Transcript Lecture 1: Rotation of Rigid Body

Chapter 14: Wave Motion
Types of mechanical waves
 Mechanical
waves
• are disturbances that travel through some material or substance
called medium for the waves.
• travel through the medium by displacing particles in the medium
• travel in the perpendicular to or along the movement of the
particles or in a combination of both
transverse waves:
longitudinal waves:
waves in a string etc. sound waves etc.
waves in water etc.
Types of mechanical waves (cont’d)
 Longitudinal
and transverse waves
sound wave = longitudinal wave
C = compression air compressed
R = rarefaction
air rarefied
Types of mechanical waves (cont’d)
 Longitudinal-transverse
waves
Types of mechanical waves (cont’d)
 Periodic
waves
• When particles of the medium in a wave undergo periodic
motion as the wave propagates, the wave is called periodic.
wavelength
l
A amplitude
t=0
x=0
t=T/4
t=T
period
x
Mathematical description of a wave
 Wave
function
• The wave function describes the displacement of particles
in a wave as a function of time and their positions:
y  y( x, t ) ; y is displacement at x, t
• A sinusoidal wave is described by the wave function:
sinusoidal wave moving in
y ( x, t )  A cos[ (t  x / v )]
+x direction
angular frequency  A cos[ ( x / v  t )]
velocity of wave, NOT of
 A cos2 f ( x / v  t ) particles of the medium
  2 f
fl  v
 A cos2 ( x / l  t / T )
wavelength
y ( x, t )  A cos[(t  v / x)]
period
f  1/ T
sinusoidal wave moving in
-x direction v->-v
phase velocity
Mathematical description of a wave
(cont’d)
 Wave
function (cont’d)
y ( x, t )  A cos2 ( x / l  t / T )
l
wavelength
 y( x  l , t )
 y ( x, t  T )
t=0
x=0
t=T/4
t=T period
x
Mathematical description of a wave (cont’d)
 Wave
number and phase velocity
wave number:
k  2 / l
y ( x, t )  A cos(kx  t )
phase
The speed of wave is the speed with which we have to
move along a point of a given phase. So for a fixed phase,
kx  t  const .
dx / dt   / k  v
phase velocity
y( x, t )  A cos(kx  t )  A cos[k ( x  vt)]
Mathematical description of a wave (cont’d)
 Particle
velocity and acceleration in a sinusoidal wave
y ( x, t )  A cos(kx  t )
u in textbook
v y ( x, t )  y ( x, t ) / t  A sin(kx  t )
velocity
a y ( x, t )   2 y ( x, t ) / t 2   2 A cos(kx  t )
  2 y ( x, t )
Also
acceleration
2 y( x, t ) / x2  k 2 Acos(kx  t )  k 2 y( x, t )
 2 y ( x, t ) / x 2  (k 2 /  2 ) 2 y ( x, t ) / t 2
  y ( x, t ) / v t
2
2
2
wave equation
Mathematical description of a wave (cont’d)
 General
solution to the wave equation
 2 y ( x, t ) k 2  2 y ( x, t )  2 y ( x , t )
 2
 2 2
2
2
x

