The relationship between mass (m) and the square of the period (T2) for a single spring is given by the formula for the period of a mass-spring system, T = 2pi sqrt((m)/(k1)), where $k1

Squaring both sides gives:

Since no experimental data is provided, the numerical value of $s_1$ cannot be determined. It represents the ratio of the spring constant to $4\pi^2$.

(xviii) On the same axes, plot $m$ on the vertical axis and $T_0^2$ on the horizontal axis.

This step was explained in the previous response (Step xvi). You would plot the data points for the parallel spring system ($m$ vs. $T_0^2$) on the same graph as the single spring system, using different symbols or colors to distinguish them. A best-fit straight line passing through the origin would then be drawn for these points.

(xix) Determine the slope, $s_2$, of the graph.

If each spring has a constant $k_1$, then $k_{eff} = k_1 + k_1 = 2k_1$.

Squaring both sides gives:

Since no experimental data is provided, the numerical value of $s_2$ cannot be determined. It represents the ratio of twice the spring constant to $4\pi^2$.

(xx) Evaluate $\gamma = \frac{s_2}{s_1}$.

The value of $\gamma$ is $\boxed{\text{2}}$.

(xxi) State two precautions taken to ensure accurate results when performing this experiment.

• Ensure oscillations are purely vertical and not swinging sideways to avoid energy loss and inaccurate period measurements.
• Avoid parallax error when reading the stopwatch or measuring the extension of the spring.

How does change in the number of springs affects the frequency of the system?

The frequency ($f$) of a mass-spring system is inversely proportional to the period ($T$), and directly proportional to the square root of the spring constant ($k$). When the number of springs in parallel increases, the effective spring constant ($k_{eff}$) of the system increases. Since $f = \frac{1}{2\pi} \sqrt{\frac{k_{eff}}{m}}$, an increase in $k_{eff}$ leads to an increase in the frequency of the system.

A spring has a stiffness of $125 \, \text{Nm}^{-1}$. If a weight of $8.0 \, \text{N}$ is suspended on it, calculate the extension, $e$ of the spring.

$F = 8.0 \, \text{N}$ and $k = 125 \, \text{Nm}^{-1}$.

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