Lecture Outlines Chapter 13 Physics, 3rd Edition James S. Walker

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1 Lecture Outlines Chapter 13 Physics, 3rd Edition James S. Walker
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2 Oscillations about Equilibrium
Chapter 13 Oscillations about Equilibrium

3 Units of Chapter 13 Periodic Motion Simple Harmonic Motion
Connections between Uniform Circular Motion and Simple Harmonic Motion The Period of a Mass on a Spring Energy Conservation in Oscillatory Motion

4 Units of Chapter 13 The Pendulum Damped Oscillations
Driven Oscillations (Tvungna svängningar) and Resonance

5 13-1 Periodic Motion Period: time required for one cycle of periodic motion Frequency: number of oscillations per unit time This unit is called the Hertz:

6 Exercise 13-1 En dator gör ett antal binära operationer per tidsenhet (som ju är en frekvens). En pc har en 1,80 GHz processor. Hur lång tid tar en binär operation, dvs hur lång är periodtiden? T = 1/f = 1/(1,80 • 109 s-1) = 0,556 ns

7 Exercise 13-2 Två tennisspelare värmer upp inför en match. Om det tar bollen 2,3 sekunder att gå från den ene spelaren till den andre, hur lång är då periodtiden och frekvensen för bollens rörelse? T = (2 • 2,57 s) = 5,14 s f = 1/T = 1/(5,14 s) = 0,195 Hz

8 Table 13-1 Typical Periods and Frequencies

9 13-2 Simple Harmonic Motion
A spring exerts a restoring force that is proportional to the displacement from equilibrium:

10 13-2 Simple Harmonic Motion
A mass on a spring has a displacement as a function of time that is a sine or cosine curve: Here, A is called the amplitude of the motion.

11 13-2 Simple Harmonic Motion
If we call the period of the motion T – this is the time to complete one full cycle – we can write the position as a function of time: It is then straightforward to show that the position at time t + T is the same as the position at time t, as we would expect. [Hint: cos(A+B) = cosA•cos B – sinA•sinB]

12 Example 13-1 Spring Time An air-track cart attached to a spring completes one oscillation every 2.4 s. At t = 0, the cart is released from rest at a distance of 0.10 m from its equilibrium position. What is the position of the cart at (a) 0.30 s, (b) 0.60 s, (c) 2.7 s, and (d) 3.0 s?

13 Example 13-1 Spring Time En luftkuddevagn är fäst vid en fjäder och gör en svängning på 2,4 s. Vid t = 0 släpps vagnen från vila vid x = 0,10 m från jämviktsläget. Vad är vagnens läge efter a) 0,30 s b) 0,60 s c) 2,7 s d) 3,0 s? En ansats är att x(t) = A cos (2πt/T) x(t) = (0,10 m) cos(2π• 0,30 s/2,4 s) = 0,071 m x(t) = (0,10 m) cos(2π• 0,60 s/2,4 s) = 0 m x(t) = (0,10 m) cos(2π• 2,7 s/2,4 s) = 0,701 m x(t) = (0,10 m) cos(2π• 3,0 s/2,4 s) = 0 m

14 13-3 Connections between Uniform Circular Motion (likformig cirkulär rörelse) and Simple Harmonic Motion An object in simple harmonic motion has the same motion as one component of an object in uniform circular motion:

15 Figure 13-5 Position versus time in simple harmonic motion
A peg rotates on the rim of a turntable of radius A. When the peg is at the angular position θ, its shadow is at x = A cos θ. Note that A cos θ is also the x component of the radius vector from the center of the turntable to the peg.

16 13-3 Connections between Uniform Circular Motion and Simple Harmonic Motion
The position as a function of time: Periodisk funktion (OBS!!! Detta är ETT val av många!!!) The angular frequency:

17 13-3 Connections between Uniform Circular Motion and Simple Harmonic Motion
The velocity as a function of time: And the acceleration: Both of these are found by taking components of the circular motion quantities.