t
v t
Solutions:
y( x, t )  f ( x  vt)
such as
wave equation
cos(kx  t )
The most general form of the solution:
y( x, t )  f ( x  vt)  g ( x  vt)
Speed of a transverse wave
 Wave
speed on a string
F2 y
•The mass of the segment is m  x.
F1x  F
F1
F2
•Consider a small segment of string whose
length in the equilibrium position is  x.
F1 y
F2 x  F • The x component of the force (tension) at both
 x x  x
x
Newton’s 2nd law
ends have equal in magnitude and opposite in
direction because this is a transverse wave.
• F1y / F  (y / x) x , F2 y / F  (y / x) xx
• The total y component of the forces is:
Fy  F1 y  F2 y  F [(y / x ) x x  (y / x ) x ]
 x( 2 y / t 2 )
mass
acceleration
Speed of a transverse wave (cont’d)
 Wave
speed on a string (cont’d)
F2 y
F
F1
• The total y component of the forces is:
F2
F
F1 y
x
x
Fy  F1 y  F2 y  F [(y / x ) x x  (y / x ) x ]
 x( 2 y / t 2 )
[(y / x) x x  (y / x) x ] / x 
(  / F )( 2 y / t 2 )
x  0
2 y / x2  (  / F )(2 y / t 2 )
v  F / 
( restoring force) /(inertia)
wave eq.
Energy in wave motion
 Total
energy of a short string segment of mass dm  dx
F2 y
F1
F a
F1 y
F2
F
• At point a, the force F1 y does work on the
work done
string segment right of point a.
• Power is the rate of work done :
P( x, t )  F1 y ( x, t )(y ( x, t ) / t )
t 0
  F (y ( x, t ) / x)(y ( x, t ) / t )
y ( x, t )  A cos(kx  t )
(y / x )  kA sin(kx  t )
(y / t )  A sin(kx  t )
x
x
P( x, t )  dE / dt  FkA2 sin 2 (kx  t )
 v 2 A2 sin 2 (kx  t )
  vk, v  F / 
2
Pmax
 F  2 A2 sin 2 (kx  t )
Energy in wave motion (cont’d)
 Maximum
power of a sinusoidal wave on a string:
Pmax  F 2 A2
 Average
power of a sinusoidal wave on a string
• The average of sin2 (kx  t ) over a period:
1
2

2
0
1
sin  d 
2
• The average power:
2
Pave  (1/ 2) F 2 A2
Wave intensity
 Wave
intensity for a three dimensional wave from a point
source:
P
I
2
4r
r1
in units of W/m
4r I  4r I
2
1 1
r2
I1 r22
 2
I 2 r1
2
2 2
2 power/unit area
Wave interference, boundary condition,
and superposition
 The
principle of superposition
• When two waves overlap, the actual displacement of any
point at any time is obtained by adding the displacement
the point would have if only the first wave were present and
the displacement it would have if only the second wave were
present:
y( x, t )  y1 ( x, t )  y2 ( x, t )
Wave interference, boundary condition,
and superposition (cont’d)
 Interference
• Constructive interference (positive-positive or negative-negative)
• Destructive interference (positive-negative)
Wave interference, boundary condition,
and superposition (cont’d)
 Reflection
• Free end
incident wave
reflected wave
y( x, t )  A cos(k x  t )  B cos(k x  t )
For x<xB
xB
At x=xB (y( x, t ) / x) x xB  0  B   A
Vertical component of the force
at the boundary is zero.
Wave interference, boundary condition,
and superposition (cont’d)
 Reflection
(cont’d)
• Fixed end
y( x, t )  A cos(k x  t )  B cos(k x  t )
For x<xB
At x=xB
y( x, t ) x xB  0  B   A
Displacement at the boundary is zero.
Wave interference, boundary condition,
and superposition (cont’d)
 Reflection
(cont’d)
• At high/low density
Wave interference, boundary condition,
and superposition (cont’d)
 Reflection
(cont’d)
• At low/high density
Standing waves on a string
 Superposition
 Superposition
of two waves moving in the same direction
of two waves moving in the opposite direction
Standing waves on a string (cont’d)
 Superposition
of two waves moving in the opposite direction
creates a standing wave when two waves have the same
speed and wavelength.
incident reflected
y( x, t )  y1 ( x, t )  y2 ( x, t )
 A cos(kx  t )  A cos(kx  t )
 2 A(sin k x)(sint )
N=node, AN=antinode
sin kx  0 when kx  n
or x  n / k  nl / 2
(n  0,1,2,..)
Normal modes of a string
 There
are infinite numbers of modes of standing waves
l1 / 2
Ln
fundamental
l
2
( n  1,2,3,...)
l2
first
ln  2 L / n
overtone
3l3 / 2
v
1
fn  n
 f1 
2L
2L
second
overtone
2 l4
third
overtone
fixed end
L
fixed end
F

