A `1.00 xx 10^(-20)`kg particle is vibrating with simple harmonic motion with a period of` 1.00 xx 10^(-5)` sec and a maximum speed of `1.00xx10^(3)` m/s . The maximum displacement of the particle is

Options

A. 1.59 mm
B. 1.00 m
C. 10m
D. 3.18 mm

Answer

To find the maximum displacement (amplitude) of a particle vibrating in simple harmonic motion (SHM), we can use the relationship between maximum speed, angular frequency, and amplitude.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Mass of the particle, \( m = 1.00 \times 10^{-20} \) kg (not needed for this calculation).
- Period of motion, \( T = 1.00 \times 10^{-5} \) seconds.
- Maximum speed, \( v_{\text{max}} = 1.00 \times 10^{3} \) m/s.

2. **Calculate Angular Frequency**:
The angular frequency \( \omega \) is given by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of \( T \):
\[ \omega = \frac{2\pi}{1.00 \times 10^{-5}} \approx 628318.53 \, \text{rad/s} \]

3. **Relate Maximum Speed to Amplitude**:
The maximum speed in SHM is given by:
\[ v_{\text{max}} = \omega A \]
where \( A \) is the amplitude (maximum displacement). Rearranging this formula gives:
\[ A = \frac{v_{\text{max}}}{\omega} \]

4. **Substitute Values**:
Now, we substitute the values of \( v_{\text{max}} \) and \( \omega \):
\[ A = \frac{1.00 \times 10^{3}}{628318.53} \approx 0.00159155 \, \text{m} \]

5. **Convert to Millimeters**:
To express the amplitude in millimeters:
\[ A \approx 0.00159155 \, \text{m} = 1.59155 \, \text{mm} \approx 1.59 \, \text{mm} \]

6. **Final Answer**:
The maximum displacement of the particle is approximately \( 1.59 \, \text{mm} \).

### Summary:
The maximum displacement (amplitude) of the particle is \( 1.59 \, \text{mm} \).

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