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Waves
andSound
Simple Harmonic Motion
period, T:
frequency, f:
time to complete one cycle
# of cycles per second (Hz)
f1 T
T1
f
(s)
wall equilibrium position
frictionless floor
x = 0 Felas = 0
xmax
–xmax
v = 0 x = max
Felas = max
period of a mass-spring system:
km
2 T m = mass (kg) k = spring constant (N/m)
A 5.5 kg cat is attached to a fixedhorizontal spring of stiffness 22.8 N/mand is set in motion on a frictionlesssurface. Find the period of motion of…
…the cat.
…a 240 g mouse, with the same spring and surface.
km
2 T m
N 22.8
kg 5.52 = 3.1 s
kg 0.24 g 10 x 1
kg 1 g 240 m 3
km
2 T m
N 22.8
kg 0.242 = 0.64 s
= 0.99 N/m
What stiffness must a spring have so that the periodof the mouse’s motion is the same as that of the cat?
km
2T
km
2T k
m4T
2
2
m4 kT 22
2
2
Tm4
k
22
1.3
24.04
s
kg
Ballpark answer: Need a “less stiff” spring; k < 22.8 N/m.
= 1.2 Hz
A 1275 kg car carries two passengerswith a combined mass of 153 kg. Thecar has four shock absorbers, each witha spring constant of 2.0 x 104 N/m. Findthe frequency of the vehicle’s motionafter it hits a pothole.
N/m) 10 x 4(2.0kg)1531275(
2
1
4
km
2 T
km2
1 f
restoring force:
simple harmonic motion (SHM):
For a mass-spring system,Hooke’s law applies:
acts to move an objectback to equilibrium
Frestore x
Frestore = Felas = k x
As displacement increases, so does Frestore.
And when x = 0… Frestore = 0.
KE = ½ m v2
energy:
kinetic energy, KE: energy of mass m having velocity v
m (kg); v (m/s)
the ability to do work
KE = max. at eq. pos.; KE = 0 at xmax.
PEelas = ½ k (x)2
potential energy, PE: stored energy
For a spring with spring constant k and “stretch” x:
k (N/m); x (m)
For a mass m at a height h above a reference line:
m (kg); h (m);g = 9.81 m/s2
For m-s sys., PEelas is max at xmax and 0 at eq. pos.
PEg = m g h
For pend., PEg is max atxmax and min. at eq. pos.
amplitude, A: maximum displacement from equilibrium
Energy of a Mass-Spring System
frictionless
A A
Period T is not affected by amplitude A.
Energy (J)
total energy
PEelas
KE PEg
km
2T
The Pendulum
For < 15o, asimple pendulum
approximates SHM. mass mof bob
length Lamplitude
Energy of a Simple Pendulum
Energy (J)
total energy
PEg
KE PEelas
period of a simple pendulum:
The period of a pendulum is 5.2 s. Find…
A. …its length
B. …the mass of the bob
g L
2 T
Period T is independent of mass and amplitude.
g L
2 T g L
2T
gL
4T
2
2
2
2
4gT
L
2
22
4
s) 2.5( s
m 9.81
= 6.7 m
NOT ENOUGH INFORMATION
Waves
vibrations moving through space and time
medium: the matter through which the energy of mechanical waves moves
Waves transmit energy, not matter.
transverse waves: particles of medium move to direction of wave travel
For a transverse wave:
amplitude A crest
wavelength
trough
Energy Amplitude2
longitudinal(compressional) wave:
particles of mediummove // to directionof wave travel
rarefaction compression
pulse wave:
periodic wave:
a single vibration
rhythmic, repeated vibrations
Wave Reflection
fixed boundary free boundary
v
v
v
v
A A
waves are reflectedand inverted
waves are reflectedand upright
Wave Interference
Two waves (unlike two objects) can occupythe same place at the same time. This
condition is called interference.
constructive interference: destructive interference: displacements arein same direction
displacements arein opposite directions
A1 A2
A1 + A2
A1 A2
A2
A1
A1 – A2
A2
A1
= 1.8 m/s
Wave Velocity
Equation: v (m/s); (m);T (s); f (Hz)
f
T v
A water wave of wavelength 8.5 mwashes past a boat at anchor every4.75 seconds. Find the wave’s velocity.
T
v
s 4.75m 8.5
vsound = 331 + 0.6Ta
The velocity of any mechanical wavedepends 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 thevelocity of sound in air:
Ta = air temp. in oC
Standing Waves
incident and reflected waves interfere so thatantinodes have a max. amplitude, while nodeshave zero amplitude
On a string, nodes remain motionless; antinodesgo from max. (+) to max. (–) displacement.
wavelength of nth
harmonic on a string:
n = 1; 1st harmonic (fundamental)
n = 2; 2nd harmonic(1st overtone)
n = 3; 3rd harmonic(2nd overtone)
n = 4; 4th harmonic(3rd overtone)
nL 2
n (n = 1,2,3,…)
1 = 2 L
2 = L
3 = 2/3 L
4 = ½ L
L
fn = n f1
Waves travel along a 96.1 cmguitar string at 492 m/s. Find thefundamental frequency of the string.
