SIMPLE HARMONIC MOTION

Transcription

1 SIMPLE HARMONIC MOTION PURPOSE The purpose of this experiment is to investigate simple harmonic motion. We will determine the elastic spring constant of a spring first and then study small vertical oscillations of a mass suspended from the spring. THEORY This kind of motion includes pendulum motion, the oscillating circuit used to tune a radio receiver, and the bobbing motion of a car with worn shock absorbers. Less apparent but just as important are the interatomic vibrations of molecules, the motion of gas molecules propagating a sound wave, and the time behavior of electric and magnetic fields (including light) traveling through space. Probably the simplest form of simple harmonic motion is the oscillation of a mass suspended from a vertical spring. It is this form we will be studying in this lab. How can so many diverse systems exhibit essentially the same kind of motion? The answer lies in the fact that each of these systems obeys the same fundamental mathematical equation. Let's quickly review how this equation comes about in the case of a mass suspended from a spring. The springs you will be using obey Hooke's Law, F s = K (x x o) (1) Here, x o is the equilibrium position of the spring and the minus sign indicates that the force exerted by the spring is in the opposite direction to the displacement, (x - x o), of the mass. This equation can be taken as the defining relation for the spring constant K. The textbook shows how to apply Hooke's Law along with Newton's Second Law to the motion of a mass M on a frictionless surface and connected to a horizontal spring of force constant K. The analysis results in the differential equation for simple harmonic motion, viz, d 2 s/dt 2 + (K / m) s = 0 (2) where s = x - x o. Every physical system that exhibits simple harmonic motion obeys an equation of this form. If you are not familiar with differential equations, don t worry. We are interested in the solution. The most general solution to this equation can be written as s(t) = A cos(ωt + φ) (3) where the constants A and f are determined from the initial position and velocity of the mass M. The constant ω is the angular frequency and is given by ω= (K / m) 1/2 (4) The angular frequency provides the time period of the motion, V-1

2 T = 2 π (m / K) 1/2 (5) In our experiment we will be working with a vertical spring-mass system. The analysis of this system yields the exact same differential equation as in the horizontal case, however, now the equilibrium position is changed by the amount the spring stretches under the weight W = M g. Since the vertical spring-mass system obeys the same equation of motion as the horizontal one, the solutions of the two systems are identical. In particular, the time period of motion is again given by Equation 5. This period relation suggests a way to experimentally determine the spring constant K. Squaring both sides of Equation 5 gives T 2 = 4 π 2 (M / K) (6) By suspending various masses from the spring and measuring the corresponding periods we can construct a graph of T 2 vs. M. Equation (6) predicts that this relationship is linear with a slope of 4 π 2 (1 / K). Having constructed this graph we can determine the line of best fit through the data points (M i, T i 2 ) and measure the slope. The spring constant is then just K = 4 π 2 (1 / SLOPE) (7) Correction for the finite mass of the spring We must also consider the mass of the spring itself. The textbook derivation of the period relation, Equation 3, is based on a spring of negligible mass. In reality, springs have finite mass and therefore possess kinetic energy while in motion. We must thus consider the finite mass of the spring in the dynamics of our spring-mass system. Let ms be the mass of the spring. The inclusion of m s in the analysis results in a corrected mass which appears in the equations above. The true mass is the suspended mass, M, plus some fraction of the mass of the spring. That is, in all the equations above, we let M true = M + R m s (8) where R is an, as yet, undetermined fraction between 0 and 1. The true period is found by using M true in Equation 3 for the time period. The result is the corrected period given as T true = 2 π (M true / K) 1/2 (9) How large is R? A moment's thought should reveal that R = 1 is too large, for it implies that the entire mass of the spring contributes to the oscillations. Not all of the mass of the spring oscillates with the same amplitude. We can determine R experimentally however. V-2

