Oscillations and Waves
IB 12
Oscillation: the vibration of an object
Wave: a transfer of energy without a transfer of matter
Examples of oscillations:
1.
2.
3.
4.
5.
6.
7.
8.
mass on spring (eg. bungee jumping)
pendulum (eg. swing)
object bobbing in water (eg. buoy, boat)
vibrating cantilever (eg. diving board)
earthquake
bouncing ball
musical instruments (eg. strings, percussion, brass, woodwinds, vocal chords)
heartbeat
Mean Position (Equilibrium Position) – position of object at rest
Displacement (x, meters) – distance in a particular direction of a particle from its mean position
Amplitude (A or x0, meters) – maximum displacement from the mean position
Period (T, seconds) – time taken for one complete oscillation
Frequency (f, Hertz) – number of oscillations that take place per unit time
Phase Difference – difference in phase between the particles of two oscillating systems
Relationship between period
and frequency:
f 
1
T
f = cycles/sec
T = sec/cycle
Angular Frequency - product of 2π times 1. A pendulum completes 10 swings in 8.0 seconds.
frequency
a) Calculate its period.
T = 0.8 s
Formula: ω = 2πf
ω = 2π/T
b) Calculate its frequency.
f = 1.25 Hz = 1.3 Hz
Symbol: ω
Units: rad/sec s-1
c) Calculate its angular frequency.
ω = 7.8 rad/s = 7.8 s-1
1
Example of an Oscillating System
IB 12
A mass oscillates on a horizontal spring without friction as shown below. At each
position, analyze its displacement, velocity and acceleration.
Force from the Spring: Fs = -kx
restoring force – tends to restore system to equilibrium
position – opposite in direction of displacement
1. When is the velocity of the mass at its maximum value?
When the displacement = 0
at equilibrium position
2. When is the acceleration of the mass at its maximum value?
When the displacement and force = max
at extreme positions
2
The Displacement Function
IB 12
A mass on a spring is allowed to oscillate up and down about its mean position without friction.
Two traces of the displacement (x) of the mass versus time (t) are shown.
Initial condition: starts at mean
position
Function: x = x0 sin ωt
Initial condition: starts at amplitude
position
Function: x = x0 cos ωt
Analyzing the Displacement Function
1. Analyze the displacement function shown at right.
a) What is the amplitude?
x0 = 0.080 m
a) What is the period?
T = 4.0 s
b) What is the frequency?
f = 0.25 Hz
c) What is the angular frequency?
ω = π/2 s-1
2. What is the displacement of the mass when:
a) t = 1.0 s?
x = (.080 m)sin (π/2)t
x = (.080 m)sin (π/2)(1) = .080 m
b) t = 2.0 s?
x = (.080 m)sin (π/2)t
x = (.080 m)sin (π/2)(2) = 0
c) t = 2.5 s?
e) Write the displacement function.
x = (.080 m)sin (π/2)t
x = (.080 m)sin (π/2)t
x = (.080 m)sin (π/2)(2.5) =
.080 sin (3.926…) = -0.057 m
(RADIAN MODE!!!)
3
Velocity and Acceleration for Simple Harmonic Motion
IB 12
a) Displacement Function
b) Velocity Function
c) Acceleration Function
Defining Equation for SHM:
a   2 ( x0 sin t )
a   2 x
a  x
Negative Sign:
1. acceleration is in opposite direction of displacement
2.
directed back towards mean position
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
4
IB 12
1. The graph shown at right shows the displacement
of an object in SHM. Use the graph to find the:
a) period of oscillation
b) amplitude of oscillation
c) displacement function
Alternate Velocity Function
d) maximum velocity
e) velocity at 1.3 seconds
f) maximum acceleration
2. Use the alternate form of the velocity function
to find the velocity of the object at 1.3 s.
g) acceleration at 1.3 seconds
5
Example of SHM – Mass on a Horizontal Spring
IB 12
A mass m oscillates horizontally on a spring
without friction, as shown. Is this SHM?
Fnet  ma
Fs  ma
kx  ma
k
a x
m
a x
Yes this is SHM since a α -x.
Angular frequency, period, and frequency for a mass on a spring
k
x
m
a   2 x
a
T
2