18 Figure 13-6 Velocity versus time in simple harmonic motion
(a) The velocity of a peg rotating on the rim of a turntable is tangential to its circular path. As a result, when the peg is at the angle θ its velocity makes an angle of θ with the vertical. The x component of the velocity, then, is –v sin θ. (b) Position, x, and velocity, v, as a function of time for simple harmonic motion. The speed is greatest when the object passes through equilibrium, x = 0. On the other hand, the speed is zero when the position is greatest; that is, at the turning points. Finally, note that as x moves in the negative direction, the velocity is negative. Similar remarks apply to the positive direction.

19 Figure 13-7 Acceleration versus time in simple harmonic motion
(a) The acceleration of a peg on the rim of a uniformly rotating turntable is directed toward the center of the turntable. Hence, when the peg is at the angle θ, the acceleration makes an angle θ with the horizontal. The x component of the acceleration is –a cos θ. (b) Position, x, and acceleration, a, as a function of time for simple harmonic motion. Note that when the position has its greatest positive value, the acceleration has its greatest negative value.

20 Example 13-2 Velocity and Acceleration
As in Example 13–1, an air-track cart attached to a spring completes one oscillation every 2.4 s. At t = 0 the cart is released from rest with the spring stretched 0.10 m from its equilibrium position. What are the velocity and acceleration of the cart at (a) 0.30 s and (b) 0.60 s?

21 Example 13-2 Velocity and Acceleration
En luftkuddevagn är fäst vid en fjäder och gör en svängning på 2,4 s. Vid t = 0 släpps vagnen från vila vid x = 0,10 m från jämviktsläget. Vad är vagnens hastighet och acceleration efter a) 0,30 s b) 0,60 s? x(t) = A cos (2πt/T) v(t) = dx/dt = - A •2π/T• sin(2πt/T) a(t) = dv/dt = - A •(2π/T)2 • cos(2πt/T) v(0,30s)= - (0,10 m)•2π/2,4s•sin(2π•0,30s/2,4s) = - 0,19 m/s a(0,30s)= - (0,10 m)•(2π/2,4s)2•cos(2π• 0,30s/2,4s)= - 0,48m/s2 v(0,60s)= - (0,10 m)•2π/2,4s•sin(2π•0,60s/2,4s) = - 0,26 m/s a(0,60s)= - (0,10 m)•(2π/2,4s)2•cos(2π•0,60s/2,4s) = 0 m/s2

22 Example 13-3 Turbulence! On December 29, 1997, a United Airlines flight from Tokyo to Honolulu was hit with severe turbulence 31 minutes after takeoff. Data from the airplane’s “black box” indicated the 747 moved up and down with an amplitude of 30.0 m and a maximum acceleration of 1.8 g. Treating the up-and-down motion of the plane as simple harmonic, find (a) the time required for one complete oscillation and (b) the plane’s maximum vertical speed.

23 Example 13-3 Turbulence │vmax│ = 2πA/T = 23 m/s
Den 29:e december 1997, råkade en 747:a ut för kraftig turbulens. Amplituden på upp- och nedåtrörelsen var 30,0 m och den maximala accelerationen var 1,8 g. Bestäm a) T och b) maximal hastighet (i “y-led”) [800 kmIh ≈ 220m/s] y(t) = A cos (2πt/T) v(t) = dy/dt = - A •2π/T•sin(2πt/T) a(t) = dv/dt = - A •(2π/T)2 • cos(2πt/T) │amax│= (30,0 m)•(2π/T)2 = 1,8•9,81 m/s2 T = 2π(A/amax)1/2 = 8,2 s │vmax│ = 2πA/T = 23 m/s

24 Active Example 13-1 Bobbing for Apples: Find the Position, Velocity, and Acceleration
A Red Delicious apple floats in a barrel of water. If you lift the apple 2.00 cm above its floating level and release it, it bobs up and down with a period of T = s. Assuming the motion is simple harmonic, find the position, velocity, and acceleration of the apple at the times (a) T/4 and (b) T/2.

