In $FIGUREEX10.26$, what is the maximum speed of a $2.0g$particle that oscillates between $$\begin{gathered} x=2.0mmand \\ x=8.0mm \end{gathered}$$?

It is the change in an object's location with regard to time.

Particle's mass $$\begin{gathered} m=2gm \\ =0.002kg \end{gathered}$$,

Formula to determine the kinetic energy of a particle,

$KE=\frac{1}{2}m{v}^2$

$KE$= kinetic energy,

$m$= particle's mass,

$v$= particle's velocity.

Consider the graph of potential energy against distance,

The highest speed will occur near the bottom of the curve, when the potential energy difference is greatest. The bottom of the graph appears at $x=4mm$in the graph above.

The difference in potential energy at $x=2mm$.

Also, at $x=4mm$is $4J$may be seen in the graph above. The potential energy differential is transformed to kinetic energy according to the rule of energy conservation. Hence,

localid="1647865168683" $$\begin{gathered} ∆PE=∆KE \\ =4J \end{gathered}$$

Calculate particle's maximum velocity,

$∆KE=\frac{1}{2}m{v}^2$

Substituting the values,

$$\begin{gathered} 4=\frac{1}{2}\times 0.002\times v^2 \\ v=63.24m/s \end{gathered}$$

Hence, particle's maximum velocity is$63.24m/s$.