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

Step 1: Determine the slope, s_1, 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. s_1 = (m_2 - m_1)/(T^2_2 - T^2_1) Using the example data from the previous response (e.g., points (1.44 s^2, 150 g) and (3.35 s^2, 350 g)): s_1 = 350 g - 150 g3.35 s^2 - 1.44 s^2 = 200 g1.91 s^2 = 104.7 g/s^2 Your actual value for s_1 will depend on your experimental data and graph. Step 2: Plot m on the vertical axis and T_0^2 on the horizontal axis (xviii). 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. Step 3: Determine the slope, s_2, of the graph (xix). 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. s_2 = (m_2 - m_1)/(T_0,2)^2 - T_0,1^2 Using the example data from the previous response (e.g., points (0.72 s^2, 150 g) and (1.69 s^2, 350 g)): s_2 = 350 g - 150 g1.69 s^2 - 0.72 s^2 = 200 g0.97 s^2 = 206.2 g/s^2 Your actual value for s_2 will depend on your experimental data and graph. Step 4: Evaluate = (s_2)/(s_1) (xx). Using the example slopes calculated above: = 206.2 g/s^2104.7 g/s^2 = 1.97 Your actual value for will depend on your calculated slopes. Step 5: State two precautions taken to ensure accurate results (xxi). • 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. Step 6: How does change in the number of springs affect the frequency of the system? 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). Step 7: Calculate the extension, e, of the spring. 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). F = ke Given: Force F = 8.0 N Stiffness k = 125 Nm^-1 Rearrange the formula to solve for extension e: e = (F)/(k) Substitute the given values: e = 8.0 N125 Nm^-1 e = 0.064 m The extension of the spring is 0.064 m. That's 2 down. 3 left today — send the next one.