A particle executes SHM of period 12s. Two sec after it passes through the centre of oscillation, the velocity is found to be 3.142 cm `s^(-1)` fin dht amplitude and the length of the path.

Answer

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

### Step 1: Identify the given values
- Period (T) = 12 seconds
- Velocity (v) = 3.142 cm/s
- Time after passing through the center (t) = 2 seconds

### Step 2: Calculate angular frequency (ω)
The angular frequency (ω) is calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the given period:
\[ \omega = \frac{2\pi}{12} = \frac{\pi}{6} \text{ rad/s} \]

### Step 3: Use the velocity formula in SHM
The velocity (v) in SHM can be expressed as:
\[ v = A \omega \cos(\omega t) \]
Where:
- A is the amplitude
- ω is the angular frequency
- t is the time

### Step 4: Substitute known values into the velocity equation
Substituting the known values into the velocity formula:
\[ 3.142 = A \left(\frac{\pi}{6}\right) \cos\left(\frac{\pi}{6} \times 2\right) \]
Calculating \(\omega t\):
\[ \omega t = \frac{\pi}{6} \times 2 = \frac{\pi}{3} \]

### Step 5: Calculate \(\cos(\frac{\pi}{3})\)
The value of \(\cos(\frac{\pi}{3})\) is:
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

### Step 6: Substitute \(\cos(\frac{\pi}{3})\) into the equation
Now substituting \(\cos(\frac{\pi}{3})\) back into the velocity equation:
\[ 3.142 = A \left(\frac{\pi}{6}\right) \left(\frac{1}{2}\right) \]
This simplifies to:
\[ 3.142 = \frac{A \pi}{12} \]

### Step 7: Solve for amplitude (A)
To find A, rearrange the equation:
\[ A = \frac{3.142 \times 12}{\pi} \]
Calculating the value:
\[ A \approx \frac{37.704}{3.142} \approx 12 \text{ cm} \]

### Step 8: Calculate the length of the path
The length of the path in SHM is given by:
\[ \text{Path Length} = 2 \times A \]
Substituting the amplitude:
\[ \text{Path Length} = 2 \times 12 = 24 \text{ cm} \]

### Final Answer
- Amplitude (A) = 12 cm
- Length of the path = 24 cm