T  2
k
m
k

m
2 
f 
1
1

T 2
m
k
k
m
1. 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.
a) Calculate the angular frequency, period and frequency of this system.
40.0
2.00
  4.47 s 1

2.00
40.0
T  1.40 s -1
T  2
1
1.40
f  0.712 Hz
f 
b) Write the displacement, velocity and acceleration functions for this system.
x  (0.500) sin(4.47t )
v  (2.24) cos(4.47t )
a  (9.99) sin(4.47t )
6
Alternate Forms of the Equations of Motion for SHM
IB 12
1. Write the equations of motion for the graphs shown below.
2. Write the equations of motion for the graphs shown below.
3. What is the difference between the motions described by the two sets of equations?
#1 - x = 0 at t = 0
#2 – x = x0 at t = 0
4. a) Write the equations of motion for the system whose
displacement is shown on the graph at right.
b) State two times when the:
ii) magnitude of the acceleration is maximum.
i) speed is maximum
7
Example of SHM – Simple Pendulum
IB 12
1. A mass is allowed to swing freely from the end of a light-weight string. Show that
the motion of this simple pendulum is approximately simple harmonic motion.
s
a  g  
L
g
a    s
L
a  s
for small angles
sx
Fnet  ma
mg sin   ma
a   g sin 
for small angles
sin   
a   g
g
a    x
L
a  -x
2. Determine the angular frequency, period and frequency for the pendulum.
g
L
g

