A particle executes a linear S.H.M. of amplitude 8 cm and period 2s. The magnitude of its maximum velocity is

Options

A. `8pi` cm/second
B. `16 pi` cm/second
C. `4pi` cm/second
D. `2pi` cm/second

Explanation

To find the magnitude of the maximum velocity of a particle executing simple harmonic motion (SHM), we can use the following steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Amplitude (A) = 8 cm
- Period (T) = 2 s

2. **Calculate Angular Frequency (ω):**
The angular frequency (ω) is given by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of T:
\[ \omega = \frac{2\pi}{2} = \pi \, \text{rad/s} \]

3. **Use the Formula for Maximum Velocity (V_max):**
The maximum velocity in SHM is given by:
\[ V_{\text{max}} = \omega A \]
Substituting the values of ω and A:
\[ V_{\text{max}} = \pi \times 8 \, \text{cm/s} \]

4. **Calculate the Maximum Velocity:**
\[ V_{\text{max}} = 8\pi \, \text{cm/s} \]

5. **Final Answer:**
The magnitude of the maximum velocity is:
\[ V_{\text{max}} \approx 25.13 \, \text{cm/s} \, (\text{using } \pi \approx 3.14) \]

### Summary:
The magnitude of the maximum velocity of the particle is approximately \( 25.13 \, \text{cm/s} \).

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