Sound waves
 Sound
• Sound is a longitudinal wave in a medium
• The simplest sound waves are sinusoidal waves which
have definite frequency, amplitude and wavelength.
• The audible range of frequency is between 20 and 20,000 Hz.
Sound waves (cont’d)
 Sound
wave (sinusoidal wave)
Sinusoidal sound wave function:
y ( x, t )  A cos(kx  t )
Change of volume:
V  S ( y2  y1 )
 S[ y( x  x, t )  y ( x, t )]
y1  y( x, t )
y2  y( x  x, t )
undisturbed
cyl. of air
disturbed
cyl. of air
Pressure:
bulk modulus
S
pressure
B   p( x, t ) /(dV / V )
x
x
dV / V  y( x, t ) / x (V  Sdx)
x+x
p( x, t )   B(y( x, t ) / x)  BkAsin(kx  t )
Pressure amplitude and ear
 Pressure
amplitude for a sinusoidal sound wave
p( x, t )  BkAsin(kx  t )
• Pressure:
• Pressure amplitude:
 Ear
pmax  BkA
Perception of sound waves
 Fourier’s
theorem and frequency spectrum
• Fourier’s theorem:
Any periodic function of period T can be written as
y(t )  n [ An sin(2f nt )  Bn cos(2f nt )]
fundamental freq.
where
f1  1 / T , f n  nf1 (n  1,2,3,...)
• Implication of Fourier’s theorem:
Perception of sound waves
 Timbre
or tone color or tone quality
Frequency spectrum
noise
music
piano
piano
Speed of sound waves (ref. only)
velocity of wave
 The
speed of sound waves in a fluid in a pipe
movable piston
pA
longitudinal momentum carried
by the fluid in motion
pA fluid in
original volume of the fluid in
equilibrium
motion
change in volume of the fluid
in motion
vt
v yt
( p  p ) A v
vy
fluid in motion
Avt
velocity
of fluid
 Avy t
 p
bulk modulus B:
-pressure change/frac. vol. change  ( Av t ) /( Avt)
y
pA
y
vy
( vtA)v y
vy
change in pressure in the fluid
in motion
fluid at rest
boundary moves at speed of wave
p  B
vy
v
Speed of sound waves (ref. only)
(cont’d)
 The
speed of sound waves in a fluid in a pipe (cont’d)
longitudinal impulse = change in momentum
pAt  B
vy
v
speed of a longitudinal
wave in a fluid
 The
At  vtAvy
v
B

speed of sound waves in a solid bar/rod
v
Y

,Y 
Young’s modulus
Speed of sound waves (cont’d)
 The
speed of sound waves in gases
B  p0

bulk modulus of a gas
p0 
In textbook
speed of a longitudinal
wave in a fluid
v
ratio of heat capacities
equilibrium pressure of gas
- P in textbook
(background
p0
pressure).
-  density


M

RT
R
gas constant 8.314472 J/(mol K)
T
temperature in Kelvin
M
molar mass
Sound level (Decibel scale)

Decibel scale
As the sensitivity of the ear covers a broad range of intensities,
it is best to use logarithmic scale:
I
12
2
Definition of sound intensity:   (10 dB) log , I 0  10 W/m
I0
( unit decibel or dB)
Sound intensity in dB
Intensity (W/m2)
Military jet plane at 30 m
140
102
Threshold of pain
120
1
Whisper
20
10-10
Hearing thres. (100Hz)
0
10-12
Standing sound waves

Sound wave in a pipe with two open ends
Standing sound waves

Standing sound wave in a pipe with two open ends
Standing sound waves

Sound wave in a pipe with one closed and one open end
Standing sound waves

Standing wave in a pipe with two closed ends
Displacement
Normal modes

Normal modes in a pipe with two open ends
2nd normal mode
ln
2L
L  n or ln 
( n  1,2,3,...)
2
n
fn  n
v
( n  1,2,3,...)
2L
Normal modes

