The amplitude and time period of a aprticle of mass `0.1kg` executing simple harmonic motion are `1m` and `6.28s`, respectively. Then its angular frequency and acceleration at a displacement fo `0.5m` are respectively

Options

A. `1 rad//s, 0.5 m//s^(2)`
B. `2rad//s, 1m//s^(2)`
C. `0.5 rad//s, 0.5m//s^(2)`
D. `1rad//s, 1m//s^(2)`

Explanation

To solve the problem, we need to find the angular frequency and the acceleration of a particle in simple harmonic motion (SHM) given its amplitude, time period, and displacement.

### Step 1: Calculate the Angular Frequency (ω)

The formula for angular frequency (ω) is given by:

\[ \omega = \frac{2\pi}{T} \]

where \( T \) is the time period.

Given:
- Time period \( T = 6.28 \, \text{s} \)

Substituting the value of \( T \):

\[ \omega = \frac{2\pi}{6.28} \]

Using \( \pi \approx 3.14 \):

\[ \omega = \frac{2 \times 3.14}{6.28} = \frac{6.28}{6.28} = 1 \, \text{rad/s} \]

### Step 2: Calculate the Acceleration (a) at Displacement (x)

The formula for acceleration in SHM is given by:

\[ a = -\omega^2 x \]

where \( x \) is the displacement.

Given:
- Displacement \( x = 0.5 \, \text{m} \)

Substituting the values of \( \omega \) and \( x \):

\[ a = - (1)^2 \times 0.5 = -0.5 \, \text{m/s}^2 \]

Since we are interested in the magnitude of acceleration, we take the absolute value:

\[ |a| = 0.5 \, \text{m/s}^2 \]

### Final Answers
- Angular frequency \( \omega = 1 \, \text{rad/s} \)
- Acceleration \( |a| = 0.5 \, \text{m/s}^2 \)

### Summary of Results
1. Angular frequency: \( 1 \, \text{rad/s} \)
2. Acceleration at \( 0.5 \, \text{m} \): \( 0.5 \, \text{m/s}^2 \)

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