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Harmonic Motion and Waves. Simple Harmonic Motion(SHM) Vibration (oscillation) Equilibrium position – position of the natural length of a spring Amplitude.

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Presentation on theme: "Harmonic Motion and Waves. Simple Harmonic Motion(SHM) Vibration (oscillation) Equilibrium position – position of the natural length of a spring Amplitude."— Presentation transcript:

1 Harmonic Motion and Waves

2 Simple Harmonic Motion(SHM) Vibration (oscillation) Equilibrium position – position of the natural length of a spring Amplitude – maximum displacement

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4 Period and Frequency Period (T) – Time for one complete cycle (back to starting point) Frequency (Hz) – Cycles per second F = 1T = 1 T f

5 Period and Frequency A radio station has a frequency of M Hz. What is the period of the wave? M Hz1X10 6 Hz= X 10 8 Hz 1M Hz T = 1/f = 1/(1.031 X 10 8 Hz) = X s

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7 Hooke’s Law F = -kx F = weight of an object k = spring constant (N/m) x = displacement when the object is placed on the spring

8 Hooke’s Law: Example 1 What is the spring constant if a kg mass causes the spring to stretch 6.0 cm? (ANS: 16 N/m)

9 Special Note: If a spring has a mass on it, and then is stretched further, equilibrium position is the starting length (with the mass on it)

10 Hooke’s Law: Example 2 A family of four has a combined mass of 200 kg. When they step in their 1200 kg car, the shocks compress 3.0 cm. What is the spring constant of the shocks? F = -kx k = -F/x k = -(200 kg)(9.8 m/s 2 )/(-0.03 m) k = 6.5 X 10 4 N/m (note that we did not include the mass of the car)

11 Hooke’s Law: Example 2a How far will the car lower if a 300 kg family borrows the car? F = -kx x = -F/k x = -(300 kg)(9.8 m/s 2 )/ 6.5 X 10 4 N/m x = 4.5 X m = 4.5 cm

12 Forces on a Spring Extreme Position (Amplitude) –Force at maximum –Velocity = 0 Equilibrium position –Force = 0 –Velocity at maximum

13 Energy and Springs KE = ½ mv 2 PE = ½ kx 2 Maximum PE = ½ kA 2 Law of conservation of Energy ½ kA 2 = ½ mv 2 + ½ kx 2

14 All PE All KE All PE Some KE and Some PE

15 Spring Energy: Example 1 A 0.50 kg mass is connected to a light spring with a spring constant of 20 N/m. Calculate the total energy if the amplitude is 3.0 cm. Maximum PE = ½ kA 2 Maximum PE = ½ (20 N/m)(0.030 m 2 ) Maximum PE = 9 X Nm (J)

16 Spring Energy: Example 1a What is the maximum speed of the mass? ½ kA 2 = ½ mv 2 + ½ kx 2 ½ kA 2 = ½ mv 2 (x=0 at the origin) 9 X J = ½ (0.50 kg)v 2 v = 0.19 m/s

17 Spring Energy: Example 1b What is the potential energy and kinetic energy at x = 2.0 cm? PE = ½ kx 2 PE = ½ (20 N/m)(0.020 m 2 ) = 4 X J ½ kA 2 = ½ mv 2 + ½ kx 2 ½ mv 2 = ½ kA 2 - ½ kx 2 KE = 9 X J - 4 X J = 5 X J

18 Spring Energy: Example 1c At what position is the speed 0.10 m/s? (Ans: cm)

19 Spring Energy: Example 2a A spring stretches m when a kg mass is suspended from it (diagrams a and b). Find the spring constant. (Ans: 19.6 N/m)

20 Spring Energy: Example 2b The spring is now stretched an additional m and allowed to oscillate (diagram c). What is the maximum velocity?

21 The maximum velocity occurs through the origin: ½ kA 2 = ½ mv 2 + ½ kx 2 ½ kA 2 = ½ mv 2 (x=0 at the origin) kA 2 = mv 2 v 2 = kA 2 /m v = \/kA 2 /m = \/(19.6 N/m)(0.100m) 2 /0.300kg v = m/s

22 Spring Energy: Example 2c What is the velocity at x = m? ½ kA 2 = ½ mv 2 + ½ kx 2 kA 2 = mv 2 + kx 2 mv 2 = kA 2 - kx 2 v 2 = kA 2 - kx 2 m v 2 = 19.6 N/m(0.100m 2 – m 2 ) = 0.49 m 2 /s kg v = m/s

23 Spring Energy: Example 2d What is the maximum acceleration? The force is a maximum at the amplitude F = maand F = kx ma = kx a = kx/m = (19.6 N/m)(0.100 m)/(0.300 kg) a = 6.53 m/s 2

24 Trigonometry and SHM Ball rotates on a table Looks like a spring from the side One rev(diameter) = 2  A T = 2  m k f = 1 T

