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

Options

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

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 1: Understand the relationship between maximum velocity, amplitude, and angular frequency
The maximum velocity (V_max) in SHM is given by the formula:
\[ V_{\text{max}} = A \cdot \omega \]
where:
- \( V_{\text{max}} \) is the maximum velocity,
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency.

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

### Step 3: Substitute the known values into the formula
We know:
- \( V_{\text{max}} = 4.4 \, \text{m/s} \)
- \( A = 7 \times 10^{-3} \, \text{m} \)

Substituting these values into the maximum velocity formula:
\[ 4.4 = (7 \times 10^{-3}) \cdot \omega \]

### Step 4: Solve for angular frequency (ω)
Rearranging the equation to solve for \( \omega \):
\[ \omega = \frac{4.4}{7 \times 10^{-3}} \]

Calculating \( \omega \):
\[ \omega = \frac{4.4}{0.007} \approx 628.57 \, \text{rad/s} \]

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

### Step 6: Solve for the period (T)
Rearranging the equation to solve for \( T \):
\[ T = \frac{2\pi}{\omega} \]

Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{628.57} \]

Calculating \( T \):
\[ T \approx \frac{6.2832}{628.57} \approx 0.01 \, \text{s} \]

### Final Answer
The period of oscillation is approximately:
\[ T \approx 0.01 \, \text{s} \]

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