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

Options

A. `100 s`
B. `0.01 s`
C. `10 s`
D. `0.1 s`

Explanation

To find the period of oscillation for a particle executing simple harmonic motion (SHM), we can use the relationship between maximum velocity, amplitude, and angular frequency. Here’s a step-by-step solution:

### Step 1: Write down the formula for maximum velocity in SHM
The maximum velocity \( v_{\text{max}} \) in simple harmonic motion is given by the formula:
\[ v_{\text{max}} = A \omega \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency.

### Step 2: Convert the amplitude to meters
The amplitude is given as \( 7 \, \text{mm} \). We need to convert this to meters:
\[ A = 7 \, \text{mm} = 7 \times 10^{-3} \, \text{m} \]

### Step 3: Substitute the values into the formula
We know that the maximum velocity \( v_{\text{max}} \) is \( 4.4 \, \text{m/s} \). Substituting the values we have:
\[ 4.4 = (7 \times 10^{-3}) \omega \]

### Step 4: Solve for angular frequency \( \omega \)
Rearranging the equation to solve for \( \omega \):
\[ \omega = \frac{4.4}{7 \times 10^{-3}} \]
Calculating this gives:
\[ \omega \approx \frac{4.4}{0.007} \approx 628.57 \, \text{rad/s} \]

### Step 5: Relate angular frequency to the period
The angular frequency \( \omega \) is related to the period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]
Rearranging this to solve for \( T \):
\[ T = \frac{2\pi}{\omega} \]

### Step 6: Substitute \( \omega \) to find the period \( T \)
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{628.57} \]
Calculating this gives:
\[ T \approx \frac{6.2832}{628.57} \approx 0.01 \, \text{s} \]

### Final Answer
The period of oscillation \( T \) is approximately \( 0.01 \, \text{s} \).

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