Determine the slope, s1, of the graph (xvii).

The graph is $m$ (vertical axis) versus $T^2$ (horizontal axis) for the single spring. To determine the slope $s_1$, choose two widely separated points $(T^2_1, m_1)$ and $(T^2_2, m_2)$ on your best-fit line.

Using the example data from the previous response (e.g., points $(1.44 \text{ s}^2, 150 \text{ g})$ and $(3.35 \text{ s}^2, 350 \text{ g})$):

Your actual value for $s_1$ will depend on your experimental data and graph.

On the same graph paper and using the same axes and scales as for the first plot, plot the points using the values of $m$ and $T_0^2$ from the "Two Springs" columns of your table. Draw the best fit straight line through these new points.

The graph is $m$ (vertical axis) versus $T_0^2$ (horizontal axis) for the two-spring system. To determine the slope $s_2$, choose two widely separated points $(T_{0,1}^2, m_1)$ and $(T_{0,2}^2, m_2)$ on your best-fit line for the two-spring system.

Using the example data from the previous response (e.g., points $(0.72 \text{ s}^2, 150 \text{ g})$ and $(1.69 \text{ s}^2, 350 \text{ g})$):

Your actual value for $s_2$ will depend on your experimental data and graph.

Using the example slopes calculated above:

Your actual value for $\gamma$ will depend on your calculated slopes.

• Ensure that the oscillations are small and purely vertical to avoid conical motion or swinging.
• Avoid parallax error when reading the stopwatch by ensuring your eye is level with the mark.
• Start and stop the stopwatch when the mass is at the same point of its oscillation (e.g., the lowest point) to ensure consistent timing.
• Minimize air currents or drafts that could affect the oscillations.

When the number of springs in parallel increases, the effective spring constant of the system increases. A higher spring constant leads to a higher frequency of oscillation (and a shorter period).

We use Hooke's Law, which states that the force $F$ applied to a spring is directly proportional to its extension $e$, where $k$ is the spring stiffness (force constant).

Given:
Force $F = 8.0 \text{ N}$
Stiffness $k = 125 \text{ Nm}^{-1}$

Rearrange the formula to solve for extension $e$:

Substitute the given values:

The extension of the spring is $\boxed{0.064 \text{ m}}$.