Copy of Module Eight Lesson One Assignment Two Lab - Word Sign in X File Home Insert Draw Design Layout References Mailings Review View Help ? Tell me what you want to do Share Breaks Indent Spacing Align Line Numbers Left: 0" I Before: 0 pt LEi Group - Margins Orientation Size Columns Position Wrap Bring Send Selection be Hyphenation Er Right: 0" After: 10 pt Text * Forward * Backward Pane Rotate Page Setup Paragraph Arrange Simple Harmonic Motion Lab Table of I of different masses Part A Period (1) T Simple Harmonic Motion (SHIM) is a periodic motion that occurs in the presence of a 0.130 0.71 linear restoring force (Atanackovic & Guram, 2013). Am oscillating mass attached to the end 0.20 0.818 0.67 Simple Harmonic Motion Lab of a spring exhibita simple harmonic motion. The period of the oscillation depends on the 0.250 0.83 Aarush Garji spring constant (k) and the mass (m). 0.300 0.943 Period (T)= 2nv- 0.350 0.1058 T.12 Where m is the mass suspended, and k is the spring constant. in this experiment, two springs were hooked to a cart and loaded with mass progressively starting from 150g. The springs were set in oscillations and time taken for 10 Spring Constant y = 0.05x + 01 oscillations tabulated for each mass. The period (T) was then calculated by dividing the total time by 10. The results were as recorded below. Table ] LO. . . 1 . . .9 . . . 1 . . . 8 . . . 1 . . . 7 . . . 1 . . . 6 . . . 1 . . . 5 . . . 1. .4 . . 1. . . 3 . . 1. . 2 01 01. . Table of Period (T) of different masses Total Mass (Kog) |Total Time (1) Period - 0. 05 0.51 1.12 0.130 0.712 0.200 From the graph, the alope is y = 0.05x + 0.1 0.230 Slope = AB 0.300 0 993 0.350 10.58 0. 1053 Slope = 1.12 -0.51 -0.35 - 0.15 0.61 To calculate the spring constant (k), we must plot a graph of period squared (T-) Slope = 02 against mass (m) in kilograms. The slope of the graph would give us the spring constant (1). Slope = 3.05 /kg Table What is your value of k? Page 1 of 8 622 words + 53% Unit 8 Pri - Googl. https://e2020.ge.. Ask a Question - . how to split micr. Sticky Notes w Copy of Module .. 4:41 PM 4/27/2022Copy of Module Eight Lesson One Assignment Two Lab - Word Sign in X File Home Insert Draw Design Layout References Mailings Review View Help ? Tell me what you want to do Share Breaks Indent Spacing Align Line Numbers Left: 0" Before: 0 pt Lei Group - Margins Orientation Size Columns Position Wrap Bring Send Selection be Hyphenation Er Right: 0" After: 10 pt Text * Forward * Backward Pane Rotate Page Setup Paragraph G Arrange I T = 201- x is the displacement in centimeters, and k is the constant of the spring. When each weight is Force Against Displacement For Spring 2 added to the spring, the downward pull balances with the upward pull of the string. The 1.5 K = 4 y= 22.087x+0 1322 equilibrium achieved is expressed as 2.5 But Slope = -, therefore, k = _tar W = kx E 1 Stope Wis the weight of the mess. The constant of the spring k can be calculated by plotting W 1.5 k - 4 x3.1423 3.05 against x on a graph. Weight is calculated by multiplying mass by the acceleration of gravity. k = 12.93Nmi W=mg 0.0G D. D 01 Displacement () From the experiment, we can note that the period increases with the weight applied The Where g is 9.8m/vec". The unit of #is given in Newtons (Atanackovic & Guram, 2013). The pring's constant is also directly proportional to the period graph is plotted by finding the best straight line fit To get the spring constant, we must Part B calculate the slope of the line. The slope of spring 1 is calculated by Table 3 A1B1 Slope = BICI Table of Force and Displacement on Spring I and Spring 2 Force Against Displacement For Spring 1 Slope = 2.5 - 0.5 Spring ! Spring 2 0.12 - 0.02 MEEE (@) Force (N) | x (@m) 6es @ Force (N) x (cm) 1.5 y = 19.958x+ 0.1299 2.0 Slope = 1,10 2.5 Slope = 20 N/m 0.030 0.49 1.6 30 0.49 1.6 Force IN 1.5 The slope of the second spring 100 3.0 4282 0.150 6.6 130 Slope = 2107 1.47 1.47 3.8 0.5 0.200 1.97 92 200 1.95 Slope = 2.75 -1 Displacement ( 0.12 - 0.04 0.250 245 TI.6 2:50 245 1.75 0.300 204 143 300 124 Slope = Slope = 21.875N/m According to Hooke's Law, a spring's restoring force is directly proportional to a small From the experiment, the length of displacement increases with weight. If the force applied displacement (Atamackovic & Guran, 2013) exceeds the elastic limit of the string, it will change its shape and retain it. This property of F = -Kx Page 4 of 8 622 words 53% Unit 8 Prj - Googl.. https://e2020.ge... Ask a Question - .. how to split micr.. Sticky Notes w Copy of Module .. 4:41 PM 4/27/2022