The time period of an oscillating body executing SHM is 0.05 sec and its amplitude is 40 cm. The maximum velocity of particle is :

Options

A. `16 pi ms^(-1)`
B. `2 pi ms^(-1)`
C. `3.1 ms^(-1)`
D. `4 pi ms//s`

Answer

To find the maximum velocity of a particle executing Simple Harmonic Motion (SHM), we can use the following 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 1: Identify the given values
From the question, we have:
- Amplitude \( A = 40 \) cm = \( 0.4 \) m (converting centimeters to meters)
- Time period \( T = 0.05 \) sec

### Step 2: Calculate the angular frequency \( \omega \)
The angular frequency \( \omega \) can be calculated using the formula:

\[ \omega = \frac{2\pi}{T} \]

Substituting the value of \( T \):

\[ \omega = \frac{2\pi}{0.05} \]

### Step 3: Calculate \( \omega \)
Now, calculate \( \omega \):

\[ \omega = \frac{2\pi}{0.05} = \frac{2 \times 3.14}{0.05} \approx \frac{6.28}{0.05} \approx 125.6 \, \text{rad/s} \]

### Step 4: Calculate the maximum velocity \( V_{\text{max}} \)
Now, substitute the values of \( A \) and \( \omega \) into the maximum velocity formula:

\[ V_{\text{max}} = A \cdot \omega = 0.4 \cdot 125.6 \]

### Step 5: Calculate \( V_{\text{max}} \)
Now, calculate \( V_{\text{max}} \):

\[ V_{\text{max}} = 0.4 \cdot 125.6 \approx 50.24 \, \text{m/s} \]

### Final Answer
The maximum velocity of the particle is approximately \( 50.24 \, \text{m/s} \).

---