Normal modes in a pipe with an open and a closed end
(stopped pipe)
ln
4L
L  n or ln 
( n  1,3,5,...)
4
n
fn  n
v
( n  1,3,5,...)
4L
Resonance

Resonance
• When we apply a periodically varying force to a system that can
oscillate, the system is forced to oscillate with a frequency equal
to the frequency of the applied force (driving frequency): forced
oscillation. When the applied frequency is close to a characteristic
frequency of the system, a phenomenon called resonance occurs.
• Resonance also occurs when a
periodically varying force is applied
to a system with normal modes.
When the frequency of the applied
force is close to one of normal
modes of the system, resonance
occurs.
Interference of waves

Two sound waves interfere each other
destructive
constructive
d1
d2
d1  d 2  nl
(constructive)
 (n  1 / 2)l
n  0,1,2,....
(destructive)
Beats

Two interfering sound waves can make beat
Two waves with different
frequency create a beat
because of interference
between them. The beat
frequency is the difference
of the two frequencies.
Beats (cont’d)

Two interfering sound waves can make beat (cont’d)
Suppose the two waves have frequencies f a and f b .
For simplicity, consider two sinusoidal waves of equal intensity:
ya (t )  Asin 2f at ;
yb (t )   Asin 2fbt
Then the resulting combined wave will be:
1
1
ya (t )  yb (t )  2 A sin[ ( 2 )( f a  f b )t ] cos[ ( 2 )( f a  f b )t ]
2
2
1
1
( sin a  sin b  2 sin ( a  b) cos (a  b))
2
2
As human ears does not distinguish negative and positive amplitude,
they hear two max. or min. intensity per cycle, so 2 x (1/2)|fa-fb|=
|fa-fb| is the beat frequency fbeat.
Doppler effect

Moving listener
Source at rest
Listener moving right
Source at rest
Listener moving left
Doppler effect (cont’d)

Moving listener (cont’d)
•The wavelength of the sound wave does not change whether
the listener is moving or not.
• The time that two subsequent wave crests pass the listener
changes when the listener is moving, which effectively changes
the velocity of sound.
freq. listener hears
freq. source generates
velocity of sound at source
velocity of listener
fL
fs
v
vL
fL 
v  vL
l
v  vL

v / fs
- for a listener moving away from
+ for a listener moving towards
the source.
Doppler effect (cont’d)

Moving source
When the source moves
Doppler effect (cont’d)

Moving source (cont’d)
• The wave velocity relative to the wave medium does not
change even when the source is moving.
• The wavelength, however, changes when the source is moving.
This is because, when the source generates the next crest, the
the distance between the previous and next crest i.e. the wavelength changed by the speed of the source.
The source at rest
v
l s
fs
When the source is moving
v vs v  vs
l  
fs fs
fs
+ for a receding source
- for a approaching source
Doppler effect (cont’d)

Moving source and listener
fL 
v  vL
l
- for a listener moving away from
+ for a listener moving towards
the source.
v  vL

fs
v  vs
+ for a receding source
- for a approaching source
The signs of vL and vS are measured
in the direction from the listener L to the
source S.
 Effect of change of source
v  vs
speed
v  vs
Doppler effect (cont’d)

Example 1
• A police siren emits a sinusoidal wave with frequency fs=300 Hz.
The speed of sound is 340 m/s. a) Find the wavelength of the waves
if the siren is at rest in the air, b) if the siren is moving at 30 m/s, find
the wavelengths of the waves ahead of and behind the source.
a) l  v / f s  340m/s /300Hz  1.13m.
b) In front of the siren:
l  (v  vs ) / f s  (340m/s - 30 m/s)/300Hz  1.03m
Behind the siren:
l  (v  vs ) / f s  (340m/s  30 m/s)/300Hz  1.23m
Doppler effect (cont’d)

