Introduction to Physics

1
Waves and Sound
2
Simple Harmonic Motion
period, T
time to complete one cycle
(s)
frequency, f
of cycles per second (Hz)
3
Dx 0
Felas 0
Dxmax
v 0
Dx max
Felas max
Dxmax
period of a mass-spring system
m mass (kg)
k spring constant (N/m)
4
A 5.5 kg cat is attached to a fixed horizontal
spring of stiffness 22.8 N/m and is set in motion
on a frictionless surface. Find the period of
motion of the cat.
3.1 s
a 240 g mouse, with the same spring and surface.
0.64 s
5
What stiffness must a spring have so that the
period of the mouses motion is the same as that
of the cat?
Ballpark answer
Need a less stiff spring k lt 22.8 N/m.
0.99 N/m
6
A 1275 kg car carries two passengers with a
combined mass of 153 kg. The car has four shock
absorbers, each with a spring constant of 2.0 x
104 N/m. Find the frequency of the vehicles
motion after it hits a pothole.
1.2 Hz
7
restoring force
acts to move an object back to equilibrium
simple harmonic motion (SHM)
As displacement increases, so does Frestore.
And when Dx 0
Frestore 0.
For a mass-spring system, Hookes law applies
Frestore Felas k Dx
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energy
the ability to do work
kinetic energy, KE energy of mass m having
velocity v
KE ½ m v2
m (kg) v (m/s)
KE max. at eq. pos. KE 0 at Dxmax.
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potential energy, PE stored energy
For a spring with spring constant k and stretch
Dx
PEelas ½ k (Dx)2
k (N/m) Dx (m)
For m-s sys., PEelas is max at Dxmax and 0 at eq.
pos.
For a mass m at a height h above a reference
line
PEg m g h
m (kg) h (m) g 9.81 m/s2
For pend., PEg is max at Dxmax and min. at eq.
pos.
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amplitude, A maximum displacement from
equilibrium
frictionless
Period T is not affected by amplitude A.
Energy of a Mass-Spring System
total energy
PEelas
KE
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The Pendulum
For Q lt 15o, a simple pendulum approximates SHM.
Energy of a Simple Pendulum
total energy
PEg
KE
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period of a simple pendulum
Period T is independent of mass and amplitude.
The period of a pendulum is 5.2 s. Find
A. its length
6.7 m
B. the mass of the bob
NOT ENOUGH INFORMATION
13
Waves
Waves transmit energy, not matter.
medium the matter through which the energy of
mechanical waves moves
14
crest
amplitude A
trough
wavelength l
For a transverse wave
15
particles of medium move // to direction of wave
travel
longitudinal (compressional) wave
compression
rarefaction
l
pulse wave
a single vibration
periodic wave
rhythmic, repeated vibrations
16

