Periodic Motion is everywhere
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1 Lecture 19 Goals: Chapter 14 Interrelate the physics and atheatics of oscillations. Draw and interpret oscillatory graphs. Learn the concepts of phase and phase constant. Understand and use energy conservation in oscillatory systes. Understand the basic ideas of daping and resonance. Phase Contrast Microscope Epithelial cell in brightfield (BF) using a 40x lens (NA 0.75) (left) and with phase contrast using a DL Plan Achroat 40x (NA 0.65) (right). A green interference filter is used for both iages. Physics 207: Lecture 19, Pg 1 Periodic Motion is everywhere Exaples of periodic otion Earth around the sun Elastic ball bouncing up an down Quartz crystal in your watch, coputer cloc, ipod cloc, etc. Physics 207: Lecture 19, Pg 2 Page 1
2 Periodic Motion is everywhere Exaples of periodic otion Heart beat In taing your pulse, you count 70.0 heartbeats in 1 in. What is the period, in seconds, of your heart's oscillations? Period is the tie for one oscillation T= 60 sec/ 70.0 = 0.86 s What is the frequency? f = 1 / T = 1.17 Hz Physics 207: Lecture 19, Pg 3 A special ind of periodic oscillator: Haronic oscillator What do all haronic oscillators have in coon? 1. A position of equilibriu 2. A restoring force, which ust be linear [Hooe s law spring F = - x (In a pendulu the behavior only linear for sall angles: sin where = s / L) ] In this liit we have: F = -s with = g/l) 3. Inertia 4. The drag forces are reasonably sall Physics 207: Lecture 19, Pg 4 Page 2
3 Siple Haronic Motion (SHM) In Siple Haronic Motion the restoring force on the ass is linear, that is, exactly proportional to the displaceent of the ass fro rest position Hooe s Law : F = -x If >> rapid oscillations <=> large frequency If << slow oscillations <=> low frequency Physics 207: Lecture 19, Pg 5 Siple Haronic Motion (SHM) We now that if we stretch a spring with a ass on the end and let it go the ass will, if there is no friction,.do soething 1. Pull bloc to the right until x = A 2. After the bloc is released fro x = A, it will A: reain at rest B: ove to the left until it reaches equilibriu and stop there C: ove to the left until it reaches x = -A and stop there D: ove to the left until it reaches x = -A and then begin to ove to the right -A 0( X eq ) A Physics 207: Lecture 19, Pg 6 Page 3
4 Siple Haronic Motion (SHM) We now that if we stretch a spring with a ass on the end and let it go the ass will. 1. Pull bloc to the right until x = A 2. After the bloc is released fro x = A, it will A: reain at rest B: ove to the left until it reaches equilibriu and stop there C: ove to the left until it reaches x = -A and stop there D: ove to the left until it reaches x = -A and then begin to ove to the right This oscillation is called Siple Haronic Motion -A 0( X eq ) A Physics 207: Lecture 19, Pg 7 Siple Haronic Motion (SHM) The tie it taes the bloc to coplete one cycle is called the period. Usually, the period is denoted T and is easured in seconds. The frequency, denoted f, is the nuber of cycles that are copleted per unit of tie: f = 1 / T. In SI units, f is easured in inverse seconds, or hertz (Hz). If the period is doubled, the frequency is A. unchanged B. doubled C. halved Physics 207: Lecture 19, Pg 8 Page 4
5 Siple Haronic Motion (SHM) The tie it taes the bloc to coplete one cycle is called the period. Usually, the period is denoted T and is easured in seconds. The frequency, denoted f, is the nuber of cycles that are copleted per unit of tie: f = 1 / T. In SI units, f is easured in inverse seconds, or hertz (Hz). If the period is doubled, the frequency is A. unchanged B. doubled C. halved Physics 207: Lecture 19, Pg 9 Siple Haronic Motion (SHM) An oscillating object taes 0.10 s to coplete one cycle; that is, its period is 0.10 s. What is its frequency f? Express your answer in hertz. f = 1/ T = 10 Hz Physics 207: Lecture 19, Pg 10 Page 5
6 Siple Haronic Motion Note in the (x,t) graph that the vertical axis represents the x coordinate of the oscillating object, and the horizontal axis represents tie. Which points on the x axis are located a displaceent A fro the equilibriu position? A. R only B. Q only C. both R and Q Position tie Physics 207: Lecture 19, Pg 11 Suppose that the period is T. Siple Haronic Motion Which of the following points on the t axis are separated by the tie interval T? A. K and L B. K and M C. K and P D. L and N E. M and P tie Physics 207: Lecture 19, Pg 12 Page 6
