Karakoram International University, Gilgit Baltistan
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FGFDH 676

Fall 2016
Source of Sound P (Pa) I dB Auditory threshold at 1 kHz 2105
1012 0 Light leaf rustling, calm breathing 6.32105 1011 10 Very
calm room 3.56104 3.161012 25 A Whisper 2103 108 40
Washing machine, dish washer 6.32103 107 50 Normal
conversation at 1 m 2102 10
Karakoram International University, Gilgit Baltistan
gsdgds
FGFDH 676

Fall 2016
y(x,t) = Asin(k4x)cos(4t).
Find the instantaneous total energy in the string in terms of M, L, n =
4, v and A (although it will simplify matters to use 4 and 4 once you
dene them in terms of the givens).
If you are a physics or math major, the word nd sho
Karakoram International University, Gilgit Baltistan
gsdgds
FGFDH 676

Fall 2016
2,000,000,000 280 Krakatoa Volcanic Explosion (1883 C.E.)
63,200,000,000 310 Tambora Volcanic Explosion (1815 C.E.)
200,000,000,000 320
Table 6: Table of (approximate) P0 and sound pressure levels in
decibels relative to the threshold of human hearing at
Karakoram International University, Gilgit Baltistan
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FGFDH 676

Fall 2016
this case only schematically in on the diagram above (assuming that
the L shown is this secondsmallest value of L1 for the f1 tuning fork),
and indicate where the nodes and antinodes are.
496 Week 12: Gravity
Optional Problems Study for the nal exam! Thi
Karakoram International University, Gilgit Baltistan
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FGFDH 676

Fall 2016
If you are a wealthy nobleman with a hobby who is generating a huge
pile of data but who also has no particular mathematical skill, what
are you going to do? You hire a lab rat, a unky, an assistant who can
do the annoying and tedious work of analyzing yo
Karakoram International University, Gilgit Baltistan
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FGFDH 676

Fall 2016
222http:/lifesnow.com/bingo/ http:/lifesnow.com/bingo/
Week 12: Gravity 505
Galileos behalf, I accepted the apology, but of course I must also
point out that Bellarmines argument is essentially correct. The
conclusions of modern science have, almost witho
Karakoram International University, Gilgit Baltistan
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FGFDH 676

Fall 2016
smaller mass); we will learn a more precisely stated version of this law
below. This law fully explained at the limit of observational resoution,
and continues to mostly explain, Keplers Laws and the motions of the
planets, moons, comets, and other visibl
Karakoram International University, Gilgit Baltistan
gsdgds
FGFDH 676

Fall 2016
= n (1069)
(for n = 0,1,2.) or
x = n (1070)
If the path dierence contains an integral number of wavelengths the
waves from the two sources arrive in phase, add, and produce sound
that has twice the amplitude and four times the intensity. This is called
co
Karakoram International University, Gilgit Baltistan
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FGFDH 676

Fall 2016
String one has mass per unit length and is at tension T and has a
travelling harmonic wave on it:
y1(x,t) = Asin(kxt) String one is also very long compared to a
wavelength: L . Identical string two has the superposition of two
harmonic travelling waves on
Karakoram International University, Gilgit Baltistan
gsdgds
FGFDH 676

Fall 2016
or
kmL = m (1049)
for m = 1,2,3. This converts to:
m =
2L m
(1050)
and
fm =
va m
=
vam 2L
(1051)
The m = 1 solution (rst harmonic) is called the principle harmonic as
it was before. The actual tone of a ute pipe with two closed ends will
be a superpositio
Karakoram International University, Gilgit Baltistan
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FGFDH 676

Fall 2016
At an open pipe end the argument is inverted. The pipe is open to the
air (at xed background/equilibrium pressure) so that there must be a
pressure node at the open end. Pressure and displacement are /2 out
of phase, so that the open end is also a displac
Karakoram International University, Gilgit Baltistan
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FGFDH 676

Fall 2016
damping. Note that you should be able to do this part even if you
cannot derive the curves that you draw or . Clearly label each curve.
Week 10: The Wave Equation 463
Problem 5.
m 4m 9m
abc
Three strings of length L (not shown) with the same mass per unit
Karakoram International University, Gilgit Baltistan
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FGFDH 676

Fall 2016
case you are confused): log(A)log(B) = logA B (1035) You should
remember this; it will be very useful next year.
Table 5 presents a number of fairly common sounds, sounds you are
likely to have directly or indirectly heard (if only from far away). Each
so
Karakoram International University, Gilgit Baltistan
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FGFDH 676

Fall 2016
for displacement waves in a pipe of length L closed at one or both
ends. This solution has a node at x = 0 (the closed end). The permitted
resonant frequencies are determined by:
kL = n (1016)
for n = 1,2. (both ends closed, nodes at both ends) or:
kL =
2
Karakoram International University, Gilgit Baltistan
gsdgds
FGFDH 676