25 Active Example 13-1 Bobbing for Apples: Find the Position, Velocity, and Acceleration [bob = hoppa, guppa] Om man lyfter upp (CM av) ett äpple (Red Delicious) som flyter i en hink vatten, 2,00 cm över vattenytan och sedan släpper det så guppar det upp och ned, med en periodtid av 0,750 s. Om svängningen är harmonisk, vad är då äpplets läge, hastighet och acceleration efter tiden a) T/4 b) T/2? x(t) = A cos (2πt/T) v(t) = dx/dt = - A •2π/T• sin(2πt/T) a(t) = dv/dt = - A •(2π/T)2 • cos(2πt/T) x(T/4) = Acos(π/2) = 0 ; x(T/2) = Acos(π) = - 2,00 cm v(T/4) = - A•2π/0,750 s•sin(π/2) = - 0,168 m/s ; v(T/2) = 0 a(T/4) = - A•(2π/0,750 s)2•cos(π/2) = 0 a(T/2) = - A•(2π/0,750 s)2•cos(π) = 1,40 m/s2

26 13-4 The Period of a Mass on a Spring
Since the force on a mass on a spring is proportional to the displacement, and also to the acceleration, we find that . Substituting the time dependencies of a and x gives

27 13-4 The Period of a Mass on a Spring
Therefore, the period is

28 Example 13-4 Spring into Motion
En 0,120 kg massa är fäst vid en fjäder och oscillerar med en amplitud av 0,0750 m och en maximal hastighet av 0,524 m/s. Bestäm a) fjäderkonstanten b) periodtiden T Eftersom v(t) = dx/dt = - A •2π/T• sin(2πt/T) så blir maxhastigheten vmax = A •2π/T = A • ω ω = 0,524 m/s /0,0750 m = 6,99 rad/s ω = (k/m)1/2 k = m • ω2 = 0,120 kg • (6,99 rad/s)2 = 5,86 N/m T = 2π/ω = 0,899 s

29 Active Example 13-2 Mass on a Spring: Find k and m
När en 0,420 kg massa är fäst vid en fjäder, oscillerar den med en periodtid av 0,350 s. När man, i stället fäster massan m2, oscillerar den med periodtiden 0,700 s. Bestäm k och m2. T = 2π/ω ω = (k/m)1/2 k = m • (2π/T)2 = 0,420 kg • (2π/0,350 s)2 = 135 N/m Eftersom T ~ √m betyder det att periodtiden fördubblades när m2 = 4 m1 = 1,68 kg

30 Example 13-5 It’s a Stretch

31 Example 13-5 It is a Stretch
När en 0,260 kg massa är fäst vid en vertical fjäder, oscillerar den med en periodtid av 1,12 s. Hur mycket drar massan ut fjädern från dess jämviktsposition? T = 2π/ω ω = (k/m)1/2 k = m • (2π/T)2 = 0,260 kg • (2π/1,12 s)2 = 8,18 N/m Eftersom F ~ Δy betyder det att Δy = mg/k = 0,260 kg • 9,81 m/s2 /8,18 N/m = 0,312 m

32 Conceptual Checkpoint 13-1 Compare Periods När en massa m fästs på en vertikal fjäder blir perioden T. Vad blir periodtiden T* om fjäderns längd halveras men belastas med samma massa? a) > b) = c) < T? Compare Periods

33 13-5 Energy Conservation in Oscillatory Motion
In an ideal system with no nonconservative forces, the total mechanical energy is conserved. For a mass on a spring: Since we know the position and velocity as functions of time, we can find the maximum kinetic and potential energies:

34 Figure 13-10 Energy as a function of position in simple harmonic motion
The parabola represents the potential energy, U, of the spring. The horizontal line shows the total energy, E, of the system, which is constant, and the distance from the parabola to the horizontal line is the kinetic energy, K. Note that the kinetic energy vanishes at the turning points, x = A and x = -A. At these points the energy is purely potential, and thus the total energy of the system is E = 1/2 kA2.

35 13-5 Energy Conservation in Oscillatory Motion
As a function of time, So the total energy is constant; as the kinetic energy increases, the potential energy decreases, and vice versa.

36 13-5 Energy Conservation in Oscillatory Motion
This diagram shows how the energy transforms from potential to kinetic and back, while the total energy remains the same.

37 Example 13-6 Stop the Block
A0.980-kg block slides on a frictionless, horizontal surface with a speed of 1.32 m/s. The block encounters an unstretched spring with a force constant of 245 N/m, as shown in the sketch. (a) How far is the spring compressed before the block comes to rest? (b) How long is the block in contact with the spring before it comes to rest?

