The maximum velocity a particle, executing simple harmonic motion with an amplitude 7 mm, 4.4 m//s. The period of oscillation is.

Options

A. (a) `0.01 s`
B. (b) `10 s`
C. (c ) `0.1s`
D. (d) `100s`

Explanation

To find the period of oscillation for a particle executing simple harmonic motion (SHM) with a given amplitude and maximum velocity, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Amplitude (A) = 7 mm = 7 × 10^(-3) m
- Maximum velocity (V_max) = 4.4 m/s

2. **Use the formula for maximum velocity in SHM:**
The maximum velocity (V_max) in simple harmonic motion is given by the formula:
\[ V_{max} = A \cdot \omega \]
where \( \omega \) is the angular frequency.

3. **Relate angular frequency to the period:**
The angular frequency \( \omega \) is related to the period (T) by the formula:
\[ \omega = \frac{2\pi}{T} \]

4. **Substitute the expression for \( \omega \) into the maximum velocity formula:**
Replacing \( \omega \) in the maximum velocity formula gives:
\[ V_{max} = A \cdot \frac{2\pi}{T} \]

5. **Rearranging the formula to find T:**
Rearranging the equation to solve for T, we have:
\[ T = \frac{A \cdot 2\pi}{V_{max}} \]

6. **Substituting the known values into the equation:**
Now substituting the values for A and V_max:
\[ T = \frac{(7 \times 10^{-3} \, \text{m}) \cdot (2\pi)}{4.4 \, \text{m/s}} \]

7. **Calculating the value:**
- First calculate \( 7 \times 2\pi \):
\[ 7 \times 2\pi \approx 7 \times 6.2832 \approx 43.9824 \, \text{m} \]
- Now divide by 4.4 m/s:
\[ T \approx \frac{43.9824}{4.4} \approx 10 \times 10^{-3} \, \text{s} = 0.01 \, \text{s} \]

8. **Final answer:**
The period of oscillation (T) is approximately **0.01 seconds**.