Transcript Physics 131: Lecture 14 Notes

Physics 151: Lecture 34
Today’s Agenda

Topic - Waves (Chapter 16 )
1-D traveling waves
Waves on a string
Wave speed
Physics 151: Lecture 34, Pg 1
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)
Stadium waves (people move up & down)
Water waves (water moves up & down)
Light waves (what moves ??)
Animation
Physics 151: Lecture 34, Pg 2
See text: 16.2
Types of Waves

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
Slinky
Animation
see Figures 16.2-5
Physics 151: Lecture 34, Pg 3
See text: 16.2
Wave Properties


Wavelength: The distance  between identical points on the wave.
Amplitude: The maximum displacement A of a point on the wave.
Wavelength

Amplitude A
A
Animation
Physics 151: Lecture 34, Pg 4
See text: 16.2
Wave Properties...

Period: The time T for a point on the wave to
undergo one complete oscillation.

Speed: The wave moves one wavelength  in one
period T so its speed is v = / T.
v
see Figure 16.6

T
Physics 151: Lecture 34, Pg 5
See text: 16.2
v=/T

Wave Properties...
We will show that the speed of a wave is a constant
that depends only on the medium, not on amplitude,
wavelength or period
 and T are related !

=vT
or  = 2 v / (since T = 2 / 
or  v / f

Recall
(since T = 1/ f )
f = cycles/sec or revolutions/sec
 rad/sec = 2f
Physics 151: Lecture 34, Pg 6
Example

The figure on the right shows a sine
wave on a string at one instant of time.

Which of the graphs on the
right shows a wave where
the frequency and wave
velocity are both doubled ?
Physics 151: Lecture 34, Pg 7
Lecture 34, Act 1
Wave Motion


The speed of sound in air is a bit over 300 m/s, and
the speed of light in air is about 300,000,000 m/s.
Suppose we make a sound wave and a light wave
that both have a wavelength of 3 meters.
What is the ratio of the frequency of the light wave
to that of the sound wave ?
(a) About 1,000,000
(b) About .000,001
(c) About 1000
Physics 151: Lecture 34, Pg 8
Lecture 34, Act 1
Solution

What are these frequencies ???
For sound having  = 3m : f 
v
300 m s

 100 Hz

3m
(low humm)
3  10 8 m s
For light having  = 3m : f  
 100 MHz

3m
v
(FM radio)
Physics 151: Lecture 34, Pg 9
See text: 16-1
Wave Forms
v

So far we have examined
“continuous waves” that go
on forever in each direction !

We can also have “pulses”
caused by a brief disturbance
of the medium:
v
v

And “pulse trains” which are
somewhere in between.
see Figure 16.3
Physics 151: Lecture 34, Pg 10
See text: 16.3
Mathematical Description

y
Suppose we have some function y = f(x):
x
0
y
f(x-a) is just the same shape moved
a distance a to the right:

0
x
x=a
y

Let a=vt Then
f(x-vt) will describe the same
shape moving to the right with
speed v.
see Figure 16.7
v
0
x=vt
x
Physics 151: Lecture 34, Pg 11
See text: 16.3
Math...
y

Consider a wave that is harmonic
in x and has a wavelength of .

A
x
If the amplitude is maximum at y  x   A cos 2  x 


x=0 this has the functional form:



y

Now, if this is moving to
the right with speed v it will be
described by:
v
x
 2
 x  vt 
y  x , t   A cos
 

Physics 151: Lecture 34, Pg 12
See text: 16-2
Math...

So we see that a simple harmonic
 2


 x  vt 
y
x
,
t

A
cos
wave moving with speed v in the x

 

direction is described by the equation:

By using v 

T


from before, and by defining
2
we can write this as:
k
2

y  x , t   A cos kx  t 
(what about moving in the -x direction ?)
Physics 151: Lecture 34, Pg 13
See text: 16-2
Movie (twave)
Math Summary


y
The formula y  x ,t   A cos kx  t 
describes a harmonic wave of
amplitude A moving in the
+x direction.

A
x
Each point on the wave oscillates in the y direction with
simple harmonic motion of angular frequency .
2
k

The wavelength of the wave is  

The speed of the wave is v 

The quantity k is often called “wave number”.

k
Physics 151: Lecture 34, Pg 14
Lecture 34, Act 2
Wave Motion


A harmonic wave moving in the positive x direction
can be described by the equation
y(x,t) = A cos ( kx - t )
Which of the following equation describes a harmonic
wave moving in the negative x direction ?
(a) y(x,t) = A sin ( kx  t )
(b) y(x,t) = A cos ( kx + t )
(c) y(x,t) = A cos (kx + t )
Physics 151: Lecture 34, Pg 15
Lecture 34, Act 3
Wave Motion

A boat is moored in a fixed location, and waves make it move up
and down. If the spacing between wave crests is 20 meters and
the speed of the waves is 5 m/s, how long Dt does it take the
boat to go from the top of a crest to the bottom of a trough ?
(a) 2 sec
(b) 4 sec
(c) 8 sec
t
t + Dt
Physics 151: Lecture 34, Pg 16
See text: 16.5
Waves on a string

What determines the speed of a wave ?

Consider a pulse propagating along a string:
v

“Snap” a rope to see such a pulse

How can you make it go faster ?
Animation
Physics 151: Lecture 34, Pg 17
See text: 16.5
Waves on a string...
Suppose:

The tension in the string is F

The mass per unit length of the string is  (kg/m)

The shape of the string at the pulse’s maximum is
circular and has radius R
F

R
Physics 151: Lecture 34, Pg 18
See text: 16.5
Waves on a string...

Consider moving along with the pulse

Apply F = ma to the small bit of string at the “top” of the pulse
which is moving with Uniform Circular Motion.
v
y
x
see Figure 16-11
Physics 151: Lecture 34, Pg 19
See text: 16.5
Waves on a string...

The total force FTOT is the sum of the tension F at
each end of the string segment.

The total force is in the -y direction.


F
F
FTOT = 2F 
y
x
(since  is small, sin  ~ )
Physics 151: Lecture 34, Pg 20
See text: 16.5
Waves on a string...

The mass m of the segment is its length (R x 2) times
its mass density .
m = R 2



R
y
x
Physics 151: Lecture 34, Pg 21
See text: 16.5
Waves on a string...

The acceleration a of the segment is v 2/ R (centripetal)
in the -y direction.
v
a
R
y
x
Physics 151: Lecture 34, Pg 22
See text: 16.5
Waves on a string...

So FTOT = ma becomes:
v2
2 F  R 2  
R
FTOT
F   v 2
a
m
v
F

v
tension F
mass per unit length 
Physics 151: Lecture 34, Pg 23
See text: 16.5
Waves on a string...

So we find:
v
Animation-1
Animation-2
F

Animation-3
v
tension F
mass per unit length 

Making the tension bigger increases the speed.

Making the string heavier decreases the speed.

As we asserted earlier, this depends only on the nature of
the medium, not on amplitude, frequency etc of the wave.
Physics 151: Lecture 34, Pg 24
Lecture 34, Act 4
Wave Motion


A heavy rope hangs from the ceiling, and a small
amplitude transverse wave is started by jiggling the
rope at the bottom.
As the wave travels up the rope, its speed will:
v
(a) increase
(b) decrease
(c) stay the same

Can you calcuate how long will it take for a pulse
travels a rope of length L and mass m ?
Physics 151: Lecture 34, Pg 25
Recap of today’s lecture

Chapter 16
Definitions
1-D traveling waves
Waves on a string
Wave speed

Next time:
Finish Chapter 16
Physics 151: Lecture 34, Pg 26