Periodic and Wave Motion

1
Periodic and Wave Motion
2
Periodic Motion

• Any behavior which repeats itself
• Oscillatory systems of any type
• Requirements of such motion
• Inertia--mass
• Restoring force
• Equilibrium or lowest potential energy state

3
Hookes Law

• Periodic motion of a mass on a spring
• Robert Hooke-who also developed an improved
  microscope-stated this law
• The mass-spring system meets the requirements for
  periodic motion since the spring provides the
  restoring force
• The formula-- F -kx where k is the spring
  constant
• The restoring force is opposite direction of
  displacement

4
Example 14.2

• 14.2 A spring with constant of 9.48 N/m and a
  block of mass 0.350 kg are placed horizontally.
  The spring is compressed 6.0 cm and released from
  rest.
• (a) What is the initial acceleration of the
  block?
• (b) What is the initial force on the block?
• (a) Compressing the spring corresponds to x
  -6.0/cm. The acceleration is then
• a - k x
• m
• a -9.48 N/m (-0.060m) 1.625 m/s2
• 0.350 kg
• (b) The initial force can be found directly from
  Newtons second law,
• F ma (0.350 kg)(1.625 m/s2) 0.57 N

5
Simple Harmonic Motion

• Any system whose acceleration is proportional to
  negative displacement is undergoing simple
  harmonic motion
• In simple harmonic motion
• xx0 cos q and a-a0 cos q where q 2pt / T
• This becomes
• xx0 cos 2pt / T and
• ax-a0 cos 2pt / T which combines to
• ax-a0(x / x0)

6
Simple Harmonic Motion

• Velocity is also a function of time
• vx -v0 sin 2pt / T
• Maximum velocity, acceleration, and displacement
  are considered originals and are the amplitudes
  of their respective quantities

7
Simple Harmonic Oscillators

• If the motion obeys the formulas noted previously
  the object is then a simple harmonic oscillator
• To make displacement start at time t0 , an
  adjustment to the angle must be used
• This makes the equation
• x x0 cos(2pt/T f ) where f is the
  adjustment, called the phase angle

8
Example 14.3

• A metal block is hung from a spring that obeys
  Hookes law. When the block is pulled down 12
  cm from the equilibrium position and released
  from rest, it oscillates with a period of 0.75 s
  passing through the equilibrium position with a
  speed of 1.0 m/s. Find displacement and velocity
  at t0.28 s
• xx0 cos 2pt / T (12)cos 2p (0.28/0.75) 12
  cos 2.35rad
• 12(-0.700) -8.4 cm
• vx -v0 sin 2pt / T -(1.0)(sin 2.35rad)
  -(1.0)(0.712)
• -0.71 m/s

9
Homework 1

• On The Website
• 3, 5, 9, 11, 15

10
Energy of Harmonic Oscillators

• From previous study, the potential energy of a
  spring is ½ kx02
• When the oscillator is moving, it also develops
  linear kinetic energy so that ½ mv02 ½ kx02
• In this, v0 x0 ?k/m
• The total energy must be constant
• E ½ mv2 ½ kx2 at any time

11
Example 14.4

• A 3.0-kg ball is attached to a spring of
  negligible mass and with a spring constant k
  40N/m. The ball is displaced 0.10 m from
  equilibrium and released from rest. What is the
  maximum speed of the ball as it undergoes simple
  harmonic motion?
• Compute the maximum speed from the energy of
  the system. The maximum speed occurs at x 0,
  when the kinetic energy is maximum and in equal
  to the initial potential energy. Then
• ½ mv20
• So
• V0 x0vk/m (0.10m) v40 N/m/ 3.0 kg 0.37 m/s

12
Period of a Harmonic Oscillator

• In periodic motion, v0T 2px0 so
• T 2px0/v0
• Since v0 x0 ?k/m , T 2p?m/k
• This implies that period depends only on mass and
  spring constant, not on initial displacement
• This period is the natural period of the system
  and its inverse is the natural frequency

13
Example 14.5

• A block of 2.4 kg oscillates on a spring of
  constant k 26 N/m with an amplitude of 17 cm.
  (a) What is the period? (b) What is the natural
  frequency? (c) What is the maximum speed? (d)
  What is the speed of the block 0.21 s after it
  passes through its maximum position?
• a) T 2p?m/k 2p?2.4/26 1.91 s
• b) f 1/T 1/1.91 0.52 Hz
• c) v0 2px0 /T 2p(0.17) / 1.91 0.56 m/s
• d) vx -v0 sin 2pt / T -(0.56)(sin
  2p(0.21/1.91)) 0.36 m/s

14
Pendulums and SHM

• Vibration
• Vibration is a wiggle in time
• Pendulum vibration is back and forth
• Length of time for pendulum swing is its period
• Pendulum vibration depends on length
• Does not depend on mass of object swinging or
  original displacement
• Depends on gravitational acceleration
• Long pendulums have long periods
• Short animals-more steps
• Tall animals-less steps