38 Example 13-6 Stop the Block
En 0,980 kg block glider på ett friktionsfritt, horisontalt golv med hastigheten 1,32 m/s. Det träffar en horisontal fjäder (i jämvikt) med fjäderkonstanten 245 N/m. a) Hur mycket pressas fjädern ihop från sin jämviktsposition? b) Hur lång tid t är blocket i kontakt med fjädern innan hastigheten = 0? mv2/2 = kA2/2 A = v • (m/k)1/2 = 1,32 m/s • (0,980 kg/245 N/m)1/2 = 0,0835 m När blocket kommit till vila har fjädern och blocket gjort en fjärdedel av en hel svängning. Eftersom T = 2π(m/k)1/2 = 0,397 s t = T/4 = 0,0993 s {Observera att a inte är konstant under förloppet som förutses i många ”användbara” ekvationer som förstås ger rätt storleksordning men inte korrekt resultat.}

39 Active Example 13-3 Bullet–Block Collision: Find the Compression and Compression Time
A bullet of mass m embeds itself in a block of mass M, which is attached to a spring of force constant k. If the initial speed of the bullet is v0, find (a) the maximum compression of the spring and (b) the time for the bullet-block system to come to rest. (See Example 9–5 for a similar system involving a pendulum.)

40 Active Example 13-3 Bullet-Block Collision: Find the Compression and the Compression Time
En kula med massan m skjuts in i ett block med massan M som är fäst vid en fjäder med fjäderkonstanten k. Om ursprungshastigheten på kulan är v0, bestäm hur mycket fjädern pressas ihop. Hur lång tid är kula-blocket i kontakt med fjädern innan deras gemensamma hastighet = 0? mv0 = (m+M)v (m+M)v2/2 = kA2/2 = m2v02/(2[m+M]) A = mv0/(k[m+M])1/2 Som tidigare visats (Ex.13-6) har blocket+kulan kommit till vila när fjädern har gjort en fjärdedel av en hel svängning. t = T/4 = π([m+M]/k)1/2/2

41 13-6 The Pendulum A simple pendulum consists of a mass m (of negligible size) suspended by a string or rod of length L (and of negligible mass). The angle it makes with the vertical varies with time as a sine or cosine.

42 Figure 13-13 The potential energy of a simple pendulum
As a simple pendulum swings away from the vertical by an angle θ its potential energy increases, as indicated by the solid curve. Near θ= 0 the potential energy of the pendulum is essentially the same as that of a mass on a spring (dashed curve). Therefore, when a pendulum oscillates with small displacements from the vertical, it exhibits simple harmonic motion—the same as a mass on a spring.

43 13-6 The Pendulum Looking at the forces on the pendulum bob (tyngd), we see that the restoring force is proportional to sin θ, whereas the restoring force for a spring is proportional to the displacement (which is θ in this case).

44 13-6 The Pendulum However, for small angles, sin θ and θ are approximately equal.

45 13-6 The Pendulum Substituting θ for sin θ allows us to treat the pendulum in a mathematically identical way to the mass on a spring. Therefore, we find that the period of a pendulum depends only on the length of the string (kx=mgsinθ≈mgθ=mgx/L):

46 Exercise 13-4 (p.415)+Conceptual Checkpoint 13-2
Ett golvur är konstruerat för att göra ett utslag (en halv period) på en sekund. Vad har det för pendellängd? T = 2π(L/g)1/2 L = g • (T/2π)2 = 9,81 m/s2 • (2s/2π)2 = 0,994 m Conceptual checkpoint 13-2 Raise or lower the weight? Ovanstående golvur går efter, hur skall pendellängden justeras? Periodtiden T måste minskas och det görs genom att pendellängden minskas något (dvs ”raise the weight”)

47 Example 13–7 Drop Time A pendulum is constructed from a string m long attached to a mass of kg. When set in motion, the pendulum completes one oscillation every 1.59 s. If the pendulum is held at rest and the string is cut, how long will it take for the mass to fall through a distance of 1.00 m?

