A particle executies linear simple harmonic motion with an amplitude `3cm` .When the particle is at `2cm` from the mean position , the magnitude of its velocity is equal to that of acceleration .The its time period in seconds is

Options

A. `sqrt(5)/(pi)`
B. `sqrt(5)/(2pi)`
C. `(4pi)/sqrt(5)`
D. `(2pi)/sqrt(3)`

Answer

To solve the problem step by step, we will use the properties of simple harmonic motion (SHM) and the relationships between displacement, velocity, and acceleration.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Amplitude (A) = 3 cm
- Displacement (x) = 2 cm

2. **Understand the Relationships:**
- The velocity (v) of a particle in SHM is given by the formula:
\[ v = \omega \sqrt{A^2 - x^2} \]
- The acceleration (a) of a particle in SHM is given by the formula:
\[ a = -\omega^2 x \]
- We are told that the magnitude of velocity is equal to the magnitude of acceleration:
\[ |v| = |a| \]

3. **Set Up the Equation:**
- From the above relationships, we can write:
\[ \omega \sqrt{A^2 - x^2} = \omega^2 x \]
- Since \(\omega\) is common in both sides, we can cancel it out (assuming \(\omega \neq 0\)):
\[ \sqrt{A^2 - x^2} = \omega x \]

4. **Substitute Values:**
- Substitute \(A = 3\) cm and \(x = 2\) cm into the equation:
\[ \sqrt{3^2 - 2^2} = \omega \cdot 2 \]
- Calculate \(A^2 - x^2\):
\[ \sqrt{9 - 4} = \omega \cdot 2 \]
\[ \sqrt{5} = 2\omega \]

5. **Solve for \(\omega\):**
- Rearranging gives:
\[ \omega = \frac{\sqrt{5}}{2} \]

6. **Find the Time Period (T):**
- The time period \(T\) is related to \(\omega\) by the formula:
\[ T = \frac{2\pi}{\omega} \]
- Substitute \(\omega\) into the equation:
\[ T = \frac{2\pi}{\frac{\sqrt{5}}{2}} = \frac{4\pi}{\sqrt{5}} \]

7. **Final Answer:**
- Thus, the time period \(T\) is:
\[ T = \frac{4\pi}{\sqrt{5}} \text{ seconds} \]