91Ó°ÊÓ

The position of the mass is given by the equation \(x=-10 \cos (1.57 t)\). Here, \(x\) represents the position of the mass (in centimeters) and \(t\) represents time (in seconds).

We need data points for the position of the mass at 0.2 second intervals for \(t=0s\) to \(t=4s\). By plugging the values of \(t\) into the equation, we can find the corresponding \(x\) values (position).

Using the equation \(x=-10 \cos (1.57 t)\), calculate the position of the mass for the required time intervals: t=0: \(x= -10 \cos (1.57 \cdot 0) = -10\) t=0.2: \(x=-10 \cos (1.57 \cdot 0.2) \approx 4.1\) t=0.4: \(x=-10 \cos (1.57 \cdot 0.4) \approx 9.2\) t=0.6: \(x=-10 \cos (1.57 \cdot 0.6) \approx 8.6\) t=0.8: \(x=-10 \cos (1.57 \cdot 0.8) \approx 1.1\) . . . t=4: \(x=-10 \cos (1.57 \cdot 4) \approx -10\)

Using the calculated data points, plot the position of the mass over time with a dot for each time interval.

Examine the plot to determine where the mass is moving the fastest and slowest. The steeper the slope of the line between two points, the faster the mass is moving, while the less steep (or more horizontal) the slope, the slower the mass is moving. From analyzing the plot, we can determine that the mass is moving fastest at the peak and trough (extreme positions) of the oscillation and slowest as it passes through the equilibrium position (x = 0). This is because as the mass moves away from the equilibrium position, the restoring force (spring force) increases, and when it reaches the extreme positions, the restoring force is at its maximum, hence the mass is moving fastest. When the mass is at the equilibrium position, the restoring force is at its minimum, and thus the mass is moving slowest.