# 11 - 3 Experiment 11 Simple Harmonic Motio.docx

1 de Dec de 2022
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### 11 - 3 Experiment 11 Simple Harmonic Motio.docx

• 1. 11 - 3 Experiment 11 Simple Harmonic Motion Questions How are swinging pendulums and masses on springs related? Why are these types of problems so important in Physics? What is a spring’s force constant and how can you measure it? What is linear regression? How do you use graphs to ascertain physical meaning from equations? Again, how do you compare two numbers, which have errors? Note: This week all students must write a very brief lab report during the lab period. It is due at the end of the period. The explanation of the equations used, the introduction and the conclusion are not necessary this week. The discussion section can be as little as three sentences commenting on whether the two measurements of the spring constant are equivalent given the propagated errors. This mini-lab report will be graded out of 50 points Concept
• 2. When an object (of mass m) is suspended from the end of a spring, the spring will stretch a distance x and the mass will come to equilibrium when the tension F in the spring balances the weight of the body, when F = - kx = mg. This is known as Hooke's Law. k is the force constant of the spring, and its units are Newtons / meter. This is the basis for Part 1. In Part 2 the object hanging from the spring is allowed to oscillate after being displaced down from its equilibrium position a distance -x. In this situation, Newton's Second Law gives for the acceleration of the mass: Fnet = m a or The force of gravity can be omitted from this analysis because it only serves to move the equilibrium position and doesn’t affect the oscillations. Acceleration is the second time- derivative of x, so this last equation is a differential equation. To solve: we make an educated guess: Here A and w are constants yet to be determined. At t = 0 this solution gives x(t=0) = A, which indicates that A is the initial distance the spring stretches before it oscillates. If friction is negligible, the mass will continue to oscillate with amplitude A. Now, does this guess actually solve the (differential) equation? A second time-derivative
• 3. gives: Comparing this equation to the original differential equation, the correct solution was chosen if w2 = k / m. To understand w, consider the first derivative of the solution: −kx = ma a = − k m ⎛ ⎝ ⎜⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟⎟ x d 2x dt 2 = − k m x x(t) = A cos(ωt)
• 4. d 2x(t) dt 2 = −Aω2 cos(ωt) = −ω2x(t) James Gering Florida Institute of Technology 11 - 4 Integrating gives We assume the object completes one oscillation in a certain period of time, T. This helps set the limits of integration. Initially, we pull the object a distance A from equilibrium and release it. So at t = 0 and x = A. (one-quarter of the total period later) the object passes the equilibrium (x=0) position. This integration yields: Canceling the A’s and w’s and evaluating these limits gives: or, and adding 1 to both sides gives cos [ (wT / 4) ] = 0
• 5. The cosine is zero only when its argument is p/2 radians. Hence this last equation implies that and rearranging gives Finally, it is clear that w is indeed the angular frequency of the object as it oscillates up and down. Earlier it was found that w2 = k/m. Putting these two relations for w together yields: or This last equation is only valid if the mass of the spring is negligible compared to m. In the case of a massive spring, the actual mass that oscillates includes a portion of the mass of the spring - the upper part of the spring is stretched by the lower part. This suggests that an “effective mass” meff should be added to m to give: or after squaring both sides Writing this last equation this way emphasizes the equation is of the form of y = Ax + B, if y is taken as T2 and x is taken as the mass m. Therefore, a graph of T2 versus mass will yield a straight line with a physically meaningful slope and y-intercept.
• 6. v(t) = dx(t) dt = −Aωsin(ωt) dx = Aω sin(ωt)dt t=0 t=T /4 ∫x =A x =0 ∫ −A = Aω cos(ωt) ω ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 0
• 7. T /4 −1 = cos( ωT 4 )−cos(0) −1 = [cos ωT 4 ⎛ ⎝ ⎜⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟⎟ −1] ωT 4 = π 2 ω =
• 8. 2π T 2π T ⎛ ⎝ ⎜⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟⎟ 2 = k m T = 2π m k ⎛ ⎝ ⎜⎜⎜⎜ ⎞
• 9. ⎠ ⎟⎟⎟⎟ 1/2 T = 2π m + m eff k ⎛ ⎝ ⎜⎜⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟⎟⎟ 1/2 T 2 = 4π2 k m + 4π2
• 10. k m eff James Gering Florida Institute of Technology 11 - 5 Procedure Part 0 Evaluations and Assessment 1) Reminder: After this experiment, find some time to use Canvas to fill out the end-of- semester course evaluation. 2) (Note: This task may be converted into a Canvas quiz or it may be something entirely different. As of this writing, things are in flux.) Instructors will reserve 35-40 minutes to administer a diagnostic exam, which is part of the Department’s year-to-year assessment of the Physics 1 lecture class (PHY 1001). Please take the time to answer the 30 questions as best you can. This exam is only for students who are taking PHY 1001 concurrently (and for the first time) with this laboratory course.