Oscillations of a Glider Moving - Lab Report Example

Insert Contents Insert List of tables and figur 2 Introduction 3 Objectives 3 Equipment 3 Background information 3 Damped oscillations 6 Procedure 8 Simple harmonic 8 Damped harmonics 8 Results and analysis 8 Conclusion and recommendation 20 List of tables and figur Figure 1 displacement 3 Figure 2 damping 7 Figure 3Run #1 11 Figure 4 Run #2 13 Figure 5 Run #3 14 Figure 6 Run #4 15 Figure 7 Run #5 16 Figure 8b versus # of magnets 19 Figure 9 Harmonics 20 Table 1 run #1 9 Table 2 Run #2 9 Table 3 Run #3 10 Table 4 Run #4 11 Table 5 Run #5 11 Table 6 b versus # of magnets 19 Introduction Objectives Laboratory activity was designed to be used in the study of oscillations of a glider moving on an air track being held by springs on each side. In so doing, the understanding of oscillation characteristics in terms of period, frequency, amplitude and phase were to be evoked. The data collected was to be analyzed fro the varying sessions with application of damping forces. Equipment Equipment utilized fro the laboratory session included air track with glider, two 7.5cm long springs, computer, PASCO PASPORT USB Link, motion sensors and six spring magnets. Background information When forces that are acting on an object are equal, the resultant effect is said to be at equilibrium; they add up to zero and has no acceleration. It is at rest for an indefinite time. The experiment made use of a glider placed in an air track and suspended on two springs. Figure 1 displacement The setup is at rest and remains at rest because the springs subject an equal force on the glider; they cancel each other; it is at a stable equilibrium position. The force applied on the setup in particular direction results a net force that acts on the glider and tries to return to the equilibrium state. The combined forces pull in a manner that the resultant force takes It to the original position; determined as restoring force. When the glider is brought to the original position, it is still under the influence of velocity and therefore goes beyond the default position by overshoot moving to the opposite direction from the earlier force. The overshoot results in the whole drag to equilibrium starting all over again. The back and forth motion is referred to as an oscillation. Mathematically known as sine and cosine function; simple harmonic motion. To analyze the problem let x be an instantaneous position of the glider while x0 is the equilibrium position. When the glider moves from the equilibrium position, the restoring force is F. without applying too much force resultant forces are linear hence force equation is; I It implies that when the glider moves away from the equilibrium, restoring force becomes bigger. F is proportional to x. k is the spring constant. The sign (-) represents the restoration character of the force in the setup. A positive different in (x-x0) indicates that the force is negative x direction. A negative (x-x0) implies a positive force direction. Given a displacement initially A = x-x0, F = -kA, and accelerates towards the negative x-axis towards equilibrium; x = x0 where the restoring force is zero. It passes beyond x0 to the opposite direction and arrives with x-x0 = -A at zero velocity and it is pulled back. This forth and back motion goes on and results to oscillations. Without friction, kinetic energy and potential energy remain the same. Applying Newton’s 2nd law, II III The equation has harmonic oscillator calculated by; IV Where; A, w and α are constants A is amplitude W is the angular frequency Α is phase angle The motion of the spring lies in between A and –A which are reached when (n=0, +- 1,…) V VI Equation (iii) is satisfied by VII The phase α is determined by the specific instant during the oscillating motion from the time timing begins. Displacement at point A, releasing it at t = t0, at this moment x = A and the equation (iv) = 1. It is due to where n = 0, hence VIII T (period) is the time taken to accomplish a cycle. From equation (iv) it can be deduced that during the cycle time T, the cosine goes through 3600. , hence IX Frequency f of the cycles is determined from X Damped oscillations Friction is present is all experiments. Study of oscillation damping can be analyzed using magnets. They generate eddy currents producing a damping force proportional to velocity XI Where b is the damping constant and hence equation (ii) becomes XII In the case of small damping XIII A is connected to b by XIV Is known as the decay constant XV When b = 0 old solution is back = non-zero, amplitude decreases exponentially with time in a motion known as damping motion. Figure 2 damping Procedure Simple harmonic The mass of the glider was measured and recorded. With glider in the air track, air source was turned on and adjusted screw till glider was