oscillation – the vibration of an object wave – a transfer...

¨  Oscillation – the vibration of an object   ¨  Wave – a transfer of energy without a

transfer of matter

¨  Equilibrium Position – position of object at rest (mean position)

  ¨  Displacement (x) – distance in a particular

direction of a particle from its mean position

  ¤ Units - meters

¨  Amplitude (A or xmax) – maximum displacement from the mean position

  ¤ Units - meters

¨  Period (T) – time taken for one complete oscillation

  ¤ Units – seconds (s / cycle)

  ¨  Frequency (f) – number of oscillations that take place

per unit time

  ¤ Units – Hertz (Hz) (cycles/s)

T = 2π

ω= 1

f

¨  Angular Frequency (ω) - a scalar measure of rotation rate

¤ Units – radians/sec or s-1

ω = 2π f =

2πT

¨  Spring Force (Fs) – restoring force – tends to restore system to equilibrium position – opposite in direction of displacement

!Fs = k

!x

x = v = F = a = x = v = F = a = x = v = F = a = x = v = F = a =

x = +A v = 0 m/s F = -Fmax (Fs = kx) a = - amax (a = Fnet/m) x = 0 m v = -vmax F = 0 N a = 0 m/s2

x = -A v = 0 m/s F = +Fmax (Fs = kx) a = + amax (a = Fnet/m) x = 0 m v = +vmax F = 0 N a = 0 m/s2

When is the magnitude of the velocity of the mass at its maximum value?  

  When is the acceleration of the mass at its maximum value?  

When is the magnitude of the velocity of the mass at its maximum value?  

At the equilibrium position   When is the acceleration of the mass at its maximum value?  

At the extreme positions

Initial condition: starts at equillibrium position   Function:

Initial condition: starts at amplitude position   Function:

x = Asin(2πft) = Asinωt

x = Acos(2πft) = Acosωt

cos

-sin

-cos

sin

cos

-sin

Displacement Function

Analyze the displacement function shown at right. ¨   What is the amplitude?

  ¨   What is the period?

¨    What is the frequency? ¨   What is the angular speed?

¨  Write the displacement function.

¨  What is the displacement of the mass when:

¨  t = 1.0 s?   ¨  t = 2.0 s?

Displacement Function

Analyze the displacement function shown at right. ¨   What is the amplitude?

 A = 0.080 m ¨   What is the period?

 T = 4.0 s ¨    What is the frequency? ¨   What is the angular speed?

¨  Write the displacement function.

¨  What is the displacement of the mass when:

¨  t = 1.0 s?   ¨  t = 2.0 s?

f = 1T

= 1

4.0 s = 0.25 Hz

ω = 2πf = 2π (0.25Hz) = 1.57 rads

x = (0.080m)sin(1.57 rads ) t

x = (0.080m)sin(1.57 rads )(1.0s) = 0.080m

x = (0.080m)sin(1.57 rads )(2.0s) = 0m

¨  Displacement Function

¨  Velocity Function

¨  Acceleration Function

x = A sin (ω t)

v = (Aω ) cos ω tv = vmax cos ω t

v = (-Aω 2) sin ω tv = -amax sin ω t

¨  Simple Harmonic Motion (SHM) – motion that takes place when the acceleration of an object is proportional to its displacement from its equilibrium position and is always directed toward its equilibrium position

Period of a Spring

Ts = 2π

mk

= 2πω

¨  A 2.00 kg mass oscillates back and forth 0.500m from its rest position on a horizontal spring whose constant is 40.0 N/m. Calculate the angular speed, period and frequency of this system.

¨  A 2.00 kg mass oscillates back and forth 0.500m from its rest position on a horizontal spring whose constant is 40.0 N/m. Calculate the angular speed, period and frequency of this system.

2πω

= 2π mk

ω = km

=40.0 N

m

2.00kg= 4.47 rad

s T = 2π m

k = 2π 2.00 kg

40.0 N/m = 1.40 s

Or T = 2π/ω

f =

1T

= 1

1.4 s = 0.712 Hz

Tp = 2π

ℓg

= 2πω

¨  A 20.0 g pendulum on an 80.0 cm string is pulled back 5.0 cm and then swings. Determine its angular speed, maximum velocity and maximum acceleration.

¨  A 20.0 g pendulum on an 80.0 cm string is pulled back 5.0 cm and then swings. Determine its angular speed, maximum velocity and maximum acceleration.

2πω

= 2π ℓg

ω = gℓ=

9.81 ms2

0.800m= 3.5 rad

s

vmax = Aω = 0.18 m/s

amax = Aω 2 = (0.050m)(3.5 rad

s )2 = 0.61 ms2

Energy & Simple Harmonic Motion x = v = F = a = x = v = F = a = x = v = F = a = x = v = F = a =

Energy & Simple Harmonic Motion x = max v = 0 m/s Us = max K = 0 J x = 0 m v = max Us = 0 J K = max x = max v = 0 m/s Us = max K = 0 J x = 0 m v = max Us = 0 J K = max

Energy & SHM

When the mass is at its mean position . . .

ET = K + Us

ET = Kmax

ET = 1/2 mvmax2

ET = 1/2 m( ω i A )2

ET = 1/2 mω 2A2

When the mass is at any position . . .

ET = K + Us

ET = 1/2 mv2 + 1/2 kx2

Energy-Displacement Function

Energy-Time Function

Waves

Pulse – single oscillation or disturbance

Continuous traveling wave – succession of oscillations (series of periodic pulses)

Both pulses and traveling waves:   Transfer energy though there is no net motion of the medium through which the wave passes.