Example 2
• If a listener l is at rest and the siren in Example 1 is moving away
from L at 30 m/s, what frequency does the listener hear?
fL 

v
340 m/s
fs 
(300 Hz)  276 Hz.
v  vs
340 m/s  30 m/s
Example 3
• If the siren is at rest and the listener is moving toward the left at 30
m/s, what frequency does the listener hear?
v  vL
340 m/s - 30 m/s)
fL 
fs 
(300 Hz)  274 Hz.
v
340 m/s
Doppler effect (cont’d)

Example 4
• If the siren is moving away from the listener with a speed of 45 m/s
relative to the air and the listener is moving toward the siren with a
speed of 15 m/s relative to the air, what frequency does the listener
hear?
v  vL
340 m/s  15 m/s
fL 

v  vs
fs 
340 m/s  45 m/s
(300 Hz)  277 Hz.
Example 5
• The police car with its 300-MHz siren is moving toward a warehouse
at 30 m/s, intending to crash through the door. What frequency does
the driver of the police car hear reflected from the warehouse?
340 m/s
Freq. reaching fW  v f s 
(300 Hz)  329 Hz.
v  vs
340 m/s  30 m/s
the warehouse
Freq. heard by
the driver
v  vL
340 m/s  30 m/s
fL 
fW 
(329 Hz)  358 Hz.
v
340 m/s
Exercises
Problem 1
A transverse wave on a rope is given by:
y( x, t )  (0.750cm) cos [(0.400cm1 ) x  (250s 1 )t ]
(a) Find the amplitude, period, frequency, wavelength, and speed of
propagation. (b) Sketch the shape of the rope at the following values
of t : 0.0005 s, and 0.0010 s. (c) Is the wave traveling in the +x or –x
direction? (d) The mass per unit length of the rope is 0.0500 kg/m.
Find the tension. (e) Find the average power of this wave.
Solution
(a)
y( x, t )  A cos2 ( x / l  t / T )
A=0.75 cm, l=2/0.400 = 5.00 cm, f=125 Hz, T=1/f=0.00800 s and
v=lf=6.25 m/s.
(b) Homework
(c) To stay with a wave front as t increases, x decreases. Therefore the
wave is moving in –x direction.
(d) v  ( F /  ) , the tension is F  v2  (0.050kg / m)(6.25m / s)2  19.6 N.
(e) Pav  (1/ 2) F2 A2  54.2 W.
Exercises
Problem 2
A triangular wave pulse on a taut string travels in the positive +x direction
with speed v. The tension in the string is F and the linear mass density of
the string is . At t=0 the shape of the pulse I given by
y (x,0) 
0
for x   L
h( L  x ) / L
h( L  x ) / L
for  L  x  0
for 0  x  L
0
for x  L
(a) Draw the pulse at t=0. (b) Determine the wave function y(x,t) at all
times t. (c) Find the instantaneous power in the wave. Show that the
power is zero except for –L < (x-vt) < L and that in this interval the
power is constant. Find the value of this constant.
Solution
y
(a)
h
-L
L
x
Exercises
Problem 2 (cont’d)
Solution
(b) The wave moves in the +x direction with speed v, so in the experession
for y(x,0) replace x with –vt:
y ( x, t ) 
0
for x   L
h( L  x  vt ) / L
h( L  x  vt ) / L
for  L  x  0
for 0  x  L
0
for x  L
(c)
y y
P( x, t )   F