L
2 
T
2
f 

T  2
L
g
f 
1
2
1
T
g
L
3. A 20.0 g pendulum on an 80.0 cm string is pulled back 5.0 cm and then swings. Determine its:
a) angular frequency
d) maximum velocity
b) displacement function
e) maximum acceleration
c) velocity function
8
Energy and Simple Harmonic Motion
IB 12
A mass oscillates back and forth on a spring. Analyze the energy in the system at each location.
When the mass is at its mean position . . .
When the mass is at any position . . .
9
IB 12
1. A 2.00 kg mass is oscillating on a spring and its
displacement function is shown.
a) At what time(s) does the mass have the most kinetic energy?
b) Determine the maximum kinetic energy of the mass.
c) At what time(s) does the mass have maximum potential energy? Determine this value.
d) What is the total energy of the system at 1.5 seconds?
e) Determine the kinetic and potential energy of the system at 1.5 seconds.
10
Energy Graphs and SHM
IB 12
Energy-Displacement Functions
EP 
1
m 2 x 2
2
1
EK  m 2  x0 2  x 2 
2
1
ET  m 2 x0 2
2
Energy-Time Functions
1
1
2
m 2 x 2  m 2  x0 sin t 
2
2
1
EP  m 2 x0 2 sin 2 t
2
EP  sin 2 t
EP 
1 2 1
mv  m(v0 cos t ) 2
2
2
1
EK  mv0 2 cos 2 t
2
EK  cos 2 t
EK 
Note that in simple harmonic motion, the energy of a system is proportional to:
1. mass
2. amplitude squared
3. frequency squared
11
Damping
Resonance
in Oscillations
IB 12
Natural
Damping:
Frequency
a dissipative
of Vibration:
force actswhen
on a asystem
systeministhe
displaced
oppositefrom equilibrium and allowed to oscillate
freely,
direction
it will
to do
theso
direction
at its natural
of motion
frequency
of theof
oscillating
vibrationparticle
Forced Oscillations – a system may be forced to oscillate at any given frequency by an outside driving
force that is applied to it
Effect of damping: system loses energy and amplitude (energy α ampl2)
Resonance – a transfer of energy in which a system is subject to an oscillating force that matches the
natural frequency of the system resulting in a large amplitude of vibration
Sketch the displacement function for a system without and with damping.
Amplitude vs. frequency graph
for forced oscillations
Factors that affect the frequency response and
sharpness of curve:
1) frequency of driving force
2) natural frequency of system
Without Damping
3) amplitude of driving force
With Damping
4) amount of damping
Degrees of Damping
Light damping (under-damping):
small resistive force so only a small
percentage of energy is removed each
1. –Sketch
frequency
for a lightly
cycle
period the
is not
affectedresponse
– can take
damped
whose
many cycles
for system
oscillations
tonatural
die out frequency is
20 Hz that experiences forced oscillations.
eg. – car shock absorbers
Heavy damping (over-damping): large resistive force –
can completely prevent any oscillations from taking place
– takes a long time for object to return to mean position
eg.- oscillations in viscous fluid
Critical damping: intermediate resistive force so time
taken for object to return to mean position is minimum –
minimal or no “overshoot”
eg. – electric meters with pointers, automatic door closers
12
IB 12
13
Waves
IB 12
Both pulses and traveling waves:
Pulse – single oscillation or disturbance
Continuous traveling wave – succession of
oscillations (series of periodic pulses)
Mechanical Waves: require a medium to transfer energy
transfer energy though there is
no net motion of the medium
through which the wave passes.
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
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).
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
Examples: light, violin
and guitar strings, ropes,
earthquake S waves
Compression: region where particles of medium are close together
Rarefaction: region where particles of medium are far apart
Note that transverse mechanical waves cannot propagate (travel) through a gas – only longitudinal waves can.
Displacement (x, meters) – distance in a particular direction of a particle
from its mean position
Amplitude (A or x0, meters) – maximum displacement from the mean
position
Period (T, seconds) – time taken for one complete oscillation
- time for one complete wave (cycle) to pass a given point
Frequency (f, Hertz) – number of oscillations that take place per unit time
Wavelength (λ, meters) – shortest distance along the wave between two points that are in phase
-the distance a complete wave (cycle) travels in one period.
Compare the motion of a single particle to
the motion of the wave as a whole (the
motion of the energy transfer).
Particle Speed:
not constant speed = SHM
Wave Speed:
constant speed
v = d/t
in time t = 1 period:
Average speed: v = d/t
in time t = 1 period:
v = 4A/T
v = λ/T
v = (1/T) λ
v=fλ
14
1.
Motion of the Wave
2.
Motion of a Particle
λ
IB 12
T
Control variable: in one medium - wave speed
Control variable: across a boundary - frequency
Wave speed depends on the properties of the medium, not
how fast the medium vibrates. To change wave speed, you
must change the medium or its properties.
As a wave crosses a boundary between two different media, the
frequency of a wave remains constant not the speed or
wavelength.
Light:
Sound:
Wavelength is proportional to speed
Wavelength is inversely proportional to frequency
Waves in Two Dimensions
Wavefront – line (or arc) joining
neighboring points that have the same
phase or displacement
Ray – line indicating direction of
wave motion (direction of energy
transfer).
At great distances,
the wavefronts are
approximately
parallel and are
known as plane
waves.
Rays are perpendicular to wavefronts.
Intensity - power received per unit area
Formula: I = P/A
1. 12 x 10-5 W of sound power pass
through each surface as shown. Surface
1 has area 4.0 m2 and surface 2 is twice
as far away from the source. Calculate
the sound intensity at each location.
Symbol: I
Units: W/m2
NOTE: for a wave, its intensity is
proportional to the square of its amplitude
15
Reflection and Refraction
IB 12
Sketch the incident and reflected rays as well as the reflected wavefront.
Law of Reflection
θi
The angle of incidence is
equal to the angle of
reflection when both
angles are measured with
respect to the normal line
(and the incident ray,
reflected ray and normal
all lie in the same plane).
θr
Mirror
Refraction: the change in direction of a wave (due to a change in speed) when it crosses a boundary between
two different media at an angle
Air to glass:
Fast to slow = bends toward the normal
n1 < n2
v1 > v2
λ1 > λ2
Glass to air:
Slow to fast = bends away from the normal
n1 > n2
v1 < v2
λ1 < λ2
Refractive Index (Index of refraction)(n):
ratio of sine of angle of incidence to sine of angle
of refraction, for a wave incident from air
sin 1 v1
n

sin  2 v2
c
n
v
Snell’s Law: the ratio of the sine of the angle of
incidence to the sine of the angle of refraction is a
constant, for a given frequency
sin 1 n2 v1 1
  
sin  2 n1 v2 2
n1 sin 1  n2 sin 2
16