(1) In an experiment of a spring oscillator as in Fig. 1, the mass is measured to be m = 0.435 kg, and the displacement as a function of time is measured to be x(t) = 0.313 sin(2.87t + 0.04) with all numbers in SI units. Please find the quantities below. 1. Oscillation amplitude A 4. Maximum elastic potential energy Umax 2. Oscillation period T 5. Maximum speed vmax 3. Spring constant k 6. Total mechanical energy U(t) + K(t) (2) Please discuss how to read out the spring constant it from the graph of data obtained in the two experiments below. (You will compare both methods in the lab.) i. Student A varies force F exerted on a spring and measures its extension Ax, and plots the best-fit line of F vs Ax. [Hint: consider Eq. (1).] ii. Student B varies mass m attached on a spring and measures the angular frequency w of its simple harmonic oscillation, and plots the best-fit line of log w vs log m. [Hint: consider Eq. (4).] (3) From the previous question, Student B records the experimental data in the table below. m (kg) 0.221+0.0005 0.271+0.0005 0.321+0.0005 0.371+0.0005 0.421+0.0005 w (rad/s) 8.44+0.03 7.48+0.02 6.87+0.02 6.53+0.05 5.98+0.03 i. Use EXCEL to find the best-fit line of (log10 w) vs (log10 m). Record the slope, y-intercept, and R2 in Pre-lab. ii. From the best-fit line, find the spring constant k. (4) In an experiment on a damped spring oscillator with spring constant k = 15.3 N/m, Student C obtains the displacement vs time curves as in Fig. 2(a), and records the maximum dis- placement data points as below. i = 1 2 3 4 5 xi (m) 0.0609 0.0488 0.0387 0.0299 0.0212 ti (s ) 1.22 2.30 3.35 4.43 5.48 i. Use Eq. (9) to calculate the total energy E; and its natural logarithm In E; from the data. ii. Use EXCEL to find the best-fit line of In E vs. t. Record the slope, y-intercept, and R in Pre-lab. iii. From the best-fit line, find the characteristic dissipation rate 1.scribes a restoring force that is proportional to the displacement, and the mass undergoes endless periodic motion at a single fre- quency. We first look at an oscillating spring with a mass m at its end, as shown in Figure 1. Hooke's law says that the elastic force acting on the mass is Figure 1: a spring oscillator. if = 4:2, (1) where k is the spring constant, and 35 is the displacement of the spring (how far it is stretched or compressed by the mass). The negative sign physically means that the restoring force acts al- ways in the opposite direction of the displacement. Considering the motion in 1D and bringing in Newton's second law, we can write, F = ma = kx => (1 = (k/m)x. (2) Because a moving objects displacement is a function of time, x = x(t), Eq. (2) tells that the force and acceleration are also a function of time, F = F(t) and a = a(t). Thus, the kinematic equations we have learned for a constant acceleration do not apply here. In fact, the simple harmonic motion has a sinusoidal form x(t) = A sin(wt). (3) Here A is the amplitude of oscillation and a) is the angular frequency given by a) = k/m. (4) From the equation of motion of Eq. (3), one can derive the velocity and acceleration as v(t) = 00A C05(wt), and a(t) = a)2A sin(wt). (5) The sinusoidal form tells that the oscillation has a period T = ZH/w, which is the time the oscil- lator takes for one full forth-and-back motion. The equation of motion can be generally de- scribed by x = A sin(wt + (3')), with a \"phase" ql) associated with the initial condition at t = 0, but we do not really need to consider it in this lab. There are two types of mechanical energy involved in an oscillating spring: the elastic poten- tial energy U(t) and kinetic energy KG), given by U(t) = %k[x(t)]2, and KR) = %m[v(t)]2. (6) The elastic potential energy and the kinetic energy convert to each other periodically during the motion, while the total mechanical energy is conserved. Of course, real world oscillators are not perfect and do not continue forever. External forces like friction always eventually bring them to a stop. This will be particularly true for the spring oscillator in this lab. Let's see how to describe such a realistic system in the next section. 3.2 Damped oscillations Damped Oscillations There is always some friction in the real world. The spring oscillations in our experiments will eventually die out due to the friction from the environment. 0 .... E to E (U U 2 a. .2 D The displacement of oscillations exhibits damped amplitude over time, as shown in Fig. 2(a). The total mechanical energyE of a damped oscillator is no longer conserved but exponentially decreases with Energy time t, as shown in Fig. 2(b), described by E(t) = BeeM. (7) ' _ ' Time Here E0 '5 the energy at t _ 0' and '1 > 0 IS the Figure 2: (a)displacement vs time andlb) total characteristic dissipation rate (in units of 1/5), which mechanical energy ustime of damped spring describes how fast the total mechanical energy dis- OSdIIations-The pom\" dam-\"e ma'dmum dis' _ _ . placements and the corresponding energies. Slpates through friction. In the lab, we are going to find the characteristic dissipation rate /l with regression analysis. Equation (7) shows an exponential relationship (it is the first time we study such relationship), so from Table 1 in Lab 2 Manual, we can take natural logarithm on both sides of Eq. (7) and de- rive [ME = (1) \"+1115, (8) Equation (8) tells that if we find the linear fit line of [HE v5 tgraph, then the negative slope is the characteristic dissipation rate it. In the experiment, we record data points for maximum dis- placements of oscillations over time, as {(t1,x1), (t2, x2), ...} in Fig. 2(a). Since the oscillator has zero speed at a maximum displacement, the total mechanical energy is the elastic potential en- ergy only, or 1 E,- = Ekxf'. (9) We use Eq. (9) to calculate the energy data points {01.51) (2,132), ...}, as in Fig. 2(b), and then perform the regression analysis to find the characteristic dissipation rate. (1) Compute the quantities below. (Express answers to 3 sig. figs.) [6] 1. Oscillation amplitudeA = m 2. Oscillation period T = s 3. Spring constant k = N/m 4. Maximum elastic potential energy Umax = 5. Maximum speed vmax = m/s 6. Total mechanical energy U(t) + K(t) = J (2) Please discuss how to read out the spring constant k. i. From the bestfit line of F vs Ax. [4] Answer: ii. From the best-fit line of log w vs log m. [4] Answer: (3) Analyze the best-fit line. (Express answers to 3 sig. figs. whenever applicable) i. (loglo W) =. X (log10 m) + ; R4 = [6] ii. Spring constant k = N/m [1](4) Analyze the damped oscillator. i. Calculate the total energy E and In El- from the data (remember to use the natural logarithm, which is the function LN in Excel; express answers to 3 sig. figs. whenever applicable) [5]2 ii. Find the bestfit line of In E vs. t. (Express answers to 3 sig. figs.) lnE= >