15
Simple Pendulum

• Simple harmonic motion has restoring force
  -mg sinq
• Displacement is arc length s, where
• s Lq
• Instantaneous calculation makes
• sin q q gives F -mgq (-mg/L) s
• which is similar to F-kx
• Period is likewise similar
• T 2p?m/k gt 2p?m/ mg/L 2p?L/g
• which means it is dependent only on length and
  the value of g for that place

16
Example 14.7

• A grandfather clock has a pendulum
• 1.0 m long. What is its period?
• T 2p?L/g 2p?1.0/9.8 2.0 s

17
Homework 2

• On The Website
• 18,21,23,25,29,35,41

18
Waves from SH Motion

• If simple harmonic motion is extended into space,
  it traces a sine curve
• Wave is a wiggle in time and space
• High points crest Low points trough
• Height of either amplitude
• Distance between corresponding pointswavelength
  ?
• How often vibrations occur frequency f
• Vibrations or waves per second
• Units called Hertz (s-1)
• Period inverse of frequency (1/f)

19
Wave Motion

• Pulses in a rope can only be transverse,
  vibrating perpendicular to wave travel
• Waves transfer energy from one place to another
  without moving their medium similarly
• Harmonic waves are a series of continuing pulses
  which have consistent period and wavelength

20
Wave Type

• Transverse waves vibrate across from direction of
  travel
• Longitudinal waves vibrate along the direction of
  travel (as in a spring)

21
Wave Motion

• Displacement in direction y y0 sin
• With respect to time y y0 sin (2pt / T)
• Combining gives y y0 sin 2p ( )
• Wave speed v l / T l f

22
Example 15.1

• The drawing represents two snapshots of a wave on
  a rope, taken
• 1/10 s apart. The wave was traveling to the
  right and moved by less than one wavelength
  between pictures. Find its (a) wavelength, (b)
  wave speed, and (c) frequency. Maximum
  displacement is 3.0 cm.
• Examining the figure, the distance between two
  successive crests, or the wavelength, is l 2.0m

(b) During the 1/10-s interval the wave moved to
the right a distance of half a wavelength, or 1
m. The wave speed is the distance traveled
divided by the time interval, or v 10 m/s to
the right. (c) Since we now know both wavelength
and speed, we can obtain the frequency from f
v/ l 10m/s / 2.0m 5.0 s-1 5.0 Hz.
23
Sound

• OriginVibrations of a Material Object
• Vibrating source sends disturbance through
  surrounding mediumDisturbance must be received
  to be soundPerceived frequency of sound is
  pitch Human perception Infrasonic
  frequencies lower than we can perceive, lt20
  Hz
• Ultrasonic frequencies higher than
• we can perceive, gt20,000 Hz

24
Nature of Sound and Air

• Compression -Pulse of compressed air Pressure
  higher in that space
• Rarefaction Disturbance in air or matter,
  molecules spread outPressure is lower in the
  space

25
Media That Transmits Sound

• Sound waves require a medium to travel
  throughThe speed of sound depends upon the
  medium Higher density higher speed High
  elasticity higher speed
• (elasticity ability to change shape and
• return)
• The speed of sound in air also varies as the
  temperature changes above 0oC vT 331.5
  0.6 TC
• where TC is Celsius temperature

26
Sound Waves

• Longitudinal rather than transversevibrate
  parallel to wave travel
• Waves travel in all directions, spherically
  spreading in air, thus losing energy as they
  travel
• Intensity is a measure of energy in sound waves
  and decreases inversely with the square of the
  distance from the source
• I2 r12
• I1 r22

27
Loudness

• Intenseness or amplitude of sound wave measured
  in decibels
• Each 10 dB is twice as loud and 10 times the
  original sound intensity

28
Example 15.3

• A loudspeaker on a tall pole standing in a
  field of tall grass generates a high-frequency
  sound at an intensity of 1.0 X 10-5 W/m2 at the
  position of the ears of a person standing 8.0 m
  directly below it. If the person walks away from
  the pole so that she is 24 m from the
  loudspeaker, what is the sound intensity at the
  new position of her ears?
• Compare the intensities at the two positions
• I2 (r12 /r22)I1 (82)/(242)(1.0 x 10-5)
• 0.11 x 10-5 W/m2

29
Forced Vibrations

• When one vibrating object is in contact with
  another, it sets the second vibrating
• Tuning fork placed on a table or piano
• Sounding board in piano works by forced
  vibration

30
Natural Frequency

• Sound perceived when an object is struck or
  dropped
• Vibration frequency which requires least forced
  vibration energy

31
Resonance

• Increase in amplitude when forced vibrations
  equal natural frequency
• Basis of tuning radio to a station
• Can cause so much forced vibration that object
  falls apart

32
The Doppler Effect

• Change in frequency due to motion of source or
  receiver
• Greater the speed of source, greater Doppler
  effect Stationary bug
• Bug swimming
• Blue Shift-Increase in frequency Red Shift-
  Decrease in frequency

