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. `2pisqrt3`
C. `(2pi)/(sqrt3)`
D. `sqrt3/(2pi)`

Answer

To solve the problem, we need to find the time period of a particle executing linear simple harmonic motion (SHM) given the amplitude and the condition that the magnitude of its velocity equals the magnitude of its acceleration when the particle is at a certain displacement from the mean position.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Amplitude, \( A = 2 \, \text{cm} \)
- Displacement from the mean position, \( x = 1 \, \text{cm} \)

2. **Use the Relationship in SHM:**
The displacement in SHM can be expressed as:
\[ x = A \cos(\omega t) \]
Substituting the known values:
\[ 1 = 2 \cos(\omega t) \]
This simplifies to:
\[ \cos(\omega t) = \frac{1}{2} \]

3. **Find \( \omega t \):**
The angle whose cosine is \( \frac{1}{2} \) is:
\[ \omega t = \frac{\pi}{3} \]

4. **Velocity and Acceleration in SHM:**
The velocity \( v \) and acceleration \( a \) in SHM are given by:
\[ v = -A \omega \sin(\omega t) \]
\[ a = -A \omega^2 \cos(\omega t) \]

5. **Substituting Values:**
Since \( \cos(\omega t) = \frac{1}{2} \):
\[ a = -A \omega^2 \left(\frac{1}{2}\right) = -2 \omega^2 \cdot \frac{1}{2} = -\omega^2 \]

6. **Using the Condition \( |v| = |a| \):**
According to the problem, the magnitudes of velocity and acceleration are equal:
\[ |A \omega \sin(\omega t)| = |A \omega^2 \cos(\omega t)| \]
Substituting \( A = 2 \):
\[ 2 \omega |\sin(\omega t)| = 2 \omega^2 \left|\frac{1}{2}\right| \]
This simplifies to:
\[ |\sin(\omega t)| = \omega \]

7. **Substituting \( \sin(\omega t) \):**
Since \( \sin(\omega t) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \):
\[ \frac{\sqrt{3}}{2} = \omega \]

8. **Relate \( \omega \) to Time Period \( T \):**
We know that:
\[ \omega = \frac{2\pi}{T} \]
Thus, substituting for \( \omega \):
\[ \frac{\sqrt{3}}{2} = \frac{2\pi}{T} \]

9. **Solving for \( T \):**
Rearranging gives:
\[ T = \frac{2\pi}{\frac{\sqrt{3}}{2}} = \frac{4\pi}{\sqrt{3}} \]

### Final Answer:
The time period \( T \) is:
\[ T = \frac{4\pi}{\sqrt{3}} \, \text{seconds} \]