7 Siple Haronic Motion Now assue that the t coordinate of point K is s. What is the period T, in seconds? How uch tie t does the bloc tae to travel fro the point of axiu displaceent to the opposite point of axiu displaceent? tie Physics 207: Lecture 19, Pg 13 Siple Haronic Motion Now assue that the t coordinate of point K is s. What is the period T, in seconds? T = s How uch tie t does the bloc tae to travel fro the point of axiu displaceent to the opposite point of axiu displaceent? t = s tie Physics 207: Lecture 19, Pg 14 Page 7
8 Siple Haronic Motion Now assue that the x coordinate of point R is What total distance d does the object cover during one period of oscillation? d = 0.48 What distance d does the object cover between the oents labeled K and N on the graph? d = 0.36 tie Physics 207: Lecture 19, Pg 15 SHM Dynaics: Newton s Laws still apply At any given instant we now that F = a ust be true. But in this case F = - x and a = d 2 x 2 dt So: - x = a = 2 d x 2 dt a x F = - x 2 d x dt 2 x = a differential equation for x(t)! Siple approach, guess a solution and see if it wors! Physics 207: Lecture 19, Pg 16 Page 8
9 SHM Solution... Try either cos ( ω t ) or sin ( ω t ) Below is a drawing of A cos ( ω t ) where A = aplitude of oscillation T = 2π/ω A π π 2π A [with ω = (/) ½ and ω = 2π f = 2π /T ] Both sin and cosine wor so need to include both Physics 207: Lecture 19, Pg 17 x(t) = Cobining sin and cosine solutions B cos ωt + C sin ωt = A cos ( ωt + φ) = A (cos ωt cos φ sin ωt sin φ ) =A cos φ cos ωt A sin φ sin ωt) Notice that B = A cos φ C = -A sin φ tan φ= -C/B π θ 2π cos sin 0 x Use initial conditions to deterine phase φ! Physics 207: Lecture 19, Pg 18 Page 9
10 Energy of the Spring-Mass Syste We now enough to discuss the echanical energy of the oscillating ass on a spring. x(t) = A cos ( ωt + φ) If x(t) is displaceent fro equilibriu, then potential energy is v(t) = dx/dt U(t) = ½ x(t) 2 = ½ A 2 cos 2 ( ωt + φ) v(t) = A ω (-sin ( ωt + φ)) And so the inetic energy is just ½ v(t) 2 K(t) = ½ v(t) 2 = ½ (Aω) 2 sin 2 ( ωt + φ) Finally, a(t) = dv/dt = -ω 2 A cos(ωt + φ) Physics 207: Lecture 19, Pg 19 Energy of the Spring-Mass Syste x(t) = A cos( ωt + φ ) v(t) = -ωa sin( ωt + φ ) a(t) = -ω 2 A cos( ωt + φ ) Kinetic energy is always K = ½ v 2 = ½ (ωa) 2 sin 2 (ωt+φ) Potential energy of a spring is, And ω 2 = / or = ω 2 U = ½ x 2 = ½ A 2 cos 2 (ωt + φ) U = ½ ω 2 A 2 cos 2 (ωt + φ) Physics 207: Lecture 19, Pg 20 Page 10
11 Energy of the Spring-Mass Syste x(t) = A cos( ωt + φ ) v(t) = -ωa sin( ωt + φ ) a(t) = -ω 2 A cos( ωt + φ ) And the echanical energy is K + U =½ ω 2 A 2 cos 2 (ωt + φ) + ½ ω 2 A 2 sin 2 (ωt + φ) K + U = ½ ω 2 A 2 [cos 2 (ωt + φ) + sin 2 (ωt + φ)] K + U = ½ ω 2 A 2 = ½ A 2 which is constant Physics 207: Lecture 19, Pg 21 Energy of the Spring-Mass Syste So E = K + U = constant =½ A 2 ω = ω 2 = At axiu displaceent K = 0 and U = ½ A 2 and acceleration has it axiu (or iniu) At the equilibriu position K = ½ A 2 = ½ v 2 and U = 0 E = ½ A 2 U~cos 2 K~sin 2 π 2π θ Physics 207: Lecture 19, Pg 22 Page 11
12 SHM So Far The ost general solution is x = A cos(ωt + φ) where A = aplitude ω = (angular) frequency φ = phase constant For SHM without friction, ω = The frequency does not depend on the aplitude! This is true of all siple haronic otion! The oscillation occurs around the equilibriu point where the force is zero! Energy is a constant, it transfers between potential and inetic Physics 207: Lecture 19, Pg 23 The Siple Pendulu A pendulu is ade by suspending a ass at the end of a string of length L. Find the frequency of oscillation for sall displaceents. Σ F y = a c = T g cos(θ) = v 2 /L Σ F x = a x = -g sin(θ) If θ sall then x L θ and sin(θ) θ dx/dt = L dθ/dt a x = d 2 x/dt 2 = L d 2 θ/dt 2 so a x = -g θ = L d 2 θ / dt 2 L d 2 θ / dt 2 - g θ = 0 and θ = θ 0 cos(ωt + φ) or θ = θ 0 sin(ωt + φ) with ω = (g/l) ½ z θ L y T g Physics 207: Lecture 19, Pg 24 x Page 12
13 Velocity and Acceleration Position: x(t) = A cos(ωt + φ) Velocity: v(t) = -ωa sin(ωt + φ) Acceleration: a(t) = -ω 2 A cos(ωt + φ) x ax = A v ax = ωa a ax = ω 2 A by taing derivatives, since: dx( t ) v( t ) = dt a( t ) = dv t dt ( ) 0 x Physics 207: Lecture 19, Pg 25 Lecture 19 Assignent HW8, Due Wednesday, Apr. 8 th Thursday: Read through Chapter 15.4 Physics 207: Lecture 19, Pg 26 Page 13
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