Oscillations Simple Harmonic Motion

Transcription

1 Oscillations Simple Harmonic Motion Lana Sheridan De Anza College Dec 1, 2017

2 Overview oscillations simple harmonic motion (SHM) spring systems energy in SHM pendula damped oscillations

3 Oscillations and Periodic Motion Many physical systems exhibit cycles of repetitive behavior. After some time, they return to their initial configuration. Examples: clocks rolling wheels a pendulum bobs on springs

4 Oscillations oscillation motion that repeats over a period of time amplitude the magnitude of the vibration; how far does the object move from its average (equilibrium) position. period the time for one complete oscillation. After 1 period, the motion repeats itself.

5 Simple Harmonic Motion The oscillations of bobs on springs and pendula are very regular and simple to describe. It is called simple harmonic motion. simple harmonic motion (SHM) any motion in which the acceleration is proportional to the displacement from equilibrium, but opposite in direction The force causing the acceleration is called the restoring force.

6 SHM and Springs If a mass is attached to a spring, the force on the mass depends on its displacement from the spring s natural length. Hooke s Law: F = kx where k is the spring constant and x is the displacement (position) of the mass. If the spring is compressed and then the mass is released, it will move outward until is stretches the spring by the same amount the spring was compressed initially before it bounces back again. (Assume no friction.) Hooke s law gives the force on the bob SHM. The spring force is the restoring force.

7 SHM and Springs How can we find an equation of motion for the block? Newton s second law: F net = F s = ma

8 SHM and Springs How can we find an equation of motion for the block? Newton s second law: F net = F s = ma Using the definition of acceleration: a = d2 x dt 2 Define d 2 x dt 2 = k m x ω = and we can write this equation as: k m d 2 x dt 2 = ω2 x

9 SHM and Springs To solve: d 2 x dt 2 = ω2 x notice that it is a second order linear differential equation. We can actually find the solutions just be inspection.

10 SHM and Springs To solve: d 2 x dt 2 = ω2 x notice that it is a second order linear differential equation. We can actually find the solutions just be inspection. A solution x(t) to this equation has the property that if we take its derivative twice, we get the same form of the function back again, but with an additional factor of ω 2.

11 SHM and Springs To solve: d 2 x dt 2 = ω2 x notice that it is a second order linear differential equation. We can actually find the solutions just be inspection. A solution x(t) to this equation has the property that if we take its derivative twice, we get the same form of the function back again, but with an additional factor of ω 2. Candidate: x(t) = A cos(ωt), where A is a constant.

12 SHM and Springs To solve: d 2 x dt 2 = ω2 x notice that it is a second order linear differential equation. We can actually find the solutions just be inspection. A solution x(t) to this equation has the property that if we take its derivative twice, we get the same form of the function back again, but with an additional factor of ω 2. Candidate: x(t) = A cos(ωt), where A is a constant. d 2 ( ) x dt 2 = ω2 A cos(ωt)

13 SHM and Springs d 2 x dt 2 = ω2 x In fact, any solutions of the form: x = B 1 cos(ωt + φ 1 ) + B 2 sin(ωt + φ 2 ) where B 1, B 2, φ 1, and φ 2 are constants. However, since sin(θ) = cos(θ + π/2), in general any solution can be written in the form: x = A cos(ωt + φ)

14 Oscillating Solutions (or equivalently, y = A sin(ωt).) x = A cos(ωt) We now call ω the angular frequency of the oscillation. 1 Figure from School of Physics webpage, University of New South Wales.

15 of 7v SHM 2 eration versus time. Notice that at A and Springs any specified time the velocity is extreme The position of 908 theout bobof atphase a givenwith time the is given position by: n agrees and the acceleration is 1808 out of phase with x = the A cos(ωt position. + φ) e define e spring and v(t) A is the amplitude of the oscillation. We could also write x max = A. x 0 A m t 0 x i A v i 0 or f, the itive and on is Figure 15.6 A block spring The speed of the particle at any point in time is: system that begins its motion from v = Aω sin(ωt + φ) rest with the block at x 5 A at t 5 0. We find that by differentiating the expression for position.

16 SHM and Springs x = A cos(ωt + φ) ω is the angular frequency of the oscillation. When t = 2π ω the block has returned to the position it had at t = 0. That is one complete cycle.

17 SHM and Springs x = A cos(ωt + φ) ω is the angular frequency of the oscillation. When t = 2π ω the block has returned to the position it had at t = 0. That is one complete cycle. Recalling that ω = k/m: Period, T = 2π m k Only depends on the mass of the bob and the spring constant. Does not depend on the amplitude.