• Wave Reflection

fixed boundary
free boundary
waves are reflected and inverted
waves are reflected and upright
17
Wave Interference
Two waves (unlike two objects) can occupy the
same place at the same time. This condition is
called interference.
constructive interference
destructive interference
displacements are in same direction
displacements are in opposite directions
A1
A2
A1
A2
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Wave Velocity
v (m/s) l (m) T (s) f (Hz)
Equation
A water wave of wavelength 8.5 m washes past a
boat at anchor every 4.75 seconds. Find the
waves velocity.
1.8 m/s
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The velocity of any mechanical wave depends only
on properties of medium through which it travels.
e.g., string tension, water depth, air
temperature, material density, type of material
An empirical equation for the velocity of sound
in air
Ta air temp. in oC
vsound 331 0.6Ta
20
Standing Waves
incident and reflected waves interfere so
that antinodes have a max. amplitude, while
nodes have zero amplitude
On a string, nodes remain motionless
antinodes go from max. () to max. ()
displacement.
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n 1 1st harmonic
l1 2 L
(fundamental)
n 2 2nd harmonic
l2 L
(1st overtone)
n 3 3rd harmonic
l3 2/3 L
(2nd overtone)
n 4 4th harmonic
l4 ½ L
(3rd overtone)
wavelength of nth harmonic on a string
(n 1,2,3,)
22
Waves travel along a 96.1 cm guitar string at 492
m/s. Find the fundamental frequency of the
string.
l1 2 L
2 (0.961 m)
1.922 m
v f l
and v f1 l1
256 Hz
Find the frequency of the 5th harmonic.
l5 2/5 L
0.3844 m
1280 Hz
fn n f1
frequency of the nth harmonic
23
Standing Waves in Open Tubes
L
n 1
n 2
n 3
l1 2 L
l2 L
l3 2/3 L
wavelength of nth harmonic of an open tube
(n 1,2,3,)
24
Closed Tubes
L
n 1
n 2
n 3
l1 4 L
l2 2 L
l3 4/3 L
wavelength of nth harmonic of a closed tube
(n 1,3,5,)
(even harmonics are not present)
25
Find the fundamental frequency for an open tube
of length 1.24 m. Assume the air temperature
to be 20.0oC.
l1 2 L
2 (1.24 m)
2.48 m
v f l
and v f1 l1
v ?
vsound 331 0.6Ta
vsound 331 0.6(20) 343 m/s
138 Hz
26
Find fundamental frequency for a closed tube of
length 1.24 m. Air temp. is 20.0oC.
l1 4 L
4 (1.24 m)
4.96 m
v f l
and v f1 l1
69.0 Hz
one octave
(three octaves)
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Sound
compression
high pressure / high density
rarefaction
low pressure / low density
audible frequencies (human hearing)
infrasonic
ultrasonic
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Fundamental frequency determines pitch.
high pitch
high f
short l
low pitch
long l
low f
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Number and intensity of an instruments
harmonics give it its unique sound quality, or
________.
timbre
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The Doppler Effect
Relative motion between wave source and
observer causes a change in the ____________
frequency.
observed
v 0
femitted
fobserved
(higher)
femitted
femitted
fobserved
(lower)
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Other examples of Doppler effect
race cars
police radar
(red-shifted)
dolphins (echolocation)
expansion of universe
32

• Traveling Very Fast

vbug 0
vbug lt vwave
bow wave
vbug gt vwave
vbug vwave
33
supersonic faster than sound (vs. subsonic)
shock wave
a 3-D bow wave
sonic boom
caused by high-pressure air, not roaring engine
lion tamers whip
cracking bullets
The Matrix
34
Sound Intensity
If a pianos power output is 0.302 W, find the
sound intensity at a distance of
A. 1.0 m
0.024 W/m2
B. 2.0 m
0.0060 W/m2
35
Intensity is related to volume (or relative
intensity)
--
how loud we perceive a sound to be
-- measured in decibels (dB)
A difference of 10 dB changes the sound
intensity by a factor of 10 and the volume by a
factor of 2.
50 dB ? 40 dB
60 dB ? 90 dB
half as loud
8X louder
1/10 as intense
1000X more intense
36
Beats
alternating loud-and-soft sounds resulting from
interference between two slightly- different
frequencies
Equation
f1 16 Hz
f1 16 Hz
f2 18 Hz
f2 17 Hz
fbeat 1 Hz
fbeat 2 Hz
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Forced Vibrations and Resonance
natural frequency
the frequency at which an object most easily
vibrates
forced vibration
a vibration due to an applied force
resonance
occurs when a force is repeatedly applied to an
object AT the objects natural frequency
-- result of resonance
large amplitude
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Examples
swing
shattering crystal wine glasses
Tacoma Narrows Bridge (1940)
British regiment (Manchester, 1831)
aeolian harps
The wind in the wires made a tattletale
sound, as a wave broke over the railing
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vsound 331 0.6Ta
Frestore Felas k Dx
KE ½ m v2
PEelas ½ k (Dx)2
fn n f1
PEg m g h
40
h
h