Find the frequency ofthe 5th harmonic.
frequency of the nth harmonic:
1 = 2 L = 2 (0.961 m) = 1.922 m
v = f and v = f1 1
11
v f
m 922.1
m/s 492 = 256 Hz
5 = 2/5 L = 0.3844 m
55
v f
m .38440
m/s 492 = 1280 Hz
nL 2
n
Standing Waves in Open Tubes
wavelength of nth
harmonic of an open tube:
n = 1
L
1 = 2 L n = 2 2 = L
n = 3 3 = 2/3 L
nL 2
n (n = 1,2,3,…)
Closed Tubes
wavelength of nth
harmonic of a closed tube:
n = 1
L
1 = 4 L n = 2
2 = 2 Ln = 3
3 = 4/3 L
nL 4
n (n = 1,3,5,…)
(even harmonicsare not present)
Find the fundamental frequencyfor an open tube of length 1.24 m.Assume the air temperature tobe 20.0oC.
1.24 m
nL 2
n
1 = 2 L = 2 (1.24 m) = 2.48 m
v = f and v = f1 1
11
v f
m 48.2m/s 343
f1 = 138 Hz
v = ?vsound = 331 + 0.6Ta
vsound = 331 + 0.6(20) = 343 m/s
Find fundamental frequency fora closed tube of length 1.24 m.Air temp. is 20.0oC.
nL 4
n
1 = 4 L = 4 (1.24 m) = 4.96 m
v = f and v = f1 1
11
v f
m 96.4
m/s 343 = 69.0 Hz
1.24 m
(same as prev. prob.)
doubling (or halving)of frequency
=
100 Hz 200 Hz 400 Hz 800 Hz (three octaves)
one octave
Sound
compression:
rarefaction:
20 Hz 20,000 Hz
high pressure / high density
low pressure / low density
audiblefrequencies
(humanhearing)infrasonic ultrasonic
Fundamental frequency determines pitch.
high pitch high f =
low f =
= short
low pitch = long
Number and intensity of an instrument’s harmonicsgive it its unique sound quality, or ________.timbre
f1 f2 f3 f4 f1 f2 f3 f4
f1 f2 f3 f4 f1 f2 f3 f4
The Doppler Effect Relative motion between wave source and observercauses a change in the ____________ frequency. observed
fobserved
fobserved
(higher)
(lower)
femitted femitted
femitted
v = 0
Other examples of Doppler effect:
police radar
race cars
expansion of universe dolphins (echolocation)
R O Y G B V
Sun
most stars (“red-shifted”)
Traveling Very Fast
vbug = 0 vbug < vwave
vbug = vwave vbug > vwave
bow wavewave barrier
supersonic: “faster than sound” (vs. subsonic)
sonic boom:
shock wave: a 3-D bow wave
caused by high-pressureair, not roaring engine
cracking bulletslion tamer’s whip
The Matrix
= 0.024 W/m2
Sound Intensity
If a piano’s power output is 0.302 W,find the sound intensity at a distance of…
A. …1.0 m
B. …2.0 m
r 4
P I 2
2r 4P
I
2m) (1.0 4 W0.302
= 0.0060 W/m22r 4
P I
2m) (2.0 4
W0.302
Intensity is related to volume (or relative intensity):
--
-- measured in decibels (dB)
A difference of 10 dB changesthe sound intensity
by a factor of 10 andthe volume by a factor of 2.
50 dB 40 dB 60 dB 90 dB
how loud we perceive a sound to be
half as loud 1/10 as intense
8X louder
1000X more intense
Beats
alternating loud-and-soft sounds resultingfrom interference between two slightly-different frequencies
Equation: f - f f 21beat
1 second elapses 1 second elapses
f 1 =
16
Hz
f 2 =
18
Hzf 1 =
16
Hz
f 2 =
17
Hz
fbeat = 1 Hz fbeat = 2 Hz
Forced Vibrations and Resonance
natural frequency:
forced vibration:
resonance:
-- result of resonance =
the frequency at which anobject most easily vibrates
a vibration due toan applied force
occurs when a force is repeatedlyapplied to an object AT the object’s natural frequency
large amplitude
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…”
f1 T
T1
f
km
2 T
Frestore = Felas = k x
KE = ½ m v2
PEelas = ½ k (x)2
PEg = m g h
g L
2 T
f
T v
vsound = 331 + 0.6Ta
nL 2
n
fn = n f1
nL 4
n
r 4
P I 2
f - f f 21beat
h
h