3 EXPERIMENTAL PROCEDURE 1. Measure the mass, M', which is the mass of the weight holder plus a 200 g mass. (Note that) Suspend one of the springs and the mass M' vertically from the support. Allow it to come to rest and measure the distance from the floor to the bottom of the weight holder. Experimental notes: The choice of mass added to the weight holder depends on the spring. The numbers stamped on the masses are not accurate. Measure the mass on the arm balance each time you use one. Before measuring the distance between the floor and the bottom of the weight holder, it is useful to allow the spring to go through a cycle of extensions by adding additional masses in steps of 100 g each (the maximum load should not exceed the weight holder plus 700 g) and contractions by removing these masses in steps of 100 g each until the weight holder plus 200 g is suspended from the spring. Also allow sufficient time for the spring to come to equilibrium and measure the vertical distance of the bottom of the weight holder from the level floor. 2. Add a 100 gm mass to the mass M' and measure the distance from the floor to the bottom of the weight holder. Experimental notes: Allow the spring to come to equilibrium before you measure the distances; a few minutes should be sufficient. 3. Add three (3) additional masses, in steps of 100 g (measuring the masses in the arm balance each time) and in each case determine the distances as above. 4. Enter the data in Table Measure the mass M" (the mass of the weight holder plus 300 g) and suspend this mass from the spring. Set the system to oscillate vertically by pushing the mass up and releasing it gently. Care should be taken to keep the amplitudes small. 6. Measure the time for 10 complete oscillations. Repeat this at least three times and each time enter your data in Table 2. Determine the average time (T ave) and the Time period (T) for one oscillation. 7. Repeat step (6) for three additional masses of 100 g each added to mass M". The mass M in Table 2 is equal to mass of the weight holder plus the mass placed on it. Find the mass (m s) of the spring and enter its value in Table 2. V-3

4 DATA ANALYSIS 1. From your data in Table 1, compute the mass added to M' and the extension in the spring for each of these masses added and enter these values in Table Plot a graph (in Excel) of mass added to M' (vertical axis) in the spring vs. the extension (horizontal axis). Note that you should label the axes, and indicate the scale used in each of the axes with appropriate units such as cm, g, etc. 3. Add a trend-line to show the best fit to these data points. Determine the slope of this straight line. The slope is in units of g cm The spring constant K = (slope) (acceleration due to gravity). Remember to convert the slope units to kg m -1. Assume acceleration due to gravity = 9.8 m sec Using data from Table 2, plot T 2 vs. Mass suspended from the spring. Origin of co-ordinates must be (0,0). 6. The slope of T 2 (vertical axis) vs. mass suspended (horizontal axis) is in s 2 g -1. This value can be converted to s 2 kg -1 and the spring constant can be determined using the equation for the elastic spring constant: K = 4 π 2 (1 / SLOPE) (7) where the slope is given in s 2 kg Determine the intercept on the negative horizontal axis. This gives the contribution from the spring to the oscillating mass. Compare this to the mass (m s) of the spring. V-4

5 NAME Sec/Group Date DATA TABLE 1 ELASTIC CONSTANT, K, OF THE SPRING # Suspended Mass ( in g ) Distance Extension Mass (in g ) (added to M') (in cm ) ( in cm ) 1 M' = 0 L 0 = L 0 - L 0 = 0 2 M' + = L 1 = L 0 - L 1 = 3 M' + = L 2 = L 0 - L 2 = 4 M' + = L 3 = L 0 - L 3 = 5 M' + = L 4 = L 0 - L 4 = 6 M' + = L 5 = L 0 - L 5 = Mass of spring = g Acceleration due to gravity g = 9.8 m s -2 Slope of mass added (y-axis ) vs. extension ( x-axis ) = g cm -1 = kg m -1 Elastic spring constant K = ( slope in kg m -1 ) ( acceleration due to gravity in m s -2 ) = N m -1 V-5

6 TABLE 2 OSCILLATIONS OF MASS FROM THE SPRING # Mass Time for 10 oscillations Period* M (g) T 1 (s) T 2 (s) T 3 (s) T (s) T 2 (s 2 ) 1. M" = 2. M" + = 3. M" + = 4. M" + = 5. M" + = *be sure to divide by 10. Slope of T 2 (vertical axis) vs. mass suspended (horizontal axis) = s 2 g -1 = s 2 kg -1 Elastic spring constant K = 4 π 2 / (slope in s 2 kg -1 ) = N m -1 Intercept on the negative x-axis = g Ratio of intercept to mass of spring, R = g V-6