Solution:

In simple harmonic motion (SHM), the maximum velocity \( v_{\text{max}} \) and maximum acceleration \( a_{\text{max}} \) are given by:

\[
v_{\text{max}} = \omega A \quad \text{and} \quad a_{\text{max}} = \omega^2 A
\]

where:
- \( \omega \) is the angular frequency,
- \( A \) is the amplitude.

 Solution:

1. Given that \( v_{\text{max}} = a_{\text{max}} \), we have:
\[
\omega A = \omega^2 A
\]

2. Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[
\omega = \omega^2
\]

3. Solving for \( \omega \):
\[
\omega = 1 \, \text{rad/s}
\]

4. Time Period \( T \):
\[
T = \frac{2\pi}{\omega} = \frac{2\pi}{1} = 2\pi \approx 6.28 \, \text{s}
\]

Answer:
The time period of the particle is **6.28 seconds**.