A particle of mass 2kg executing SHM has amplitude 20cm and time period 1s. Its maximum speed is

Options

A. 0.314 m/s
B. 0.628 m/s
C. 1.256 m/s
D. 2.512 m/s

Answer

To find the maximum speed of a particle executing simple harmonic motion (SHM), we can follow these steps:

### Step 1: Identify the given values
- Mass of the particle (m) = 2 kg (not needed for maximum speed calculation)
- Amplitude (A) = 20 cm = 0.2 m (convert cm to m)
- Time period (T) = 1 s

### Step 2: Calculate the angular frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of T:
\[ \omega = \frac{2\pi}{1} = 2\pi \, \text{rad/s} \]

### Step 3: Calculate the maximum speed (Vmax)
The maximum speed in SHM can be calculated using the formula:
\[ V_{\text{max}} = \omega A \]
Substituting the values of ω and A:
\[ V_{\text{max}} = (2\pi)(0.2) = 0.4\pi \, \text{m/s} \]

### Step 4: Calculate the numerical value of Vmax
Now, we can calculate the numerical value:
\[ V_{\text{max}} \approx 0.4 \times 3.14 \approx 1.256 \, \text{m/s} \]

### Conclusion
The maximum speed of the particle executing SHM is approximately:
\[ V_{\text{max}} \approx 1.256 \, \text{m/s} \]