18 SHM and Springs Question A mass-spring system has a period, T. If the mass of the bob is quadrupled (and everything else is unchanged), what happens to the period of the motion? (A) halves, T /2 (B) remains unchanged, T (C) doubles, 2T (D) quadruples, 4T

19 SHM and Springs Question A mass-spring system has a period, T. If the mass of the bob is quadrupled (and everything else is unchanged), what happens to the period of the motion? (A) halves, T /2 (B) remains unchanged, T (C) doubles, 2T (D) quadruples, 4T

20 SHM and Springs Question A mass-spring system has a period, T. If the spring constant is halved (and everything else is unchanged), what happens to the period of the motion? (A) halves, T /2 (B) is reduced by a factor of 2, so, T / 2 (C) remains unchanged, T (D) is increased by a factor of 2, so, 2T

21 SHM and Springs Question A mass-spring system has a period, T. If the spring constant is halved (and everything else is unchanged), what happens to the period of the motion? (A) halves, T /2 (B) is reduced by a factor of 2, so, T / 2 (C) remains unchanged, T (D) is increased by a factor of 2, so, 2T

22 Oscillations and Waveforms Any oscillation can be plotted against time. eg. the position of a vibrating object against time. The result is a waveform.

23 Oscillations and Waveforms Any oscillation can be plotted against time. eg. the position of a vibrating object against time. The result is a waveform. From this wave description of the motion, a lot of parameters can be specified. This allows us to quantitatively compare one oscillation to another. Examples of quantities: period, amplitude, frequency.

24 Measuring Oscillations frequency The number of complete oscillations in some amount of time. Usually, oscillations per second. f = 1 T Units of frequency: Hertz. 1 Hz = 1 s 1 If one oscillation takes a quarter of a second (0.25 s), then there are 4 oscillations per second. The frequency is 4 s 1 = 4 Hz. ω = 2πf

25 Period and frequency question What is the period, in seconds, that corresponds to each of the following frequencies? 1 10 Hz Hz 3 60 Hz 1 Hewitt, page 350, Ch 18, problem 2.

26 Waveform x = A cos(ωt + φ) x T x i A t ø f = 1 T 1 Figure from Serway & Jewett, 9th ed.

27 460 Chapter 15 Oscillatory Motion Energy in SHM a b c d e f S a max S v max S v max A 0 A x S v S a max S a max x % % % % % % Kinetic energy Potential energy Total energy t 0 T 4 T 2 3T 4 T t

28 Energy in SHM Potential Energy: U = 1 2 kx 2 = 1 2 ka2 cos 2 (ωt + φ) Kinetic Energy: K = 1 2 mv 2 = 1 2 ma2 ω 2 sin 2 (ωt + φ) Using ω 2 = k/m K + U = 1 2 ka2( cos 2 (ωt + φ) + sin 2 (ωt + φ) ) = 1 2 ka2 This does not depend on time!

29 Energy in SHM 15.3 Energy of the Simple H In either plot, notice that K U constant. U K U kx 2 K mv 2 K, U K, U 1 2 ka2 1 2 ka2 a T 2 T t A b O A x K + U = 1 see that K and U are always positive quantities 2 ka2 or zero. Because v 2 5 k/m, we can ress the 1 total Figure mechanical from Serway energy & Jewett, of 9th the ed. simple harmonic oscillator as

30 Pendula and SHM pendulum a massive bob attached to the end rod or string that will oscillate along a circular arc under the influence of gravity A pendulum bob that is displaced to one side by a small amount and released follows SHM to a good approximation. Gravity and the tension in the string provide the restoring force.

31 Pendula and SHM

32 Pendula and SHM modeled as simple harmonic motion about the equilibrium position u 0. L u T S s m g sin u The simple pendulum is anot It consists of a particle-like b that is fixed at the upper end vertical plane and is driven b the angle u is small (less than harmonic oscillator. The forces acting on the bo tational force m g S. The tange always acts toward u 5 0, oppo tion. Therefore, the tangenti Newton s second law for moti mg S where the negative sign indic The net force on thefigure pendulum A bob simple is mg sin θrium along (vertical) the arc position s, and s when the string is atpendulum. an angle θ to the vertical. expressed the tangential acc Because s 5 Lu (Eq. 10.1a wit u m m g cos u Therefore, Newton s second law for rotations (τ net = Iα) analyzing about the pivot gives: I d2 θ 2 = mg sin θ dt F t 5

33 Pendula and SHM For a simple pendulum I = ml 2 (string is massless). ml 2 d2 θ 2 = mgl sin θ dt Dividing by ml 2 : d 2 θ dt 2 = g L sin θ

34 Pendula and SHM For a simple pendulum I = ml 2 (string is massless). ml 2 d2 θ 2 = mgl sin θ dt Dividing by ml 2 : d 2 θ dt 2 = g L sin θ This is a non-linear second order differential equation, and the exact solution is a bit ugly. However, we can approximate this motion as SHM. (Remember, for SHM acceleration is proportional to displacement but in the opposite direction.)