25 Period depends only on mass and spring constant Amplitude does not affect period v o = 2  Aforv o = 2  A T v o is the initial (and maximum) velocity

26 Period: Example 1 What is the period and frequency of a 1400 kg car whose shocks have a k of 6.5 X 10 4 N/m after it hits a bump? T = 2  m= 2  (1400 kg/6.5 X 10 4 N/m) 1/2 k T = 0.92 s f = 1/T = 1/0.92 s = 1.09 Hz

27 Period: Example 2a An insect (m=0.30 g) is caught in a spiderweb that vibrates at 15 Hz. What is the spring constant of the web? T = 1/f = 1/15 Hz = s T = 2  m k T 2 = (2  ) 2 m k k = (2  ) 2 m = (2  ) 2 (3.0 X kg) = 2.7 N/m T 2 (0.0667) 2

28 Period: Example 2b What would be the frequency for a lighter insect, 0.10 g? Would it be higher or lower? T = 2  m k T = 2  (m/k) 1/2 T = 2  (1.0 X kg/2.7 N/m) 1/2 = s f = 1/T = 1/0.038 s = 26 Hz

29 Cosines and Sines Imagine placing a pen on a vibrating mass Draws a cosine wave

30 x = A cos2  torx = A cos2  ft T A = Amplitude t = time T = period f = frequency

31 x = A cos2  ft v = -v o sin2  ft a = -a o cos2  ft Velocity is the derivative of position Acceleration is the derivative of velocity

32 Cos: Example 1a A loudspeaker vibrates at 262 Hz (middle C). The amplitude of the cone of the speaker is 1.5 X m. What is the equation to describe the position of the cone over time? x = A cos2  ft x = (1.5 X m) cos2  (262 s -1 )t x = (1.5 X m) cos(1650 s -1 )t

33 Cos: Example 1b What is the position at t = 1.00 ms (1 X s) x = A cos2  ft x = (1.5 X m) cos2  (262 s -1 ) (1 X s) x = (1.5 X m) cos(1.65 rad) = -1.2 X m

34 Cos: Example 1c What is the maximum velocity and acceleration? v o = 2  Af v o = 2  (1.5 X m)(262 s -1 ) = 0.25 m/s

35 F = ma kx = ma a = kx/mBut we don’t know k or m a = k xSolve for k/m m T = 2  m k T 2 = (2  ) 2 m k k = (2  ) 2 = (2  ) 2 f 2 mT2mT2

36 a = k x m a = (2  f) 2 x = (2  f) 2 A a = [(2  )(262 Hz)] 2 (1.5 X m) = 410 m/s 2

37 Cos: Example 2a Find the amplitude, frequency and period of motion for an object vibrating at the end of a spring that follows the equation: x = (0.25 m)cos  t 8.0

38 x = A cos2  ft x = (0.25 m)cos  t 8.0 Therefore A = 0.25 m 2  ft =  t 8.0 2f =  8.0 f = 1/16 HzT = 1/f = 16 s

39 Cos: Example 2b Find the position of the object after 2.0 seconds. x = (0.25 m)cos  t 8.0 x = (0.25 m)cos  4.0 x = 0.18 m

40 The Pendulum Pendulums follow SHM only for small angles (<15 o ) The restoring force is at a maximum at the top of the swing.  F r = restoring Force

41 Remember the circle (360 o = 2  rad)  = x L F r = mgsin  at small angles sin  =   F r = mg   L x  mg FrFr

42 F r = mg  F r = mgx(Look’s like Hook’s Law F = -kx) L k = mg L T = 2  m k T = 2  mL mg

43 T = 2  L g f = 1= 1 g T 2  L The Period and Frequency of a pendulum depends only on its length

44 Swings and the Pendulum To go fast, you need a high frequency Short length (tucking and extending your legs) f = 1 g 2  Ldecrease the denominator

45 Example 1: Pendulum What would be the period of a grandfather clock with a 1.0 m long pendulum? T = 2  L g Ans: 2.0 s

46 Example 2: Pendulum Estimate the length of the pendulum of a grandfather clock that ticks once per second (T = 1.0 s). T = 2  L g Ans: 0.25 m

47 Damped Harmonic Motion Most SHM systems slowly stop For car shocks, a fluid “dampens” the motion

48 Resonance: Forced Vibrations Can manually move a spring (sitting on a car and bouncing it) Natural or Resonant frequency (f o ) When the driving frequency f = f o, maximum amplitude results –Tacoma Narrows Bridge –1989 freeway collapse –Shattering a glass by singing

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50 Wave Medium Mechanical Waves –Require a medium –Water waves –Sound waves –Medium moves up and down but wave moves sideways Electromagnetic Waves –Do not require a medium –EM waves can travel through the vacuum of space

