Physics 1C
Lecture 14B
Today:
End of Chapter 14
Start of Chapter 24
Wave Interference
Example
Two strings with linear densities of 5.0g/m are stretched
over pulleys, adjusted to have vibrating lengths of 50cm,
and attached to hanging blocks. The block attached to
string 1 has a mass of 20kg and the block attached to
string 2 has mass M. Listeners hear a beat frequency of
2.0Hz when string 1 is excited at its fundamental
frequency and string 2 at its third harmonic. What is one
possible value for M?
Answer
Here we need to make a diagram of the
situation and list the known quantities.
Wave Interference
Answer
Making a diagram
we note that the
mass creates the
tension to
increase the wave
speed:
Listing the quantities that we know:
fbeat = 2.0Hz
L1 = L2 = 0.50m
μ1 = μ2 = 5.0x10-3kg/m
m1 = 20kg
Wave Interference
Answer
An observer hears a beat frequency of 2Hz so:
f beat  2Hz  f 2  f1
The frequency for the first string
will be:
 v 
f n  n 
2L 
 F 
T

 v1 
1 
f1  1  

 2L1   2L1 


 m g 
1

1 
f1  

 2L1 


Wave Interference
Answer
The frequency for the second
string will be:
 v 
f n  n 
2L 
 F

T
 v 2  
2 
f 2  3  3

2L2   2L2 


 Mg 

2 
f 2  3

 2L2 


Put these into the beat frequency
equation to find out the mass, M
(remember that L1 = L2 and
μ1 = μ2).
Wave Interference
Answer
f beat  2Hz  f 2  f1
 Mg   m g
1





2Hz  3
 
2L   2L



 





 g 
  
2Hz  
 3 M  20kg
 2L 





 g   9.8 N kg

3 kg 



5.0
10
The factor out in front
m

 44.3



is:
20.5m

 2L  


 


Eliminating the
absolute value sign:

0.0452
4.52
kg
kg
 3 M  4.472
3 M
kg
M  2.27kg
1
s kg
Beats: Math Model
Consider two waves with equal amplitudes and slightly
different frequencies f1 and f2.
Represent the displacement of an element of the medium
associated with each wave at a fixed position, x = 0:
y1 = A cos(2pf1t)
and
y2 = A cos(2pf2t)
Resultant position (use superposition principle):
y = y1 + y2 = A [cos(2pf1t) + cos(2pf2t)]
Use trigonometric identity:
cos b = cos[(a-b)/2] cos[(a+b)/2]
cos a +
To get: y = 2Acos {2p[(f1-f2)/2]t} cos {2p[(f1+f2)/2]t}
Vibrations
Nearly every object will vibrate at a certain
frequency which is known as the natural
frequency.
You can stimulate vibrations to an object at its
natural frequency. This is known as
resonance.
When resonance occurs, the amplitude of
vibrations, for a given energy as stimulus,
tends to increase dramatically.
Concept Question
You are tuning a guitar by comparing the sound of the
string with that of a standard tuning fork. You notice a
beat frequency of 5Hz when both sounds are present.
You tighten the string and the beat frequency rises to
8Hz. To tune the string exactly to the tuning fork, what
should you do?
A) Continue to tighten the string
B) Loosen the string
C) It is impossible to determine
FT
fn  n

2L
Start of Chapter 24
EM Waves
A “special” type of wave is the electromagnetic
wave.
Accelerated charges or changing magnetic fields
produce electromagnetic waves (radiation).
In EM waves, the electric and magnetic fields are
perpendicular to each other.
Both fields are also
perpendicular to the
direction of
propagation.
This makes EM
waves transverse
waves.
EM Waves
Electromagnetic waves propagate through the
vacuum at the speed of light:
where the electric constant εo = 8.85x10–12 C2/(Nm2) and the
magnetic constant μo = 1.26x10-6 (Tm/A).
In a medium, EM waves usually do not propagate at a
speed equal to c. Different types of light will travel at
different speeds.
Maxwell concluded that light is an electromagnetic
wave.
EM waves carry energy as they travel
through space.
EM Waves
Energy carried by EM waves is shared equally by
the electric and magnetic fields.
The average power per unit area of the EM wave
is given by the intensity, I:
When light waves strike a surface of area, A, in a
time interval, Δt, the intensity, I, of the light will yield
a certain amount of energy, U:
EM Waves
Electromagnetic waves transport linear momentum
as well as energy.
Even if waves are reflected from a surface,
momentum is transferred to the surface.
If the surface is a perfect absorber, then the
momentum delivered to the surface will become:
P = U/c
If the surface is a perfect
reflector, then the momentum
delivered to the surface will
become:
P = 2U/c
Spectrum of EM Waves
There are distinct forms of EM waves at different
frequencies (and wavelengths).
Recall that the wave speed is given by:
vwave = c = l f.
Wavelengths for visible light range between
400nm (violet) and 700nm (red).
There is no sharp division between one kind of EM
wave and the next.
For example, you can have an X-ray and a Gamma
Ray with the exact same wavelength.
EM Spectrum
Note the overlap between
types of waves (such as UV
and X-rays).
All EM waves have the
same speed in a vacuum,
what distinguishes the types
are their frequencies or
wavelengths.
Note that the visible section
is a quite small portion of
the spectrum.
EM Spectrum
Wavelengths of light can
range from very long (radio,
~100km) to very short
(gamma, ~1fm).
Frequencies have an
equally long range of
possible values: (gamma,
~1022Hz) to (radio, ~10Hz).
Visible light ranges from
Red (700nm, 4x1014Hz) to
Violet (400nm, 7x1014Hz)
EM Spectrum
Radio waves have a long
wavelength (~100m) and
thus are good for use as a
communication tool (TV,
AM, FM).
Microwaves are smaller
(~1cm) and interfere easily
with common things
(μwave oven grates).
Infrared waves are
produced by hot objects.
EM Spectrum
Visible light (~500nm)
is detected by the
human eye. We are
most sensitive to
yellow-green (560nm).
UV light (~100nm) that
comes from the Sun is
mostly absorbed by the
Earth’s ozone layer.
EM Spectrum
X-rays (~0.1nm) are
associated with fast
electrons hitting off of a
metal target (medical
applications).
Gamma rays (~1fm) are
emitted by radioactive
nuclei. They can cause
serious damage to living
tissue as they penetrate
deeply into most matter.
Cell Phone Intensity
Example
A cell phone emits 0.60Watts of 1.9GHz radio
waves. What are the amplitudes of the electric and
magnetic fields at a distance of 10cm?
Answer
Assume the cell phone is a point source of
electromagnetic waves (or r = 0).
Cell Phone Intensity
Answer
The intensity of the radio waves at 10cm is:
We want the maximum values
(amplitudes) for the electric and
magnetic fields.
Cell Phone Intensity
Answer
For magnetic field we can turn to:
For Next Time (FNT)
Continue reading Chapter 24
Start working on the homework for
Chapter 24