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

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

5. Period and Frequency
A radio station has a frequency of M Hz. What is the period of the wave?
103.1 M Hz 1X106 Hz = X 108 Hz
1M Hz
T = 1/f = 1/(1.031 X 108 Hz) = X 10-9 s

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/s2)/(-0.03 m)
k = 6.5 X 104 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/s2)/ 6.5 X 104 N/m
x = 4.5 X 10-2 m = 4.5 cm

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

13. Energy and Springs KE = ½ mv2 PE = ½ kx2 Maximum PE = ½ kA2
Law of conservation of Energy
½ kA2 = ½ mv2+ ½ kx2

14. All PE
All KE
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 kA2
Maximum PE = ½ (20 N/m)(0.030 m2)
Maximum PE = 9 X 10-3 Nm (J)

16. Spring Energy: Example 1a
What is the maximum speed of the mass?
½ kA2 = ½ mv2+ ½ kx2
½ kA2 = ½ mv2 (x=0 at the origin)
9 X 10-3 J = ½ (0.50 kg)v2
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 kx2
PE = ½ (20 N/m)(0.020 m2) = 4 X 10-3 J
½ kA2 = ½ mv2 + ½ kx2
½ mv2 = ½ kA2 - ½ kx2
KE = 9 X 10-3 J - 4 X 10-3 J = 5 X 10-3 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: 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 kA2 = ½ mv2+ ½ kx2
½ kA2 = ½ mv2 (x=0 at the origin)
kA2 = mv2
v2 = kA2/m
v = \/kA2/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 kA2 = ½ mv2+ ½ kx2
kA2 = mv2+ kx2
mv2 = kA2 - kx2
v2 = kA2 - kx2 m
v2 = 19.6 N/m(0.100m2 – m2) = 0.49 m2/s2
0.300 kg
v = m/s

23. Spring Energy: Example 2d
What is the maximum acceleration?
The force is a maximum at the amplitude
F = ma and F = kx
ma = kx
a = kx/m = (19.6 N/m)(0.100 m)/(0.300 kg)
a = 6.53 m/s2

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

25. Period depends only on mass and spring constant
Amplitude does not affect period
vo = 2pAf or vo = 2pA T
vo 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 104 N/m after it hits a bump?
T = 2p m = 2p (1400 kg/6.5 X 104 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 = 2p m k
T2 = (2p)2m
k = (2p)2m = (2p)2(3.0 X 10-4 kg) = 2.7 N/m
T (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 = 2p m k
T = 2p (m/k)1/2
T = 2p (1.0 X 10-4 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 cos2pt or x = A cos2pft T
A = Amplitude
t = time
T = period
f = frequency

31. x = A cos2pft v = -vosin2pft a = -aocos2pft
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 10-4 m. What is the equation to describe the position of the cone over time?
x = A cos2pft
x = (1.5 X 10-4 m) cos2p(262 s-1)t
x = (1.5 X 10-4 m) cos(1650 s-1)t

33. Cos: Example 1b What is the position at t = 1.00 ms (1 X 10-3 s)
x = A cos2pft
x = (1.5 X 10-4 m) cos2p(262 s-1) (1 X 10-3 s)
x = (1.5 X 10-4 m) cos(1.65 rad) = -1.2 X 10-5 m

34. Cos: Example 1c What is the maximum velocity and acceleration?
vo = 2pAf
vo = 2p(1.5 X 10-4 m)(262 s-1) = 0.25 m/s

35. F = ma
kx = ma
a = kx/m But we don’t know k or m
a = k x Solve for k/m m
T = 2p m k
T2 = (2p)2m
k = (2p)2 = (2p)2f2
m T2

36. a = k x m
a = (2pf)2x = (2pf)2A
a = [(2p)(262 Hz)]2(1.5 X 10-4 m) = 410 m/s2

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 p t
8.0

38. x = A cos2pft
x = (0.25 m)cos p t
8.0
Therefore A = 0.25 m
2pft = p t
2f = 1
f = 1/16 Hz T = 1/f = 16 s

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

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

41. Remember the circle (360o = 2p rad) q = x L
Fr = mgsinq
at small angles sinq = q
Fr = mgq q L x q mg Fr

42. Fr = mgq
Fr = mgx (Look’s like Hook’s Law F = -kx) L
k = mg
T = 2p m k
T = 2p mL mg

43. T = 2p L g
f = 1 = g
T 2p 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 = g
2p L decrease the denominator

45. Example 1: Pendulum
What would be the period of a grandfather clock with a 1.0 m long pendulum?
T = 2p 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 = 2p L g
Ans: 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 (fo)
When the driving frequency f = fo, maximum amplitude results
Tacoma Narrows Bridge
1989 freeway collapse
Shattering a glass by singing

50. Wave Medium Mechanical Waves Electromagnetic 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

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

55. Velocity of Waves in a String
Depends on:
Tension (FT) [tighter string, faster wave]
Mass per unit length (m/L) [heavier string, more inertia]
v = FT
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 = N = 140 m/s
15 kg/300 m
v = lf f = v/l = 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

60. Longitudinal Waves: Velocity
Wave moving along a long solid rod
Wire
Train track
vlong= E Elastic modulus r
Wave moving through a liquid or gas
vlong = B Bulk 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?
vlong= E = (2.0 X 1011/7800 kg/m3)1/2 r
vlong = 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 pr2
Comparing two distances:
I1r12 = I2r22

64. Intensity: Example 1
The intensity of an earthquake wave is 1.0 X 106 W/m2 at a distance of 100 km from the source. What is the intensity 400 km from the source?
I1r12 = I2r22
I2 = I1r12/r22
I2 = (1.0 X 106 W/m2)(100 km)2/(400 km)2
I2 = 6.2 X 104 W/m2

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
Destructive Constructive
Interference Interference

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 l
First overtone (Second Harmonic)
Second overtone (Third Harmonic)

71. Resonance: Equations L = nln n = 1, 2, 3….. 2 f = nv = nf1 2L v = lf
v = FT
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 = nln 2
l1 = 2L = m 1
v = lf = (2.20 m)(131 Hz) = 288 m/s

73. v = FT
m/L
v2 = FT
FT = v2m = (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?
f1 = 131 Hz 1st Harmonic
f2 = 262 Hz 2nd Harmonic 1st Overtone
f3 = 393 Hz 3rd Harmonic 2nd Overtone
f4 = 524 Hz 4th Harmonic 3rd Overtone

75. Hitting a Boundary Both reflection and refraction occur
Angle of incidence = angle of reflection q1
q2 q1 = q2
Reflected wave
air
water
Refracted wave

76. Refraction Velocity of a wave changes when crossing between substances
Soldiers slow down marching into mud
sin q1 = v1
sin q2 = v2

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 30o, what is the angle of refraction?
sin q1 = v1 sin 30o = km/s
sin q2 = v2 sin q2 8.0 km/s
q2 = 38o

78. Example 2: Refraction
A sound wave travels through air at 343 m/s and strikes water at an angle of 50. If the refracted angle is 21.4o, what is the speed of sound in water?
(Ans: 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