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Holt Physics Chapter 11
Vibrations and Waves
Simple Harmonic Motion
Simple Harmonic Motion – vibration
about an equilibrium position in which
a restoring force is proportional to
the displacement from equilibrium.
Springs, pendulums, etc…
The spring force always pushes or
pulls the mass back toward its original
equilibrium position (sometimes called
a restoring force).
See figure 12-1, page 438
Vibrations and Waves
Simple Harmonic Motion
Example:
gravity acting
on a mass
hanging from a
string.
Example:
gravity acting
on a mass
hanging from a
spring.
*
When the restoring force is linearly
proportional to the amount of the
displacement from equilibrium, the force is
said to be a Hooke’s Law force.
Hooke’s Law
For mass-spring systems
Formula Felastic = -kx
 Spring force=-(spring constantXdisplacement)
The negative sign in the equation refers
to the force being opposite to the
displacement.
k is a measure of stiffness and the unit
is N/m
Remember, stretching or compressing
a spring stores elastic potential
energy that can be converted to
kinetic energy.
Book Example, page 366
If a mass of 0.55 kg attached to
a vertical spring stretches the
spring 2.0 cm from its original
equilibrium position, what is the
spring constant?
m=0.55 kg
g=9.81m/s2
x=-2.0cm = -0.020m
k=?
If a mass of 0.55 kg attached to a vertical spring
stretches the spring 2.0 cm from its original
equilibrium position, what is the spring constant?
m=0.55 kg
g=9.81m/s2
Felastic
Fg
x=-2.0cm = -0.020m
k=?
Felastic = |mg|
Felastic = -kx
mg=-kx
k=-mg/x
k=-(0.55kg)(9.81m/s2)/(-.020m)
k=270N/m
Example 2. How much force will a vertical spring
with a spring constant of 350 N/m exert when
it is stretched to 22 cm past its equilibrium
point?
FE  - kx  - (350 N/m)(-0.22 m)  77 N
Example 3. When a 1.5 kg mass is
suspended from a spring hanging vertically,
How much does it stretch if the spring constant
Is 98N/m?
Remember: The spring force (FE) is equal and
opposite to the weight (Fw).
FE  | Fw |
- kx  mg
- mg - (1.5 kg)(9.81 m/s 2 )
x

 - 0.15m
k
98 N/m
Practice page 367
Simple Harmonic Motion, pg. 371
Some basic elements of simple
harmonic motion
At equilibrium position,
velocity = maximum
At maximum displacement, the
spring force and acceleration =
maximum
This motion repeats itself in
simple harmonic motion.
See page 371 for a comparison of
spring-mass systems and pendulum
motion.
The pendulum stores gravitational
potential energy, and the spring-mass
system stores elastic potential energy.
 See page 370
Measuring Simple Harmonic Motion
Amplitude – maximum displacement
from equilibrium position.
 Unit is radians or meters
Period (T) time it takes to complete
one cycle of motion.
 Unit is seconds
Frequency (f) is the number of cycles
per second. T=1/f or f=1/T
 Unit for f is s-1 or Hertz (Hz)
The time for one cycle is the period, T.
The number of cycles per second is the frequency.
Calculating the Period of a
Simple Pendulum
T = 2
L/g
L = length of pendulum
g = free-fall
T = Period of pendulum
Amplitude and mass do not affect
the period of a pendulum
Measuring
Simple Harmonic Motion
Example 1. A chandelier exhibiting simple
harmonic motion has a period of 1.3 s. How
long is the cord it hangs from?
2
2
2
T g (1.3 s) (9.81 m/s )
L 2 
 0.42 m
2
4
4
Example 2. An astronaut on a foreign
planet notes that a pendulum with a length
of 2.5 m has a period of 1.4 s. Find g on
this planet.
L4
(2.5 m)(4  )
2
g 2 
 50 m/s
2
T
(1.4 s)
2
2
Example 3. Find the period (T) and the
frequency (f) for wave (B) below.
From the image, we see that
the period, or time to complete
one cycle, is 0.60 s.
1
1
f  
 1.66 Hz
T 0.60 s
Example 4. Find the period (T) and the
frequency (f) for wave (b) below.
From the image, we see that
the period, or time to complete
one cycle, is 0.80 s.
1
1
f  
 1.25 Hz
T 0.80 s
Homework: Practice page 375 and
Formative Assessment pg. 371
Proof of
Earth’s
rotation?
Foucault
Pendulum
Mass-Spring Systems
The period of a mass-spring
system depends on mass and
spring constant.
T = 2 m/k
Book example
The body of a 1275 kg car is supported on a frame by
four springs. Two people riding in the car have a
combined mass of 153 kg. When driven over a pothole
in the road, the frame vibrates with a period of 0.840 s.
For the first few seconds, the vibration approximates
simple harmonic motion. Find the spring constant of a
single spring.
m car=1275 kg
T=0.840s
mtotal
4 wheels
=
mpeople = 153 kg
k=?
4 wheels
1275kg + 153 = 357kg/wheel
4
T = 2 m/k
Square both sides…and solve for k
k=4 2m/T2
k=42357kg/(0.84s)2
k=2.00X104N/m
Practice and Formative
Assessment pg. 377
Example 1. A mass of 0.50 kg is attached
to a spring and is set into vibration with a
period of 0.42 s. What is the spring constant
of the spring?
m
Ts  2
k
4 m 4 (0.50 kg)
k

