A block of mass 0.01 kg is attached to a spring of force constant `10^(-1)Nm^(-1)` . If the energy of oscillation is `10^(-5)` J, find the angular frequency and amplitude of oscillation.

Explanation

To solve the problem step by step, we need to find the angular frequency and amplitude of oscillation for a block attached to a spring.

### Given Data:
- Mass of the block, \( m = 0.01 \, \text{kg} \)
- Spring constant, \( k = 10^{-1} \, \text{N/m} \)
- Energy of oscillation, \( E = 10^{-5} \, \text{J} \)

### Step 1: Calculate Angular Frequency
The formula for angular frequency \( \omega \) in simple harmonic motion (SHM) is given by:

\[ \omega = \sqrt{\frac{k}{m}} \]

Substituting the values of \( k \) and \( m \):

\[ \omega = \sqrt{\frac{10^{-1}}{0.01}} = \sqrt{\frac{10^{-1}}{10^{-2}}} = \sqrt{10^{1}} = \sqrt{10} \]

Thus, the angular frequency \( \omega \) is:

\[ \omega = \sqrt{10} \, \text{rad/s} \]

### Step 2: Calculate Amplitude
The total energy in SHM is given by the formula:

\[ E = \frac{1}{2} k A^2 \]

Where \( A \) is the amplitude. Rearranging the formula to solve for \( A \):

\[ A^2 = \frac{2E}{k} \]

Substituting the values of \( E \) and \( k \):

\[ A^2 = \frac{2 \times 10^{-5}}{10^{-1}} = \frac{2 \times 10^{-5}}{0.1} = 2 \times 10^{-4} \]

Now, taking the square root to find \( A \):

\[ A = \sqrt{2 \times 10^{-4}} = \sqrt{2} \times 10^{-2} \]

Calculating \( \sqrt{2} \):

\[ A \approx 1.414 \times 10^{-2} \, \text{m} \]

### Final Results:
- Angular frequency \( \omega = \sqrt{10} \, \text{rad/s} \)
- Amplitude \( A \approx 1.414 \times 10^{-2} \, \text{m} \)