Waves

¨  Mechanical Waves: require a medium to transfer energy ¤ eg. – sound waves, water waves, waves on strings,

earthquake waves   ¨  Electromagnetic Waves: do not require a medium to

transfer energy

¤ eg. – light waves, all EM waves

Transverse Waves

¨  A transverse wave is one in which the direction of the oscillation of the particles of the medium is perpendicular to the direction of travel of the wave (the energy).    ¨  Examples: light, violin and guitar

strings, ropes, earthquake S waves

Longitudinal Waves

¨  A longitudinal wave is one in which the direction of the oscillation of the particles of the medium is parallel to the direction of travel of the wave (the energy).   ¨  Example: sound, earthquake

P waves

¨  Displacement (x) – distance in a particular direction of a particle from its mean position

  ¨  Amplitude (A or xmax) – maximum displacement from the mean position   ¨  Period (T) – time taken for one complete oscillation - time for one complete wave (cycle)

to pass a given point   ¨  Frequency (f) – number of oscillations that take place per unit time   ¨  Wavelength (λ) – shortest distance along the wave between two points that are in

phase -the distance a complete wave (cycle) travels in one period.

¨  Compression: region where particles of medium are close together   ¨  Rarefaction: region where particles of medium are

far apart

Waves in Motion

¨  Compare the motion of a single particle to the motion of the wave as a whole (the motion of the energy transfer).

Waves in Motion

¨  Compare the motion of a single particle to the motion of the wave as a whole (the motion of the energy transfer).

Particle motion is perpendicular to the wave motion

Particle Speed Wave Speed

average speed = v = dt

v = 4 AT

Speed is not constant = SHM

constant speed = v = dt

v = λT

= 1T

⎛⎝⎜

⎞⎠⎟

λ

v = fλ λ =

vf

Sound

v =

3kTm

Sound

¨  Mechanical Wave ¤ Longitudinal

¨  Amplitude ¤ Volume ¤ Energy

¨  Frequency ¤ Pitch

Light

Speed = c = 3.00 x 108 m/s

¨  Electromagnetic Wave ¤ Transverse

¨  Amplitude ¤ Brightness / Intensity

¨  Frequency ¤ Color / Type ¤ Energy

Phase

¨  Phase – the position of any particle in its cycle of oscillation.

¨  In phase: (A,E,I) (B,F) (D,H) (C,G)

¨  The phase difference between any points in phase is 0° or λ

¨  Completely out of phase: (A,C) (B,D) (A,G) (B,H)

¨  The phase difference between any points completely out of phase is π or 180° or λ/2

Reflection at a Boundary Between Two Media

Reflection at a Boundary Between Two Media

Superposition & Interference

¨  Principle of Linear Superposition – When two or more waves (pulses) meet, the resultant displacement is the vector sum of the individual displacements.

Constructive Interference

Destructive Interference

Beats

¨  Beats – periodic variations in loudness resulting from the superposition of two sound waves of slightly different frequencies

¨  What frequency is heard when the two tuning forks pictured above are sounded together?

¨  What is the frequency of the beats that they produce?

¨  What frequency is heard when the two tuning forks pictured above are sounded together?

¨  What is the frequency of the beats that they produce?

fsound =

f1 + f2

2= 440Hz +438Hz

2= 439Hz

fbeat = f1 − f2 = 440Hz −438Hz = 2Hz

Standing Waves

¨  A traveling wave moving in one direction in a medium is reflected off the end of the medium.

This sends a reflected wave traveling in the opposite direction in the medium. This second wave is (nearly) identical with the first traveling wave. (same frequency, same wavelength, almost same amplitude)

¨  Node: location of constant complete destructive interference  Anti-Node: location of maximum constructive interference

Transverse Standing Wave

1st Harmonic (fundamental)

3rd Harmonic (2nd overtone)

2nd Harmonic (1st overtone)

6 meters

v = 1200 m/s L = λ1 = f1 = v = 1200 m/s L = λ2 = f2 = v = 1200 m/s L = λ3 = f3 =

Transverse Standing Wave

1st Harmonic (fundamental)

3rd Harmonic (2nd overtone)

2nd Harmonic (1st overtone)

6 meters

v = 1200 m/s L = ½ λ1 λ1 = 2 L = 12 m f1 = v / λ1 = 1200 / 12 = 100 Hz v = 1200 m/s L = λ2 λ2 = L = 6 m f2 = v / λ2 = 1200 / 6 = 200 Hz = 2 f1 v = 1200 m/s L = (1.5) 3/2 λ3 λ3 = 2/3 L = 4 m f3 = v / λ3 = 1200 / 4 = 300 Hz = 3 f1

Fundamental wavelength and frequency:

L = ½ λ1 so λ1 = 2L

f1 =

vλ1

= v

2L

Other natural frequencies (Resonant modes):

fn = n f1 = n

v2L

⎛⎝⎜

⎞⎠⎟

where n = 1,2,3,4,…

Waves on a String Under Tension

linear density = µ =

mL

v =

FT

µ

Pipe open at both ends

Boundary conditions for a pipe open at both ends: 2 free ends – antinode at each end

Fundamental wavelength and frequency:

L = ½ λ1 so λ1 = 2L

f1 =

vλ1

= v

2L

Other natural frequencies (Resonant modes):

fn = n f1 = n

v2L

⎛⎝⎜

⎞⎠⎟

where n = 1,2,3,4,…

Pipe open at both ends

Boundary conditions for a pipe closed at one end: 1 fixed and one free end – one node and one antinode

Fundamental wavelength and frequency:

L = ¼ λ1 so λ1 = 4L

Other natural frequencies (Resonant modes):

where n = 1,3,5,… f1 =

vλ1

= v

4L

fn = n f1 = n

v4L

⎛⎝⎜

⎞⎠⎟