sin(45^\circ) \]
Since \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ 2 = A \cdot \frac{1}{\sqrt{2}} \]
Thus, we can solve for \( A \):
\[ A = 2 \sqrt{2} \]

### Step 8: Relate \( A \) to the given amplitude
We are given that the amplitude can be expressed as \( X \sqrt{2} \):
\[ A = X \sqrt{2} \]
Setting \( 2 \sqrt{2} = X \sqrt{2} \), we can divide both sides by \( \sqrt{2} \):
\[ X = 2 \]

### Final Answer
The value of \( X \) is:
\[ \boxed{2} \]