physics 1c - ucsdtier2€¦ · its fundamental frequency and string 2 at its third harmonic. what...

Physics 1C�Lecture 14B �

���

Today: �End of Chapter 14 �

Start of Chapter 27 ���

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: �

!   The frequency for the first string will be: �

!

fbeat = 2Hz = f2 " f1

!

fn = n v2L"

# $

%

& '

!

f1 = 1( ) v12L1

"

# $

%

& ' =

FTµ1

2L1

"

#

$ $ $

%

&

' ' '

!

f1 =

m1gµ1

2L1

"

#

$ $ $

%

&

' ' '

Wave Interference�! Answer�!   The frequency for the second

string will be: �

!

fn = n v2L"

# $

%

& '

!

f2 = 3( ) v22L2

"

# $

%

& ' = 3

FTµ2

2L2

"

#

$ $ $

%

&

' ' '

!

f2 = 3Mg

µ22L2

"

#

$ $ $

%

&

' ' '

!   Put these into the beat frequency equation to find out the mass, M (remember that L1 = L2 and μ1 = μ2).�

Wave Interference�! Answer�

!   The factor out in front is: �

!

fbeat = 2Hz = f2 " f1

!

2Hz = 3Mg

µ2L

"

#

$ $ $

%

&

' ' ' (

m1gµ

2L

"

#

$ $ $

%

&

' ' '

!

2Hz =

gµ

2L

"

#

$ $ $

%

&

' ' ' 3 M ( 20kg( )

!

gµ

2L

"

#

$ $ $

%

&

' ' '

=

9.8Nkg5.0 (10)3 kgm2 0.5m( )

"

#

$ $ $ $

%

&

' ' ' '

= 44.3 1s kg

!   Eliminating the absolute value sign: �

!

0.0452 kg = 3 M " 4.472 kg

!

M = 2.27kg

!

4.52 kg = 3 M

Math of Beats�!   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(2πf1t) and y2 = A cos(2πf2t) �

!   Resultant position (use superposition principle): �

y = y1 + y2 = A [cos(2πf1t) + cos(2πf2t)] �

!   Use trigonometric identity: cos a + cos b = cos[(a-b)/2] cos[(a+b)/2]�

!   To get: y = 2A cos{2π[(f1-f2)/2]t} cos{2π[(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�fn = n

FTµ

2L

Start of Chapter 27 �

Interference�!   We just discussed interference effects with

sound.�

!   Chapter 27 follows up on this to discuss interference effects with light.�

!   We thus change the order from the order in the book, and jump from Chapter 14 to Chapter 27, and get back to Chapters 24,25,26 later in the quarter.�

!   The hope is that this change in order will make it easier for you to grasp the concepts.�

Interference�

!   In order to create sustained interference, you need two sources with identical wavelengths (monochromatic) that are coherent.�

Interference�

!   Coherence means that the waves must maintain a constant phase with respect to each other.�

Double Slit Experiment �!   Two narrow slits, S1 and S2, can act as sources

of waves.�!   The waves emerging from the

slits originate from the same wavefront and therefore are always in phase (coherence).

!   The light from the two slits forms a visible pattern on a screen.

!   The pattern consists of a series of bright and dark parallel bands called fringes.

Double Slit Experiment �!   The fringe pattern formed by

a Double Slit Experiment would look like the picture to the right.�

!   Alternating bright and dark fringes are created.�

!   Constructive interference occurs where a bright fringe appears.�

!   Destructive interference results in a dark fringe.�

Double Slit Experiment �!   Constructive interference occurs at the center, O.�!   There is no path length difference between these two

waves.�

!   Therefore, they arrive in phase with each other.�

!   This will result in a bright area on the screen.�

!   This bright spot is called the central maximum (zeroth order maximum).�

Double Slit Experiment �!   If we moved to a spot, P, above the center spot, O, we find that the lower wave, S2, travels farther than the upper wave, S1.�

!   If the lower wave travels one wavelength farther than the upper wave; the waves will still arrive in phase.�

!   A bright spot will occur.�

!   This bright spot is called the first order maximum.�

Double Slit Experiment �!   Looking at the distance travelled by the light,

we see that for a point above the center spot the distance travelled by the lower wave, r2, is longer than the upper wave, r1.�

!   The path length difference, δ, will be: �

!

" = r2 # r1

Double Slit Experiment �!   What does the path length difference have to

be in order to get constructive interference?�!   δ = 0 or λ or 2λ or ...�!   Basically any integer times the wavelength.�

!   So for constructive interference to occur: �

!

" = m#!   where m is

0, ±1, ±2, ...�

Double Slit Experiment �!   We define the line between the middle of the two slits

and the central maximum as the centerline.�!   We also define an angle θ as the angle from the

centerline to the fringe you are looking at.�

!   The distance between the slits is labeled, d.�

!   The central maximum has an angle of 0o.�

Double Slit Experiment �!   Using trigonometry we find that the path length

difference, δ, is related to the slit distance, d, and the angle θ by: �

!   Thus, for constructive interference to occur: �

!   where m is 0, ±1, ±2, ...�

!   m is called the order number.�

!

" = m# = d sin$!

" = d sin#

Double Slit Experiment �!   There is a spot, R, between P and Q, where find a

dark area.�!   The lower wave travels λ/2 farther than the upper

wave to reach R; the waves arrive at R 180o out of phase.�

!   This results in destructive interference.�

!   In this exact case the path length difference would be: �

!

" = r2 # r1 = 12 $

Double Slit Experiment �!   What does the path length difference have to

be in order to get destructive interference?�!   (1/2)λ or (3/2)λ or ...�!   Basically any half integer times the wavelength.�!   So for destructive interference to occur: �

!   where m is 0, ±1, ±2, ...�

!   And we can extend this to: �!

" = m + 12( )#

!

" = m + 12( )# = d sin$

Double Slit Experiment �!   The perpendicular distance, y, from the

centerline to the fringe you are observing can be related by: �

!   at small angles: �

!

y = L tan" # Lsin"

!

" # sin" # tan"

Concept Question �!   Suppose the viewing screen in a double-slit experiment is

moved closer to the double slit apparatus. What happens to the interference fringes?�

!   A) They get brighter but otherwise do not change.�

!   B) They get brighter and closer together.�

!   C) They get brighter and farther apart.�

!   D) No change will occur.�!   E) They fade out and disappear.�

Double Slit Experiment �!   For small angles these equations can be combined

and for bright fringes we get: �

!   Note y is then linear in the order number m, so the bright fringes are equally spaced.�

( 0, 1, 2 )brightmy L md!" #

= = ± ±$ %& '

…

Double Slit Experiment �!   Double Slit Experiments provides a method for

measuring wavelength of the light.�!   This experiment gave the wave model of light a

great deal of credibility.�!   The bright fringes in the interference pattern do

not have sharp edges.�!   The derived equations give the location of only the

centers of the bright and dark fringes.�!   We can also calculate the distribution of light

intensity associated with the double-slit interference pattern.�

Double Slit Experiment �

2max

sincos

dI I ! "#

$ %= & '

( )

!   The interference pattern consists of equally spaced fringes of equal intensity.�

!   This result is valid only if L >> d and for small values of θ.�

!   The phase difference between the two waves at P depends on their path difference.�

!   The time-averaged light intensity at a given angle θ is: �

!"#

=$"#

=% sind22

For Next Time (FNT)�

!  Continue reading Chapter 27 �

!  Start working on the homework for Chapter 27 