33
Doppler Effect Shifts

• This effect causes a new wavelength and thus a
  new frequency depending on whether the source is
  approaching or receding
• New frequency equation

where vs is the source speed and v is the speed
of wave This also works for light, as noted
previously
34
Doppler Effect

• The equation is different for an observer moving
  with respect to the source

where vo is observer speed This equation is used
in law enforcement of speed laws
35
Doppler Effect

• What if both source and observer are moving? It
  takes a little logic to figure the signs, but the
  equation is just a combination

36
Examples 15.6 15.7

• A police car horn emits a 250-Hz tone when
  sitting still. What frequency does a stationary
  observer hear if the police car sounds its horn
  while approaching at a speed of 27.0 m/s (60
  mi/h)? What frequency is heard if the horn is
  sounded as the car is leaving at 27.0 m/s?
• f 272 Hz gives the
  apparent frequency for the
• 
  approaching car.
• f 232 Hz gives the
  frequency as the car passes
• If the police car mentioned above were sounding
  its horn while stationary, what frequency would
  be heard by an observer who was approaching it at
  a speed of 27.0 m/s (60 mi/h)?
• In this case, the observer is moving toward a
  stationary source. Therefore
• f
• f 270 Hz

37
Bow waves

• An additional Doppler effect when the source is
  moving at or above wave speed on surface of the
  medium
• V-shaped pattern made by overlapping crests

38
Shock Waves

• Produced when source is within the medium,
  three-dimensional cone shaped
• Sonic boom sharp crack heard when conical
  shell of compressed air that sweeps behind a
  supersonic aircraft reaches listeners on the
  ground below.

39
Homework 3

• p437 34,35
• p476 ff 4, 6, 18

40
Reflection of Wave Pulses

• When a wave reaches the end of its medium, it is
  reflected
• The way it is reflected depends on whether the
  end is fixed or free to move
• Newtons 3rd law determines the direction of the
  reflected wave pulse
• Fixedrope pulls up on wall, wall pulls down on
  string, creating reversed pulse
• Freemotion simply continues in opposite
  direction, with no reversal

41
Standing Waves

• Occurs when a wave reflects upon itself and
  interference causes the pattern
• Nodes remain stationary Anti nodes-occur half
  way between nodes

42
Principle of Superposition

• Standing waves get their shape due to the
  principle of superposition
• At any instant, the resultant combination of the
  intersection of two waves is the algebraic sum of
  the components of both(or all) waves
• The sum is then the basis for constructive and
  destructive interference

43
Standing Waves
Change the frequency in a standing wave and more
nodes/antinodes appear in the event
44
Standing Waves and Harmonics in Strings

• The lowest resonant frequency of a string (one
  loop) is the fundamental frequency
• Resonant frequencies which are whole number
  multiples of the fundamental are harmonic
  frequencies overtones are any multiple of the
  fundamental
• Harmonics in a standing wave make more loops in
  the string

45
Standing Waves in Strings

• In a string, the points which do not vibrate are
  called nodes . The ends are counted as well as
  nodes in between
• The points of greatest amplitude are called
  antinodes
• The fundamental frequency can be calculated
• f v / l v / 2L where L is string length
• and v is velocity of the wave
• in the string
• Other relationships resonant wavelengths ln
  2L/n
• where n 1,2,3,4,.
• fn v / ln n v / 2L
• v where T is the tension and
• m is the linear density
• This makes fn n /2L ? T/m

46
Example 15.8

• What is the fundamental frequency of a guitar
  string whose tension is adjusted to give a wave
  speed of 143 m/s if the length of string free to
  vibrate is 0.65 m?
• The fundamental wavelength of a string fixed
  on both ends is twice the length of the string.
  The frequency is found from the ratio of sound
  speed to wavelength. We set n 1 for the
  fundamental frequency.
• f
  110 Hz

47
Waves in a Vibrating Column of Air

• In an open end pipe, waves vibrate in a similar
  fashion to a string, except that there is no node
  at the ends, only within the pipe
• The equations for frequency and wavelength are
  the same as in a string
• In a pipe closed at one end, a node is made there
  and no other node at the fundamental. The
  wavelength is therefore 4 times the length of the
  pipe. Closed pipes always have a lower frequency
  than open pipes. In addition, only odd-numbered
  harmonics are possible. Equations are similar,
  but with 4L rather than 2L.

48
Example 15.9

• The second overtone of standing waves in a pipe
  closed at one end is 512 Hz. How long is
  the pipe?
• The fundamental resonant wavelength of a pipe
  closed at one end is four times the length of the
  pipe. For such a pipe, only odd harmonics occur.
  The second overtone in such a series is the
  fifth harmonic, which has a wavelength
• The frequency of the second overtone is
• Solving for L, we have

49
Beats and Interference

• Beats occur when two closely tuned but not
  identical notes are sounded together
• These beats occur due to interference between the
  frequencies
• The beat frequency can be calculated by the
  formula fB f1- f2
• Beats of less than 10 Hz can be distinguished
  easily, above that with much difficulty

50
Homework 4

• p477ff 19, 28, 29, 38, 39

51
Important Formulas