A particle executes linear simple harmonic motion with an amplitude of 2 cm . When the particle is at 1 cm from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

Options

A. `(1)/(2pisqrt(3))`
B. `2pisqrt(3)`
C. `(2pi)/(sqrt(3))`
D. `(sqrt(3))/(2pi)`

Explanation

To solve the problem step by step, we can follow these calculations:

### Step 1: Understand the problem
We have a particle in simple harmonic motion (SHM) with an amplitude \( A = 2 \, \text{cm} \). We need to find the time period \( T \) when the particle is at a position \( x = 1 \, \text{cm} \) from the mean position, and at this position, the magnitude of its velocity \( v \) is equal to the magnitude of its acceleration \( a \).

### Step 2: Write the equations for velocity and acceleration in SHM
The formulas for velocity \( v \) and acceleration \( a \) in SHM are:
- \( v = \omega \sqrt{A^2 - x^2} \)
- \( a = -\omega^2 x \)

Where \( \omega \) is the angular frequency.

### Step 3: Set up the equation based on the given condition
Since the magnitudes of velocity and acceleration are equal at \( x = 1 \, \text{cm} \), we can write:
\[ |\omega \sqrt{A^2 - x^2}| = |-\omega^2 x| \]

Substituting \( A = 2 \, \text{cm} \) and \( x = 1 \, \text{cm} \):
\[ \omega \sqrt{2^2 - 1^2} = \omega^2 \cdot 1 \]

### Step 4: Simplify the equation
Now, simplifying the left side:
\[ \omega \sqrt{4 - 1} = \omega^2 \]
\[ \omega \sqrt{3} = \omega^2 \]

### Step 5: Cancel out \( \omega \) (assuming \( \omega \neq 0 \))
Dividing both sides by \( \omega \):
\[ \sqrt{3} = \omega \]

### Step 6: Find the time period \( T \)
The time period \( T \) is given by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting \( \omega = \sqrt{3} \):
\[ T = \frac{2\pi}{\sqrt{3}} \]

### Final Answer
Thus, the time period \( T \) of the particle is:
\[ T = \frac{2\pi}{\sqrt{3}} \, \text{seconds} \]

---