48 Exemple 13-7 Drop Time Ett pendelur är konstruerat med en pendellängd = 0,627 m och en massa på 0,250 kg. Den får då periodtiden 1,59 s. Om pendel är i vila och pendellinan plötsligt skärs av, hur lång tid tar det för massan att falla 1,00 m? T = 2π(L/g)1/2 g = L • (2π/T)2 = 0,627 m • (2π/1,59s)2 = 9,79 m/s2 x = x0 + v0t + gt2/2 t = (2•1,00/9,79)1/2 = 0,452 s

49 Photo 13-4 Gravitational strength for the state of Ohio
A map of gravitational strength for the state of Ohio. The purple areas are those where the gravitational field is weakest. Areas where the field is strongest (red) represent regions where denser rocks lie near the surface.

50 13-6 The Pendulum A physical pendulum is a solid mass that oscillates around its center of mass, but cannot be modeled as a point mass suspended by a massless string. Examples:

51 13-6 The Pendulum In this case, it can be shown that the period depends on the moment of inertia (tröghetsmomentet I): Substituting the moment of inertia of a point mass a distance l from the axis of rotation gives, as expected,

52 Figure 13-17 The leg as a physical pendulum
As a person walks, each leg swings much like a physical pendulum. A reasonable approximation is to treat the leg as a uniform rod about 1 m in length.

53 Figure 13-17 The leg as a physical pendulum*
Om benet aproximeras som en homogen cylinder med tröghetsmomentet I = mL2/3 med pendellängden λ=L/2. Om benets längd L ≈ 1,0 m T = 2π(λ/g)1/2 (I/mλ2)1/2 = 2π(L/g)1/2 (2/3)1/2 = 2,0•0.82 ≈ 1,6 s På en halv periodtid flyttas foten ungefär sträckan s = Lθ där vinkeln är ungefär en radian. Därför är en människas gånghastighet ungefär v = 1,0 m •1,0 rad/0,8 s = 1,25 m/s (4,5 km/h)

54 13-7 Damped Oscillations In most physical situations, there is a nonconservative force of some sort, which will tend to decrease the amplitude of the oscillation, and which is typically proportional to the speed: This causes the amplitude (för en oscillerande massa m) to decrease exponentially with time:

55 13-7 Damped Oscillations This exponential decrease is shown in the figure:

56 13-7 Damped Oscillations The previous image shows a system that is underdamped – it goes through multiple oscillations before coming to rest. A critically damped system is one that relaxes back to the equilibrium position without oscillating and in minimum time; an overdamped system will also not oscillate but is damped so heavily that it takes longer to reach equilibrium.

57 13-8 Driven Oscillations and Resonance
An oscillation can be driven by an oscillating driving force; the frequency of the driving force may or may not be the same as the natural frequency of the system.

58 13-8 Driven Oscillations and Resonance
If the driving frequency is close to the natural frequency (systemets egenfrevens, som tex f = 1/T där T = 2π(L/g)1/2 pendel T = 2π(k/m)1/2 fjäder) the amplitude can become quite large, especially if the damping is small. This is called resonance.

59 Summary of Chapter 13 Period: time required for a motion to go through a complete cycle Frequency: number of oscillations per unit time Angular frequency: Simple harmonic motion occurs when the restoring force is proportional to the displacement from equilibrium.

60 Summary of Chapter 13 The amplitude is the maximum displacement from equilibrium. Position as a function of time: Velocity as a function of time:

61 Summary of Chapter 13 Acceleration as a function of time:
Period of a mass on a spring: Total energy in simple harmonic motion:

62 Summary of Chapter 13 Potential energy as a function of time:
Kinetic energy as a function of time: A simple pendulum with small amplitude exhibits simple harmonic motion

63 Summary of Chapter 13 Period of a simple pendulum:
Period of a physical pendulum*:

64 Summary of Chapter 13 Oscillations, where there is a nonconservative force, are called damped. Underdamped: the amplitude decreases exponentially with time: Critically damped: no oscillations; system relaxes back to equilibrium in minimum time Overdamped: also no oscillations, but slower than critical damping

65 Summary of Chapter 13 An oscillating system may be driven by an external force This force may replace energy lost to friction, or may cause the amplitude to increase greatly at resonance Resonance occurs when the driving frequency is equal to the natural frequency of the system