Answer
It takes $~~1.1~ms~~$ to move between $2.0~mm$ and $-2.0~mm$
Work Step by Step
We can find the period:
$T = \frac{2\pi}{\omega}$
$T = \frac{2\pi}{600~rad/s}$
$T = 0.01047~s$
A displacement of $2.0~mm$ is $\frac{1}{3}$ of the amplitude of $6.0~mm$
We can find the angle $\theta_1$ when $sin(\theta_1) = \frac{1}{3}$:
$sin(\theta_1) = \frac{1}{3}$
$\theta_1 = 2.80~rad$
We can find the angle $\theta_2$ when $sin(\theta_2) = -\frac{1}{3}$:
$sin(\theta_2) = -\frac{1}{3}$
$\theta_2 = 3.48~rad$
We can find the time it takes to move between $2.0~mm$ and $-2.0~mm$:
$t = (\frac{3.48~rad-2.80~rad}{2\pi~rad})(0.01047~s)$
$t = 0.0011~s$
$t = 1.1~ms$
It takes $~~1.1~ms~~$ to move between $2.0~mm$ and $-2.0~mm$
