Answer
0.47 m/s
Work Step by Step
When the kinetic energy equals the potential energy, each must be half the total energy.
The speed at its maximum, $V_{max}=A\omega$. At this time energy is all kinetic. So we can write the total energy E,
$E=\frac{1}{2}mV_{max}^{2}=\frac{1}{2}mA^{2}\omega^{2}$
Let's take the speed = V when the kinetic energy has half this value.
$K=\frac{1}{2}mV^{2}=\frac{1}{2}\times\frac{1}{2}mA^{2}\omega^{2}$
$V^{2}=\frac{A^{2}\omega^{2}}{2}=\gt V=\frac{A\omega}{\sqrt 2}$
Let's plug known values into this equation.
$V=\frac{28.2\times10^{-2}\times2\pi\times 0.377\space s^{-1}}{\sqrt 2}=0.47\space m/s$