in place. PASPort USB link was connected to computer ensuring the motion sensor was in place. Launched PASCO Capstone and selected table and Graph. Glider was slid towards rod and motion sensor ensured was fastened into place. Then preceded by rotating motion sensor about horizontal and vertical axis till it was parallel with reflector. Set sample rate at 20 hz. Glider was attached to the spring and the displacement applied up to a value of 40cm. Recording of data output was done. Damped harmonics Weight of the mass and color of ring magnets was taken. Lunched PASCO Capstone. Placed magnets on each leg of the glider. Taking a set of positions versus of the attached magnets, the experiment was run by displacing the glider and clicking on run. Measured two magnets and added them to glider and noting the position, the run #3 was performed. Repeated the procedure the third, fourth, and fifth time. Results and analysis Investigation 1 Mass measured = 375.5grams Investigation 2 Run #1 equilibrium position Run #2 Masses measured = 375.5 + 2 yellow magnets (31.4g) Run #3 masses measured = 375.5 + 2 yellow magnets (31.4g) + 2 green magnets (31.4g) Run #4 masses measured = 375.5 + 2 yellow magnets (31.4g) + 2green magnets (31.4g) + 2 orange magnets (33.2g) Run#1 SAMPLED DATA Table 1 run #1 Run #1 Position (cm) Position corrected Time (s) 63.677 -0.2854 0 63.607 -0.3554 0.05 63.572 -0.3904 0.1 63.537 -0.4254 0.15 63.521 -0.4414 0.2 63.485 -0.4774 0.25 63.485 -0.4774 0.3 63.485 -0.4774 0.35 Run #2 sampled data Table 2 Run #2 Run #2 Position (cm) Position corrected Time (s) 80.71 16.7476 0 82.752 18.7896 0.05 84.487 20.5246 0.1 85.894 21.9316 0.15 86.995 23.0326 0.2 87.735 23.7726 0.25 88.134 24.1716 0.3 88.136 24.1736 0.35 87.756 23.7936 0.4 87.07 23.1076 0.45 Run #3 sampled data Table 3 Run #3 Run #3 Position (cm) Position Corrected Time (s) 37.995 -25.9674 0 37.978 -25.9844 0.05 37.977 -25.9854 0.1 37.994 -25.9684 0.15 38.011 -25.9514 0.2 38.05 -25.9124 0.25 37.885 -26.0774 0.3 38.186 -25.7764 0.35 38.845 -25.1174 0.4 40.034 -23.9284 0.45 41.104 -22.8584 0.5 Table 3 Run #3 Run #4 sample data Table 4 Run #4 Run #4 Auto Position (cm) Position Corrected Time (s) 38.184 -25.7784 0 38.184 -25.7784 0.05 38.166 -25.7964 0.1 38.189 -25.7734 0.15 37.96 -26.0024 0.2 37.957 -26.0054 0.25 38.104 -25.8584 0.3 37.819 -26.1434 0.35 38 -25.9624 0.4 38.523 -25.4394 0.45 Run #5 sample data Table 5 Run #5 Run #5 Auto Position (cm) Position Corrected Time (s) 37.507 -26.4554 0 37.793 -26.1694 0.05 38.371 -25.5914 0.1 39.225 -24.7374 0.15 40.358 -23.6044 0.2 41.51 -22.4524 0.25 43.39 -20.5724 0.3 44.97 -18.9924 0.35 47.256 -16.7064 0.4 49.469 -14.4934 0.45 51.801 -12.1614 0.5 54.255 -9.7074 0.55 56.778 -7.1844 0.6 Figure 3Run #1 Run #1 average X0 = 63.96 Mass m = 375.5gramms = 3.68N K = 1.1N/m = = 0.546 Then, (phase angle) α = -wtp tp= 1.55 (from the result table) Hence, = - 0.8463 = (0.8463*180/π) = 48.40 Figure 4 Run #2 Run #2 X0 = 64.00 tp = 0.35 mass m = (375.5 + 31.4) = 406.9g = 3.99N W = =0.524 α = -wtp = - 0.524*0.35 = 0.1834 = (0.1834*180/π) =10.50 Figure 5 Run #3 Run #3 X0 = 63.67 tp = 1.66 mass m = (375.5 + 31.4 + 31.2) = 438.1g = 4.3N W = =0.506 α = -wtp = - 0.506*1.66 = 0.83996rads = (0.8399*180/π) =48.10 Figure 6 Run #4 Run #4 X0 = 62.8 tp= 1.75 mass m = (375.5 + 31.4 + 31.2 + 33.2) = 471.3g = 4.62N W = =0.487 α = -wtp = - 0.487*1.75 = 0.8522rads =(0.8522*180/π) =48.80 Figure 7 Run #5 Run #5 X0 = 63.95 tp= 1.4 mass m = (375.5 + 31.4 + 31.4 + 31.4 + 33.2 + 33.2) = 504.3g = 4.95N W = =0.471 α = -wtp = - 0.471*1.4rads = (0.6594*180/π) = 37.780 Plotting b versus no. of magnets Zero magnets α = 0.843 b = (0.843 * 2*3.68) b = 6.21 2 magnets =0.1834 b = (0.183*2*3.99) =1.42 4 magnets =0.83 b = (0.83*2*4.3) b= 7.14 6 magnets =0.65 =(0.65*2*4.6) =5.9 Table 6 b versus # of magnets Number of magnets Value of b 0 6.21 2 1.42 4 7.14 6 5.9 8 6.4 Figure 8b versus # of magnets Figure 9 Harmonics Conclusion and recommendation Experiment objectives were met with a higher degree of accuracy. It was determined that the value of k kept on changing with the addition of magnets. The increase was not systematic because of measurement errors. It resulted from timing issues of releasing the spring from displacement and start of recording. Harmonics was witnessed in the data analyzed as is shown from the graphs above. The experiment equipped me with skills of measuring oscillations on a computer terminal using a motion sensor, a set of springs and magnet pairs. To increase accuracy, timing should have an automatic switch that starts recording the same time the displaced spring are released to start oscillating. Accuracy in the measurement of spring constant is key in order to determine and project the accuracy of results. Work cited Hansen, Mark. "Engineering Systems." 2000. Modern systems. 17 August 2014. Appendix Read More