The period of oscillation of a particle in `SHM` is `4s` and its amplitude of vibration is `4cm`. The distance of the particle `(1)/(3)s` after passing the mean poistion is

Options

A. `1.33cm`
B. `2cm`
C. `3cm`
D. `2.33cm`

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)** = \( \frac{1}{3} \) seconds

### Step 2: Calculate the angular frequency (ω)
The angular frequency \( ω \) is given by the formula:
\[ ω = \frac{2\pi}{T} \]
Substituting the value of T:
\[ ω = \frac{2\pi}{4} = \frac{\pi}{2} \text{ radians/second} \]

### Step 3: Write the equation of motion for SHM
The position of a particle in SHM can be described by the equation:
\[ X(t) = A \sin(ωt) \]
Substituting the values of A and ω:
\[ X(t) = 4 \sin\left(\frac{\pi}{2} t\right) \]

### Step 4: Substitute the value of time (t)
Now, we substitute \( t = \frac{1}{3} \) seconds into the equation:
\[ X\left(\frac{1}{3}\right) = 4 \sin\left(\frac{\pi}{2} \cdot \frac{1}{3}\right) \]
This simplifies to:
\[ X\left(\frac{1}{3}\right) = 4 \sin\left(\frac{\pi}{6}\right) \]

### Step 5: Calculate the sine value
We know that:
\[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]
So,
\[ X\left(\frac{1}{3}\right) = 4 \cdot \frac{1}{2} = 2 \text{ cm} \]

### Step 6: Conclusion
The distance of the particle after \( \frac{1}{3} \) seconds after passing the mean position is **2 cm**.

---