 110 N/m
2
2
T
(0.42s)
2
2
Example 2. A 0.60 kg mass attached to
a vertical spring stretches the spring 0.25 m.
A)Find the spring constant.
B) The system is now placed on a horizontal
surface and set to vibrate. Find the period.
FE  - kx
m
Ts  2
k
- mg - (0.60 kg)(9.81 m/s 2 )
A) k 

 24 N/m
x
(-0.25 m)
m
0.60 kg
B) Ts  2
 2
 0.99 s
k
24 N/m
What is a wave ?
A definition of a wave:
A wave is a traveling disturbance that
transports energy but not matter.
Examples:
Sound waves (air moves back & forth)
Water waves (water moves up & down)
Light waves (do not require a medium)
Mechanical waves require a medium
 Electromagnetic waves (like light) do
not
Measuring Waves
Frequency (f) – number of cycles per
second
Period – time in seconds required for
one complete cycle of motion
Velocity(v) – speed of wave in m/s
Formulas
v = /T
f = 1/T
v = f 
Wave Properties
The distance  between
identical points on the wave.
 Amplitude: The maximum displacement
 Wavelength:
A of a point on the wave.
Wavelength

Amplitude A
A
Example 1. A transverse wave has a
wavelength of 0.80 m and a period of 0.25 s.
Find the speed of the wave.

0.80 m
 
 3.2 m/s
T 0.25 s
Example 2. A transverse wave has a
wavelength of 40 cm and a frequency of
30.0 Hz. Find the speed of the wave.
  f  (30.0 Hz)(0.40 m)  12 m/s
Example 3. An astronaut broadcasts radio
waves with a speed (c) of 3.00 x 108 m/s
and a frequency (f) of 92.0 MHz. Calculate
the wavelength of these waves.
8
c 3.00 x 10 m/s
 

3.26
m
6
f 92.0 x 10 Hz
Example 4. A certain laser emits light of
wavelength 633 nm. What is the
Frequency of this light in a vacuum?
c
8
3.00 x 10 m/s
14
f 
 4.74 x 10 Hz
-9
 633 x 10 m
Homework pg. 383 practice and
384 formative assessment
Wave Types
Pulse wave – single, non-periodic
disturbance.
Periodic wave – a wave whose
source is some kind of periodic
motion.
Transverse: The medium
oscillates perpendicular to the
direction the wave is moving.
Water (more or less)
String waves
Longitudinal: The medium oscillates
in the same direction as the wave is
moving.
Sound
Wave Interactions
Because mechanical waves are not
matter, but are displacements of
matter, they can occupy the same
space at the same time.
Superposition – combination of
two overlapping waves
Wave Interactions
Interference – interaction of
waves as a result of them passing
through each other.
When the displacements are in the
same direction, this produces
Constructive Interference (see figure
4.3, pg. 386)
The resultant amplitude of the
resultant wave is the sum of the
amplitudes of the individual waves.
Each wave maintains it’s own
characteristics after the interference
When the displacements are in the
opposite direction, this produces
Destructive Interference (see figure
4.4, pg. 387)
The resultant amplitude of the
resultant wave is the difference
between the two pulses.
If the difference is zero, then it is
called Complete Destructive
Interference (fig 4.5)
Again, each wave maintains it’s own
characteristics after the interference
Wave Motion
Waves and springs animation.
Wave pulse traveling down
string, striking “fixed”
boundary.
Wave pulse traveling down
string, striking “free” boundary.
Reflection
At a free boundary (fig 4.6a pg.
388) waves are reflected.
At a fixed boundary (fig 4.6b pg.
388) waves are reflected and
inverted.
Fixed Boundary…
Free Boundary…
pulse
travels down
string/slinky
reflects
(& inverts)
meets another
pulse on the
way back
Standing Waves
A standing wave is produced when two
waves with the same amplitude,
wavelength and frequency travel in
opposite directions and interfere (fig.
4.7, pg. 389).
Node – point on a standing wave that
always undergoes complete destructive
interference and therefore is
stationary.
Antinode – a point in a standing wave
half-way between two nodes at which
the largest amplitude occurs.
If both ends of the string
are fixed, a standing wave
results.
If the string is “infinitely”
long and the up-and-down
pulses repeat, we get a
traveling wave.
Only certain frequencies will
produce standing wave patterns.
The ends have to be nodes, so this
includes a wavelength of 2L (twice
the length of the string or tube
where the vibration is taking place),
L, and 2/3 L. (see figure 4.8, pg.
390)
Formative Assessment pg. 390
Review Problems