The maximum speed of a particle executing SHM is 10 m/s and maximum acceleration is `31.4m//s^(2)`. Its periodic time is

Options

A. 2 s
B. 4 s
C. 6 s
D. 1 s

Explanation

To find the periodic time (T) of a particle executing Simple Harmonic Motion (SHM) given the maximum speed (V_max) and maximum acceleration (A_max), we can use the following relationships:

1. **Maximum Speed (V_max)**:
\[ V_{\text{max}} = A \cdot \omega \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

2. **Maximum Acceleration (A_max)**:
\[ A_{\text{max}} = A \cdot \omega^2 \]

From the problem, we know:
- \( V_{\text{max}} = 10 \, \text{m/s} \)
- \( A_{\text{max}} = 31.4 \, \text{m/s}^2 \)

### Step 1: Relate V_max and A_max
We can divide the equation for maximum speed by the equation for maximum acceleration:

\[ \frac{V_{\text{max}}}{A_{\text{max}}} = \frac{A \cdot \omega}{A \cdot \omega^2} \]

This simplifies to:

\[ \frac{V_{\text{max}}}{A_{\text{max}}} = \frac{1}{\omega} \]

### Step 2: Solve for ω
Rearranging the equation gives:

\[ \omega = \frac{A_{\text{max}}}{V_{\text{max}}} \]

Substituting the known values:

\[ \omega = \frac{31.4}{10} = 3.14 \, \text{rad/s} \]

### Step 3: Relate ω to the Periodic Time (T)
The angular frequency \( \omega \) is related to the periodic time \( T \) by:

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

### Step 4: Solve for T
Rearranging gives us:

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

Substituting the value of \( \omega \):

\[ T = \frac{2\pi}{3.14} \]

Calculating this gives:

\[ T \approx 2 \, \text{seconds} \]

### Final Answer
Thus, the periodic time \( T \) is approximately **2 seconds**.

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