Solution:

Given:
- Amplitude, \( A = 0.01 \, \text{m} \)
- Frequency, \( f = 60 \, \text{Hz} \)

Solution:

1. Angular Frequency \( \omega \):
\[
\omega = 2 \pi f = 2 \pi \times 60 = 120 \pi \, \text{rad/s}
\]

2. Maximum Acceleration \( a_{\text{max}} \):
Maximum acceleration in SHM is given by:
\[
a_{\text{max}} = \omega^2 A
\]

Substitute values of \( \omega \) and \( A \):
\[
a_{\text{max}} = (120 \pi)^2 \times 0.01 = 14400 \pi^2 \times 0.01 = 144 \pi^2 \, \text{m/s}^2
\]

Answer:
The maximum acceleration of the particle is \( 144 \pi^2 \, \text{m/s}^2 \).