35 Pendula and SHM For small values of θ (measured in radians!), sin θ θ. That means for small oscillations of a pendulum, the restoring force on the bob is roughly: and the motion is SHM. F = mgθ Our equation of motion becomes: where ω = g L d 2 θ dt 2 = ω2 θ

36 Pendula and SHM Equation of motion: d 2 θ dt 2 = ω2 θ We already know the solutions should be of the form θ = θ max cos(ωt + φ) where θ max is the amplitude and T = 2π ω is the period.

37 Pendula and SHM Equation of motion: d 2 θ dt 2 = ω2 θ We already know the solutions should be of the form θ = θ max cos(ωt + φ) where θ max is the amplitude and T = 2π ω is the period. Recalling that ω = g/l, for the simple pendulum: Period, T = 2π L g

38 Problem An astronaut on the Moon attaches a small brass ball to a 1.00 m length of string and makes a simple pendulum. She times 15 complete swings in a time of 75 seconds. From this measurement she calculates the acceleration due to gravity on the Moon. What is her result? 1 1 Hewitt, Conceptual Physics, problem 8, page 350.

39 Problem An astronaut on the Moon attaches a small brass ball to a 1.00 m length of string and makes a simple pendulum. She times 15 complete swings in a time of 75 seconds. From this measurement she calculates the acceleration due to gravity on the Moon. What is her result? m/s 2 1 Hewitt, Conceptual Physics, problem 8, page 350.

40 Physical Pendulum O θ h d The swinging rigid object has a moment of inertia I about the pivot. F g sin θ t brings the to center. C θ F g cos θ As before, for small angles Newton s second law for rotations (τ net = Iα) analyzing about the pivot gives: I d2 θ dt 2 = mgdθ F g

41 Physical Pendulum O θ h d The swinging rigid object has a moment of inertia I about the pivot. F g sin θ t brings the to center. C θ F g cos θ As before, for small angles Newton s second law for rotations (τ net = Iα) analyzing about the pivot gives: I d2 θ dt 2 = mgdθ Solving yields θ = θ max cos(ωt + φ), with ω = I mgd/i ; T = 2π mgd F g

42 force. In many real s also act and retard th Imagine now a system with a restoring force, but alsothe a resistive system diminishe force. energy of the system ing medium. Figure and submersed in a v in air. After being se air resistance. The o in practice. The spri transform mechanic One common ty the force is proport m tion opposite the ve force is often obser the retarding force c damping coefficient) a Damped Oscillations Figure One example of ton s second law as In the picture, that would be a fluid resistance force, R = bv. a damped oscillator is an object attached to a spring and submersed in a viscous liquid. F net = kx b dx dt = m d2 x dt 2

43 Damped Oscillations F net = kx b dx dt = m d2 x dt 2 d 2 x dt 2 + b dx m dt + k m x = 0 Solutions 1 are where ω = x = A e (b/2m)t cos(ωt + φ) k m ( b 2m) 2 and A and φ are constants. 1 See the appendix to these slides for proof.

44 Damped Oscillations 15.7 Forced x = Oscillations A e (b/2m)t cos(ωt φ) object oscillating in x all, the oscillatory es exponentially in A ectable. Any system ashed black lines in represent the expoe amplitude decays nt and object mass, tarding force. hat b/2m, v 0, the resented by Figure creases, the amplib reaches a critical d is said oscillating to be criti- solution. The amplitude decreases as Ae (b/2m)t. 0 t The decaying exponential Figure e15.21 (b/2m)t Graph shows of position versus (energy time for lost). a damped The cosine is that the amplitude of the waveform is decreasing an oscillator.

45 Summary oscillations simple harmonic motion (SHM) spring and pendulum systems damped harmonic motion (Uncollected) Homework Serway & Jewett, Ch 15, onward from page 472. OQs: 13; CQs: 5, 7; Probs: 1, 3, 9, 35, 41

46 Appendix: Damped Oscillations Solution Derivation d 2 x dt 2 + b dx m dt + k m x = 0 Suppose an exponential function is the solution to this equation: r and B are constants. Then x = B e rt B e rt (r 2 + b m r + k m r) = 0 The exponential function is not zero for any finite t, so the other factor must be zero. We must find the roots for r that make this equation true.

47 Damped Oscillations Solution Derivation This is called the characteristic equation The roots are: r 2 + b m r + k m r = 0 ( r = b ) b 2 2m ± k 2m m This means the solutions are of the form: x = e b/(2m)t ( B 1 e iωt + B 2 e iωt) where ω = k m ( ) b 2 2m

48 Damped Oscillations Solution Derivation This means the solutions are of the form: x = e b/(2m)t ( B 1 e iωt + B 2 e iωt) where ω = k m ( ) b 2 2m Recall that cos(x) = 1 2 (eix + e ix ). (If you haven t seen this, try to prove it using the series expansions of cosine and the exponential function.) We can write the solution as x = A e (b/2m)t cos(ωt + φ) where B 1 = A e iφ and B 2 = A e iφ.