Answer:
The maximum acceleration of the system is 359.970 centimeters per square second.
Explanation:
The motion of the mass-spring system is represented by the following formula:

Where:
- Position of the mass with respect to the equilibrium position, measured in centimeters.
- Amplitude of the mass-spring system, measured in centimeters.
- Angular frequency, measured in radians per second.
- Time, measured in seconds.
- Phase, measured in radians.
The acceleration experimented by the mass is obtained by deriving the position equation twice:

Where the maximum acceleration of the system is represented by
.
The natural frequency of the mass-spring system is:

Where:
- Spring constant, measured in newtons per meter.
- Mass, measured in kilograms.
If
and
, the natural frequency is:


Lastly, the maximum acceleration of the system is:


The maximum acceleration of the system is 359.970 centimeters per square second.