Fall 2016
conversion factor that scales microscopic displacement to pressure).
Note that:
P(x,t) = Z
d dt
s(x,t)
or, the displacement wave is a scaled derivative of the pressure wave.
The pressure waves represent the oscillation of the pressure around
the baseline
Karakoram International University, Gilgit Baltistan
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FGFDH 676

Fall 2016
supported frequencies and wavelengths match those of the string xed
at both ends, or pipe closed at both ends.
11.6: Beats
If you have ever played around with a guitar, youve probably noticed
that if two strings are ngered to be the same note but are real
Karakoram International University, Gilgit Baltistan
gsdgds
FGFDH 676

Fall 2016
diminishes as 1/r2 with the distance from the source. Electrostatic eld
also diminishes as 1/r2. There seems to be a shared connection
between symmetric propagation and spherical geometry; this will form
the basis for Gausss Law in electrostatics and much
Karakoram International University, Gilgit Baltistan
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FGFDH 676

Fall 2016
does not have a mass per unit length , rather it has a mass per unit
volume, .
The velocity of waves in air is given by va =sB 343m/sec (1020)
The approximately here is fairly serious. The actual speed varies
according to things like the air pressure (whi
Karakoram International University, Gilgit Baltistan
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FGFDH 676

Fall 2016
Week 11: Sound 485
an experience (although it can denitely and annoyingly occur if you
hook up speakers in your stereo out of phase). Interference will be
treated next semester in the context of coherent light waves. Just to
give you a head start on that,
Karakoram International University, Gilgit Baltistan
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FGFDH 676

Fall 2016
Decibels: Audible sound waves span some 20 orders of magnitude in
intensity. Indeed, the ear is barely sensitive to a doubling of intensity
this is the smallest change that registers as a change in audible
intensity. This motivate the use of sound inten
Karakoram International University, Gilgit Baltistan
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FGFDH 676

Fall 2016
waves and for other waves including water waves, light waves, even
waves on strings. Our standing wave solution can be rederived as
the superposition of a left and righttravelling harmonic wave, for
example. You can have interference from more than one
Karakoram International University, Gilgit Baltistan
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FGFDH 676

Fall 2016
displaced at t = 0. Verify that it is a solution to the ODE in a).
Remember that the string is xed at both ends!
c) Find kn,n,fn,n for the rst three modes supported by the string.
Sketch them in on the axes below, labelling nodes and antinodes. Note
that
Karakoram International University, Gilgit Baltistan
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FGFDH 676

Fall 2016
However, it is quite possible that over the next few decades the bright
and motivated physics students of today will help create the
bioelectronic and/or stem cell replacements of key organs and nervous
tissue that will relegate agerelated deafness to th
Karakoram International University, Gilgit Baltistan
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FGFDH 676

Fall 2016
or
kmL =
2m2 2
(1054)
for m = 1,2,3. (odd halfintegral multiples of . As before, you will
see dierent conventions used to name the harmonics, with some
books asserting that only odd harmonics are supported, but I prefer to
make the harmonic index do exa
Karakoram International University, Gilgit Baltistan
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FGFDH 676

Fall 2016
Catholic Church as heretical at the same time that Galileo was both
persecuted and prosecuted. Wealthy Tycho Brahe accumulated data
and his paid assistant, Johannes Kepler, t that data to specic orbits
and deduced Keplers Laws. All Brahe got for his eorts
Karakoram International University, Gilgit Baltistan
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FGFDH 676

Fall 2016
motion () can easily be determined from an examination of gure
139 above: = 0 vsT (1036)
We would really like the frequency of the doppler shifted sound. We
can easily nd this by using
480 Week 11: Sound
= va/f and = va/f. We substitute and use f = 1/T:
Karakoram International University, Gilgit Baltistan
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FGFDH 676

Fall 2016
given by A (perpendicular to the direction of the waves propagation)
and vt (in the direction of the waves propagation. All the energy in
this box crosses through A in time t. That is:
E = (
12
2s2)Avt (1024)
or
I=
E At
=
12
2s2v (1025)
which looks very m
Karakoram International University, Gilgit Baltistan
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FGFDH 676

Fall 2016
that the object arrives at innity at rest. Since U() = 0 by
denition, the escape energy for a particle is: Eescape = K() + U()
= 0 + 0 = 0 (1085) Since mechanical energy is conserved moving
through the (presumed) vacuum of space, the total energy must be
Karakoram International University, Gilgit Baltistan
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FGFDH 676

Fall 2016
center. The resonant frequency series for the pipe is the same,
however, as for a pipe closed at both ends!
This is a panpipe, one of the most primitive (and beautiful) of musical
instruments. A panpipe is nothing more than a tube, such as a piece of
holl