## Maximum Force Experienced By a Body Experiencing Simple Harmonic Motion

The maximum acceleration experienced by a body undergoing simple harmonic motion is

\[a_{MAX} = \omega^2 A\]

where \[A , \: \omega\]

are the amplitude and the angular frequency of the motion.Suppose a body of mass 0.3 kg executes simple harmonic motion with amplitude

\[A=2.1 m\]

and frequency \[f=20 Hz\]

.Then

\[\omega = 2 \pi f = 2 \pi \times 20 =40 \pi\]

.Then

\[a_{MAX} = \omega^2 A = (40 \pi)^2 \times 2.1=33161 m/s^2\]

Now use Newtons Second Law

\[F=ma\]

.The maximum force is

\[F_{MAX} =0.3 \times 33162=9948.6 N\]

.