Plot a graph against M and T

The period $T$ is calculated by dividing the time for 10 oscillations ($t$) by 10. Then, $T^2$ is calculated by squaring the period.

• For $m = 150.0$ g, $t = 6.82$ s: $T = \frac{6.82}{10} = 0.682$ s $T^2 = (0.682)^2 = 0.465124 \approx 0.465$ s$^2$
• For $m = 200.0$ g, $t = 7.81$ s: $T = \frac{7.81}{10} = 0.781$ s $T^2 = (0.781)^2 = 0.609961 \approx 0.610$ s$^2$
• For $m = 250.0$ g, $t = 8.70$ s: $T = \frac{8.70}{10} = 0.870$ s $T^2 = (0.870)^2 = 0.7569 \approx 0.757$ s$^2$
• For $m = 300.0$ g, $t = 9.62$ s: $T = \frac{9.62}{10} = 0.962$ s $T^2 = (0.962)^2 = 0.925444 \approx 0.925$ s$^2$
• For $m = 350.0$ g, $t = 10.31$ s: $T = \frac{10.31}{10} = 1.031$ s $T^2 = (1.031)^2 = 1.062961 \approx 1.063$ s$^2$

Here is the completed table with the calculated values:

Mass $m$ (g)	Time $t$ for 10 oscillations (s)	Period $T$ (s)	$T^2$ (s$^2$)
150.0	6.82	0.682	0.465
200.0	7.81	0.781	0.610
250.0	8.70	0.870	0.757
300.0	9.62	0.962	0.925
350.0	10.31	1.031	1.063

To plot the graph of $m$ against $T^2$:

• Horizontal Axis ($x$-axis): Label this axis "$T^2$ (s$^2$)". Choose a scale that accommodates values from approximately 0.4 s$^2$ to 1.1 s$^2$.
• Vertical Axis ($y$-axis): Label this axis "$m$ (g)". Choose a scale that accommodates values from 150 g to 350 g.
• Plot each pair of $(T^2, m)$ values from the completed table as a point on your graph.
• Draw a line of best fit through these plotted points. Since the relationship between $m$ and $T^2$ for a spring-mass system is linear ($T^2 \propto m$), your line of best fit should be a straight line.