The maximum velocity for particle in SHM is 0.16 m/s and maximum acceleration is `0.64 m//s^(2)`. The amplitude is

Options

A. `4xx10^(-2)m`
B. `4xx10^(-1)m`
C. `4xx10m`
D. `4xx10^(0)m`

Explanation

To find the amplitude of a particle in Simple Harmonic Motion (SHM) given the maximum velocity (V_max) and maximum acceleration (a_max), we can use the following relationships:

1. The maximum velocity (V_max) in SHM is given by:
\[ V_{max} = \omega A \]
where \( \omega \) is the angular frequency and \( A \) is the amplitude.

2. The maximum acceleration (a_max) in SHM is given by:
\[ a_{max} = \omega^2 A \]

Given:
- \( V_{max} = 0.16 \, \text{m/s} \)
- \( a_{max} = 0.64 \, \text{m/s}^2 \)

### Step 1: Express \( \omega \) in terms of \( A \) using the maximum velocity formula.
From the first equation, we can express \( \omega \) as:
\[ \omega = \frac{V_{max}}{A} \]

### Step 2: Substitute \( \omega \) into the maximum acceleration formula.
Substituting \( \omega \) into the second equation gives:
\[ a_{max} = \left(\frac{V_{max}}{A}\right)^2 A \]
This simplifies to:
\[ a_{max} = \frac{V_{max}^2}{A} \]

### Step 3: Rearrange the equation to solve for \( A \).
Rearranging the equation to find \( A \):
\[ A = \frac{V_{max}^2}{a_{max}} \]

### Step 4: Substitute the known values into the equation.
Now substituting the known values:
\[ A = \frac{(0.16)^2}{0.64} \]

### Step 5: Calculate \( A \).
Calculating the numerator:
\[ (0.16)^2 = 0.0256 \]
Now substituting this value:
\[ A = \frac{0.0256}{0.64} \]
Calculating this gives:
\[ A = 0.04 \, \text{m} \]

### Conclusion
Thus, the amplitude \( A \) is:
\[ A = 4 \times 10^{-2} \, \text{m} \]