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52 Parts of a wave Crest Trough Amplitude Wavelength Frequency (cycles/s or Hertz (Hz)) Velocity v = f

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55 Velocity of Waves in a String Depends on: Tension (F T ) [tighter string, faster wave] Mass per unit length (m/L) [heavier string, more inertia] v = F T m/L

56 Example 1: Strings A wave of wavelength 0.30 m is travelling down a 300 m long wire whose total mass is 15 kg. If the wire has a tension of 1000 N, what is the velocity and frequency? v = 1000 N= 140 m/s 15 kg/300 m v = ff = v/  = 140 m/s/0.30 m = 470 Hz

57 Transverse and Longitudinal Waves Transverse Wave – Medium vibrates perpendicular to the direction of wave –EM waves –Water waves –Guitar String Longitudinal Wave – Medium vibrates in the same direction as the wave –Sound

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60 Longitudinal Waves: Velocity Wave moving along a long solid rod –Wire –Train track v long = EElastic modulus  Wave moving through a liquid or gas v long = BBulk modulus 

61 Ex. 1: Longitudinal Waves: Velocity How fast would the sound of a train travel down a steel track? How long would it take the sound to travel 1.0 km? v long = E = (2.0 X /7800 kg/m 3 ) 1/2  v long = 5100 m/s (much fast than in air) v = x/t t = x/v  1000m/5100m/s = 0.20 s

62 Earthquakes Both Transverse and Longitudinal waves are produced S(Shear) –Transverse P(Pressure) – Longitudinal In a fluid, only p waves pass Center of earth is liquid iron

63 Energy Transported by Waves Intensity = Power transported across a unit area perpendicular to the wave’s direction I = Power = P Area 4  r 2 Comparing two distances: I 1 r 1 2 = I 2 r 2 2

64 Intensity: Example 1 The intensity of an earthquake wave is 1.0 X 10 6 W/m 2 at a distance of 100 km from the source. What is the intensity 400 km from the source? I 1 r 1 2 = I 2 r 2 2 I 2 = I 1 r 1 2 /r 2 2 I 2 = (1.0 X 10 6 W/m 2 )(100 km) 2 /(400 km) 2 I 2 = 6.2 X 10 4 W/m 2

65 Reflection of a Wave Hard boundary inverts the wave Exerts an equal and opposite force Loose rope returns in same direction

66 Continue in same direction if using another rope boundary

67 Constructive and Destructive Interference DestructiveConstructiveInterference

68 Constructive and Destructive Interference : Phases Waves “in phase” “out of phase”in between

69 Resonance Standing Wave – a wave that doesn’t appear to move Node – Point of destructive interference Antinode – Point of constructive interference (think “Antinode,Amplitude) “Standing waves are produced only at the natural (resonant) frequencies.”

70 Resonance: Harmonics Fundamental Lowest possible frequency “first harmonic” L = ½ First overtone (Second Harmonic) Second overtone (Third Harmonic)

71 Resonance: Equations L = n n n = 1, 2, 3….. 2 f = nv = nf 1 2L v = f v = F T m/L

72 Example 1: Resonance A piano string is 1.10 m long and has a mass of 9.00 g. How much tension must the string be under to vibrate at 131 Hz (fund. freq.)? L = n n 2 1 = 2L = 2.20 m 1 v = f = (2.20 m)(131 Hz) = 288 m/s

73 v = F T m/L v 2 = F T m/L F T = v 2 m= (288 m/s) 2 (0.009 kg) = 676 N L(1.10 m)

74 What are the frequencies of the first four harmonics of this string? f 1 = 131 Hz1 st Harmonic f 2 = 262 Hz2 nd Harmonic1 st Overtone f 3 = 393 Hz3 rd Harmonic2 nd Overtone f 4 = 524 Hz4 th Harmonic3 rd Overtone

75 Hitting a Boundary Both reflection and refraction occur Angle of incidence = angle of reflection air water 11  2  1 =  2 Refracted wave Reflected wave

76 Refraction Velocity of a wave changes when crossing between substances Soldiers slow down marching into mud sin  1 = v 1 sin  2 = v 2

77 Example 1: Refraction An earthquake p-wave crosses a rock boundary where its speed changes from 6.5 km/s to 8.0 km/s. If it strikes the boundary at 30 o, what is the angle of refraction? sin  1 = v 1 sin 30 o = 6.5 km/s sin  2 = v 2 sin  km/s  2 = 38 o

78 Example 2: Refraction A sound wave travels through air at 343 m/s and strikes water at an angle of 5 0. If the refracted angle is 21.4 o, what is the speed of sound in water? (Ans: 1440 m/s)

79 Diffraction Note bending of wave into “shadow region”

80 Diffraction Bending of waves around an object Only waves diffract, not particles The smaller the obstacle, the more diffraction in the shadow region


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