Answer
The maximum speed of the particle is 63.2 m/s
Work Step by Step
We know the total energy $E$ is equal to the sum of the potential energy $U$ and the kinetic energy $K$. If the particle oscillates between x = 2.0 mm and x = 8.0 mm, then the kinetic energy at those two points is zero. We can consider the point x = 2.0 mm to find the total energy;
$E = U+K$
$E = 5.0~J+0$
$E = 5.0~J$
The total energy is 5.0 J.
The maximum speed occurs when the kinetic energy is at a maximum. This occurs when the potential energy is at a minimum. From the graph, we can see that the potential energy is at a minimum at the point x = 4.0 mm. We can find the kinetic energy at the point x = 4.0 mm;
$K+U = E$
$K = E-U$
$K = 5.0~J - 1.0~J$
$K = 4.0~J$
We can find the speed of the particle at point x = 4.0 mm;
$K = 4.0~J$
$\frac{1}{2}mv^2 = 4.0~J$
$v^2 = \frac{(2)(4.0~J)}{m}$
$v = \sqrt{\frac{(2)(4.0~J)}{m}}$
$v = \sqrt{\frac{(2)(4.0~J)}{0.0020~kg}}$
$v = 63.2~m/s$
The maximum speed of the particle is 63.2 m/s.