x t
 F (0)0  0
for x   L
 F ( h / L)(hv / L)  Fv(h / L)2
for  L  x  0
 F ( h / L)(hv / L)  Fv(h / L)2
for 0  x  L
 F (0)(0)  0
for x  L
Thus the instantaneous power is zero except for –L < (x-vt) < L where
It has the constant value Fv(h/L)2.
Exercises
Problem 3
The sound from a trumpet radiates uniformly in all directions in air. At a
distance of 5.00 m from the trumpet the sound intensity level is 52.0 dB.
At what distance is the sound intensity level 30.0 dB?
Solution
The distance is proportional to the reciprocal of the square root of the
intensity and hence to 10 raised to half of the sound intensity levels
divided by 10:
I  P / 4d 2 ,   10log(I / I 0 )  I  I 010 /10 , I  d 2
I1 / I 2  101 /102 /10  (d2 / d1 )2  d110( 1 /102 /10) / 2  d2
(5.00m)10(5.203.00) / 2  62.9 m.
Exercises
Problem 4
An organ pipe has two successive harmonics with frequencies 1,372 and
1,764 Hz. (a) Is this an open or stopped pipe? (b) What two harmonics are
these? (c) What is the length of the pipe?
Solution
(a) For an open pipe, the difference between successive frequencies is
the fundamental, in this case 392 Hz, and all frequencies are integer
multiples of this frequency. If this is not the case, the pipe cannot be
an open pipe. For a stopped pipe, the difference between the successive
frequencies is twice the fundamental, and each frequency is an odd
integer multiple of the fundamental. In this case, f1 = 196 Hz, and
1372 Hz = 7f1 , 1764 Hz = 9f1 . So this is a stopped pipe.
(b) n=7 for 1,372 Hz, n=9 for 1,764 Hz.
(c) f1  v /(4 L), so L  v /(4 f1 )  (344m / s) /(784 Hz)  0.439 m.
Exercises
Problem 5
Two identical loudspeakers are located at
points A and B, 2.00 m apart. The loudA
speakers are driven by the same amplifier
and produce sound waves with a frequency
of 784 Hz. Take the speed of sound in air to
be 344 m/s. A small microphone is moved out
from Point B along a line perpendicular to the B
line connecting A and B. (a) At what distances
from B will there be destructive interference?
(b) At what distances from B will there be
constructive interference? (c) If the frequency
is made low enough, there will be no positions
along the line BC at which destructive
interference occurs. How low must the
frequency be for this to be the case?
2.00 m
C
x
Exercises
Problem 5
Solution
(a) If the separation of the speakers is denoted by h, the condition for
destructive interference is
x 2  h 2  x  l,
where  is an odd multiple of one-half. Adding x to both sides, squaring,
canceling the x2 term from both sides and solving for x gives:
x  [h2 /(2l)  l / 2]. Using l  v / f and h from the given data yields:
9.01m for   1 / 2, 2.71m for   3 / 2, 1.27 m for   5 / 2, 0.53m for   7 / 2, 0.026m for   9 / 2.
(b) Repeating the above argument for integral values for  , constructive
interference occurs at 4.34 m, 1.84 m, 0.86 m, 0.26 m.
(c) If h  l / 2 , there will be destructive interference at speaker B.
If h  l / 2, the path difference can never be as large as l / 2 .
The minimum frequency is then v/(2h)=(344 m/s)/(4.0 m)=86 Hz.
Exercises
Problem 6
A 2.00 MHz sound wave travels through a pregnant woman’s abdomen
and is reflected from fetal heart wall of her unborn baby. The heart wall is
moving toward the sound receiver as the heart beats. The reflected sound
is then mixed with the transmitted sound, and 5 beats per second are
detected. The speed of sound in body tissue is 1,500 m/s. Calculate the
speed of the fetal heart wall at the instance this measurement is made.
Solution
Let f0=2.00 MHz be the frequency of the generated wave. The frequency
with which the heart wall receives this wave is fH=[(v+vH)/v]f0, and this is
also the frequency with which the heart wall re-emits the wave. The detected
frequency of this reflected wave is f’=[v/(v-vH )]fH, with the minus sign indicating
that the heart wall, acting now as a source of waves, is moving toward the
receiver. Now combining f’=[(v+vH)/(v-vH)]f0, and the beat frequency is:
fbeat  f ' f0  [(v  vH ) /(v  vH )] f0  2vH f0 /(v  vH ). Solving for vH ,
vH  v[ fbeat /(2 f0  fbeat )]  (1500m / s){85Hz /[2(2.00106 Hz)  85Hz)}  3.19 102 m / s.