The position velocity and acceleration of a particle executihg simple harmonic motionn asre found to have magnitudes 2cm, 1ms^-1 and 10ms^-2 at a certain instant. Find the amplitude and the time period of the motion.

Explanation

To solve the problem, we need to find the amplitude (A) and the time period (T) of the simple harmonic motion (SHM) given the position (x), velocity (v), and acceleration (a) of a particle at a certain instant.

### Step-by-Step Solution:

1. **Convert Position to Meters**:
The position \( x \) is given as 2 cm. We need to convert this to meters:
\[ x = 2 \, \text{cm} = 0.02 \, \text{m} \]

2. **Identify Given Values**:
We have:
- Position \( x = 0.02 \, \text{m} \)
- Velocity \( v = 1 \, \text{m/s} \)
- Acceleration \( a = 10 \, \text{m/s}^2 \)

3. **Use the Relationship Between Acceleration, Angular Frequency, and Position**:
In SHM, the acceleration \( a \) is related to the position \( x \) and angular frequency \( \omega \) by the formula:
\[ a = -\omega^2 x \]
Since we are interested in magnitudes, we can write:
\[ a = \omega^2 x \]
Rearranging gives:
\[ \omega^2 = \frac{a}{x} \]

4. **Calculate Angular Frequency \( \omega \)**:
Substitute the known values of \( a \) and \( x \):
\[ \omega^2 = \frac{10 \, \text{m/s}^2}{0.02 \, \text{m}} = 500 \, \text{s}^{-2} \]
Therefore,
\[ \omega = \sqrt{500} = 10\sqrt{5} \, \text{s}^{-1} \]

5. **Use the Velocity Formula in SHM**:
The velocity in SHM can be expressed as:
\[ v = \omega \sqrt{A^2 - x^2} \]
Rearranging gives:
\[ \sqrt{A^2 - x^2} = \frac{v}{\omega} \]

6. **Calculate Amplitude \( A \)**:
Substitute \( v \), \( \omega \), and \( x \):
\[ \sqrt{A^2 - (0.02)^2} = \frac{1}{10\sqrt{5}} \]
Squaring both sides:
\[ A^2 - (0.02)^2 = \left(\frac{1}{10\sqrt{5}}\right)^2 \]
\[ A^2 - 0.0004 = \frac{1}{500} \]
\[ A^2 = 0.0004 + 0.002 = 0.0024 \]
\[ A = \sqrt{0.0024} \approx 0.049 \, \text{m} \approx 4.9 \, \text{cm} \]

7. **Calculate Time Period \( T \)**:
The time period \( T \) is related to the angular frequency \( \omega \) by:
\[ T = \frac{2\pi}{\omega} \]
Substitute \( \omega = 10\sqrt{5} \):
\[ T = \frac{2\pi}{10\sqrt{5}} = \frac{\pi}{5\sqrt{5}} \approx 0.141 \, \text{s} \]

### Final Answers:
- Amplitude \( A \approx 4.9 \, \text{cm} \)
- Time Period \( T \approx 0.141 \, \text{s} \)