The period of oscillation of a particle in `SHM` is `4sec` and its amplitude of vibration is `4cm`. The distance of the particle `0.5s` after passsing the mean position is

Options

A. `1.33 cm`
B. `2 cm`
C. `2sqrt(2)`
D. `2sqrt(3)cm`

Explanation

To solve the problem step by step, we will use the concepts of Simple Harmonic Motion (SHM).

### Step 1: Identify the given values
- Time period (T) = 4 seconds
- Amplitude (A) = 4 cm
- Time (t) = 0.5 seconds

### Step 2: Write the equation for displacement in SHM
The displacement \( x \) of a particle in SHM can be expressed as:
\[ x = A \sin(\omega t) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( t \) is the time.

### Step 3: Calculate the angular frequency \( \omega \)
The angular frequency \( \omega \) can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of \( T \):
\[ \omega = \frac{2\pi}{4} = \frac{\pi}{2} \text{ radians/second} \]

### Step 4: Substitute the values into the displacement equation
Now, substitute \( A \), \( \omega \), and \( t \) into the displacement equation:
\[ x = 4 \sin\left(\frac{\pi}{2} \cdot 0.5\right) \]
This simplifies to:
\[ x = 4 \sin\left(\frac{\pi}{4}\right) \]

### Step 5: Calculate \( \sin\left(\frac{\pi}{4}\right) \)
We know that:
\[ \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Thus, substituting this value back into the equation gives:
\[ x = 4 \cdot \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \text{ cm} \]

### Final Answer
The distance of the particle from the mean position after 0.5 seconds is:
\[ \boxed{2\sqrt{2} \text{ cm}} \]