A student hangs different masses on one spring and times 20 oscillations each time: Since T² =…

Step 1. Periods. $T = \dfrac{14.1\ \text{s}}{20} = 0.705\ \text{s}$, and so on. The squares are $0.497$, $0.980$, $1.488$ and $1.974$, all in $\text{s}^2$.

Step 2. Slope. The best line through the origin has slope $\dfrac{\sum mT^2}{\sum m^2} = 2.47\ \text{s}^2/\text{kg}$.

Step 3. Spring constant. $k = \dfrac{4\pi^2}{2.47\ \text{s}^2/\text{kg}} = 16.0\ \text{N/m}$.

Why time 20 oscillations? A stopwatch error of about $0.2\ \text{s}$ is spread over 20 periods. One period alone would carry the whole error.