The displacement of a particle executing shm is x = 0.5 cos (314t - 0.3) m. Find (i) amplitude (i) period.

Explanation

To solve the problem, we will analyze the given equation of motion for a particle executing simple harmonic motion (SHM) and extract the required parameters step by step.

### Given:
The displacement of the particle is given by the equation:
\[ x = 0.5 \cos(314t - 0.3) \, \text{m} \]

### Step 1: Identify the Amplitude
The standard form of the equation for SHM can be written as:
\[ x = A \cos(\omega t + \phi) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase constant.

From the given equation, we can see that:
\[ A = 0.5 \, \text{m} \]

### Step 2: Calculate the Period
The angular frequency \( \omega \) is given as:
\[ \omega = 314 \, \text{rad/s} \]

The relationship between angular frequency and the time period \( T \) is given by:
\[ \omega = \frac{2\pi}{T} \]

We can rearrange this equation to solve for \( T \):
\[ T = \frac{2\pi}{\omega} \]

Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{314} \]

### Step 3: Calculate the Numerical Value of the Period
Now, we can compute \( T \):
\[ T = \frac{2 \times 3.14}{314} \]
\[ T \approx \frac{6.28}{314} \]
\[ T \approx 0.02 \, \text{s} \]

### Final Results:
- **Amplitude**: \( A = 0.5 \, \text{m} \)
- **Period**: \( T \approx 0.02 \, \text{s} \)