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ScanSolve AI Answer
Here are the calculations for the period T and its square T^2, followed by instructions for plotting the graph. Step 1: Calculate T and T^2 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 = (6.82)/(10) = 0.682 s T^2 = (0.682)^2 = 0.465124 ≈ 0.465 s^2 For m = 200.0 g, t = 7.81 s: T = (7.81)/(10) = 0.781 s T^2 = (0.781)^2 = 0.609961 ≈ 0.610 s^2 For m = 250.0 g, t = 8.70 s: T = (8.70)/(10) = 0.870 s T^2 = (0.870)^2 = 0.7569 ≈ 0.757 s^2 For m = 300.0 g, t = 9.62 s: T = (9.62)/(10) = 0.962 s T^2 = (0.962)^2 = 0.925444 ≈ 0.925 s^2 For m = 350.0 g, t = 10.31 s: T = (10.31)/(10) = 1.031 s T^2 = (1.031)^2 = 1.062961 ≈ 1.063 s^2 Step 2: Complete the Table 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 | Step 3: Plot the Graph 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 m), your line of best fit should be a straight line. That's 2 down. 3 left today — send the next one.